Exercises from Folland:

• Chapter 1: Exercises 3, 7, 10, 12, 14 (with the sets in 3(a) being non-empty) Exercises 15, 17, 18, 19, 22(a), 24, 28 Exercises 26, 30 (also check out 31)

• Chapter 2: Exercises 2, 3, 7, 9 (in 9(c) you can use Exercise 1.29 without proof Exercises 10, 12, 13, 14, 16, 19 Exercises 24, 25, 28(a,b), 33, 34, 35, 38, 41 (note that 24 shows that upper sums are not needed in the definition of integrals, and the extra hypotheses also show that they are not desired either) Exercises 40, 44, 47, 49, 50, 51, 52, 54, 56, 58, 59

• Chapter 3: Exercises 3(b,c), 5, 6, 9, 12, 13, 14, 16, 20, 21, 22

# 1 Basics

Notation Definition
\begin{align*}{\left\lVert {f} \right\rVert}_\infty \mathrel{\vcenter{:}}=\sup_{x\in \operatorname{dom}(f)} {\left\lvert {f(x)} \right\rvert}\end{align*} The Sup norm
\begin{align*} {\left\lVert {f} \right\rVert}_{L^\infty} \mathrel{\vcenter{:}}=\inf\left\{{M \geq 0 {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} \leq M \text{ for a.e. } x }\right\} \end{align*} The $$L^ \infty$$ norm
\begin{align*} f_n \overset{n \to \infty }\to f \end{align*} Convergence of a sequence
\begin{align*} f(x) \overset{{\left\lvert {x} \right\rvert} \to \infty}\to 0 \end{align*} Vanishing at infinity
\begin{align*} \int_{{\left\lvert {x} \right\rvert} \geq N} f \overset{N\to \infty}\to 0 \end{align*} Having small tails
\begin{align*} H, \mathcal{H} \end{align*} A Hilbert space
\begin{align*} X \end{align*} A topological space

## 1.1 Useful Techniques

• General advice: try swapping the orders of limits, sums, integrals, etc.

• Limits:

• Take the $$\limsup$$ or $$\liminf$$, which always exist, and aim for an inequality like \begin{align*} c \leq \liminf a_n \leq \limsup a_n \leq c .\end{align*}
• $$\lim f_n = \limsup f_n = \liminf f_n$$ iff the limit exists, so to show some $$g$$ is a limit, show \begin{align*} \limsup f_n \leq g \leq \liminf f_n \qquad (\implies g = \lim f) .\end{align*}
• A limit does not exist if $$\liminf a_n > \limsup a_n$$.
• Sequences and Series

• If $$f_n$$ has a global maximum (computed using $$f_n'$$ and the first derivative test) $$M_n \to 0$$, then $$f_n \to 0$$ uniformly.
• For a fixed $$x$$, if $$f = \sum f_n$$ converges uniformly on some $$B_r(x)$$ and each $$f_n$$ is continuous at $$x$$, then $$f$$ is also continuous at $$x$$ .
• Equalities

• Split into upper and lower bounds: \begin{align*} a=b \iff a\leq b \text{ and } a\geq b .\end{align*}
• Use an epsilon of room: \begin{align*} \qty{ \forall \epsilon, \,\,a < b + \varepsilon} \implies a\leq b .\end{align*}
• Showing something is zero: \begin{align*} \qty{ \forall \epsilon, \,\, {\left\lVert {a} \right\rVert} < \varepsilon} \implies a = 0 .\end{align*}
• Continuity / differentiability: show it holds on $$[-M, M]$$ for all $$M$$ to get it to hold on $${\mathbb{R}}$$.

• Simplifications:

• To show something for a measurable set, show it for bounded/compact/elementary sets/
• To show something for a function, show it for continuous, bounded, compactly supported, simple, indicator functions, $$L^1$$, etc
• Replace a continuous sequence ($$\varepsilon\to 0$$) with an arbitrary countable sequence ($$x_n \to 0$$)
• Intersect with a ball $$B_r(\mathbf{0})\subset {\mathbb{R}}^n$$.
• Integrals

• Calculus techniques: Taylor series, IVT, MVT, etc.
• Break up $${\mathbb{R}}^n = \left\{{{\left\lvert {x} \right\rvert} \leq 1}\right\} \coprod \left\{{{\left\lvert {x} \right\rvert} > 1}\right\}$$.
• Or break integration region into disjoint annuli.
• Break up into $$\left\{{f>g}\right\} {\coprod}\left\{{f=g}\right\} {\coprod}\left\{{f< g}\right\}$$.
• Tail estimates!
• Most of what works for integrals will work for sums.
• Measure theory:

• Always consider bounded sets, and if $$E$$ is unbounded write $$E = \cup_{n} B_{n}(0) \cap E$$ and use countable subadditivity or continuity of measure.

• $$F_\sigma$$ sets are Borel, so establish something for Borel sets and use this to extend it to Lebesgue.

• $$s = \inf\left\{{x\in X}\right\} \implies$$ for every $$\varepsilon$$ there is an $$x\in X$$ such that $$x \leq s + \varepsilon$$.

• Approximate by dense subsets of functions

• Useful facts about compactly supported ($$C_c({\mathbb{R}})$$) continuous functions:

• Uniformly continuous
• Bounded almost everywhere

## 1.2 Definitions

$$f$$ is uniformly continuous iff

\begin{align*} &\forall \varepsilon \quad \exists \delta(\varepsilon) \mathrel{\Big|}\quad \forall x, y, \quad {\left\lvert {x - y} \right\rvert} < \delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < \varepsilon \\ \iff &\forall \varepsilon \quad \exists \delta(\varepsilon) \mathrel{\Big|}\quad \forall x, y, \quad {\left\lvert {y} \right\rvert} < \delta \implies {\left\lvert {f(x-y) - f(y)} \right\rvert} < \varepsilon .\end{align*}

A set $$S$$ is nowhere dense iff the closure of $$S$$ has empty interior iff every interval contains a subinterval that does not intersect $$S$$.

A set is meager if it is a countable union of nowhere dense sets.

An $$F_\sigma$$ set is a union of closed sets, and a $$G_\delta$$ set is an intersection of opens.1

\begin{align*} \limsup_n a_n = \lim_{n\to \infty} \sup_{j\geq n} a_j &= \inf_{n\geq 0} \sup_{j\geq n} a_j \\ \liminf_n a_n = \lim_{n\to \infty} \inf_{j\geq n} a_j &= \sup_{n\geq 0} \inf_{j\geq n} a_j .\end{align*}

Let $$X$$ be a metric space and $$A$$ a subset. Let $$A'$$ denote the limit points of $$A$$, and $$\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu \mathrel{\vcenter{:}}= A\cup A'$$ to be its closure.

• A neighborhood of $$p$$ is an open set $$U_p$$ containing $$p$$.

• An $$\varepsilon{\hbox{-}}$$neighborhood of $$p$$ is an open ball $$B_r(p) \mathrel{\vcenter{:}}=\left\{{q {~\mathrel{\Big|}~}d(p, q) < r}\right\}$$ for some $$r>0$$.

• A point $$p\in X$$ is an accumulation point of $$A$$ iff every neighborhood $$U_p$$ of $$p$$ contains a point $$q\in Q$$

• A point $$p\in X$$ is a limit point of $$A$$ iff every punctured neighborhood $$U_p\setminus\left\{{p}\right\}$$ contains a point $$q\in A$$.

• If $$p\in A$$ and $$p$$ is not a limit point of $$A$$, then $$p$$ is an isolated point of $$A$$.

• $$A$$ is closed iff $$A' \subset A$$, so $$A$$ contains all of its limit points.

• A point $$p\in A$$ is interior iff there is a neighborhood $$U_p \subset A$$ that is strictly contained in $$A$$.

• $$A$$ is open iff every point of $$A$$ is interior.

• $$A$$ is perfect iff $$A$$ is closed and $$A\subset A'$$, so every point of $$A$$ is a limit point of $$A$$.

• $$A$$ is bounded iff there is a real number $$M$$ and a point $$q\in X$$ such that $$d(p, q) < M$$ for all $$p\in A$$.

• $$A$$ is dense in $$X$$ iff every point $$x\in X$$ is either a point of $$A$$, so $$x\in A$$, or a limit point of $$A$$, so $$x\in A'$$. I.e., $$X\subset A\cup A'$$.

• Alternatively, $$\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu = X$$, so the closure of $$A$$ is $$X$$.

\begin{align*} (\forall \varepsilon>0)\left(\exists n_{0} = n_0(\varepsilon) \right)(\forall x \in S)\left(\forall n>n_{0}\right)\left(\left|f_{n}(x)-f(x)\right|<\varepsilon\right) .\end{align*} Negated:2 \begin{align*} (\exists \varepsilon>0)\left(\forall n_{0} = n_0 (\varepsilon) \right)(\exists x = x(n_0) \in S)\left(\exists n>n_{0}\right)\left(\left|f_{n}(x)-f(x)\right| \geq \varepsilon\right) .\end{align*}

A sequence of functions $$\left\{{ f_j }\right\}$$ is said to converge pointwise to $$f$$ if and only if \begin{align*} (\forall \varepsilon>0)(\forall x \in S)\left(\exists n_{0} = n_0(x, \varepsilon) \right)\left(\forall n>n_{0}\right)\left(\left|f_{n}(x)-f(x)\right|<\varepsilon\right) .\end{align*}

The outer measure of a set is given by \begin{align*} m_*(E) \mathrel{\vcenter{:}}=\inf_{\substack{\left\{{Q_{i}}\right\} \rightrightarrows E \\ \text{closed cubes}}} \sum {\left\lvert {Q_{i}} \right\rvert} .\end{align*}

\begin{align*} \limsup_{n} A_{n} \mathrel{\vcenter{:}}=\cap_{n} \cup_{j\geq n} A_{j}&= \left\{{x {~\mathrel{\Big|}~}x\in A_{n} \text{ for inf. many $n$}}\right\} \\ \liminf_{n} A_{n} \mathrel{\vcenter{:}}=\cup_{n} \cap_{j\geq n} A_{j} &= \left\{{x {~\mathrel{\Big|}~}x\in A_{n} \text{ for all except fin. many $n$}}\right\} \\ .\end{align*}

A subset $$E\subseteq {\mathbb{R}}^n$$ is Lebesgue measurable iff for every $$\varepsilon> 0$$ there exists an open set $$O \supseteq E$$ such that $$m_*(O\setminus E) < \varepsilon$$. In this case, we define $$m(E) \mathrel{\vcenter{:}}= m_*(E)$$.

$$f\in L^+$$ iff $$f$$ is measurable and non-negative.

A measurable function is integrable iff $${\left\lVert {f} \right\rVert}_1 < \infty$$.

\begin{align*} {\left\lVert {f} \right\rVert}_\infty &\mathrel{\vcenter{:}}=\inf_{\alpha \geq 0} \left\{{\alpha {~\mathrel{\Big|}~}m\left\{{{\left\lvert {f} \right\rvert} \geq \alpha}\right\} = 0}\right\} .\end{align*}

A function $$f:X \to {\mathbb{C}}$$ is essentially bounded iff there exists a real number $$c$$ such that $$\mu(\left\{{{\left\lvert {f} \right\rvert} > x}\right\}) = 0$$, i.e. $${\left\lVert {f} \right\rVert}_\infty < \infty$$.

\begin{align*} L^\infty(X) \mathrel{\vcenter{:}}=\left\{{f: X\to {\mathbb{C}}{~\mathrel{\Big|}~}f \text{ is essentially bounded }}\right\} \mathrel{\vcenter{:}}=\left\{{f: X\to {\mathbb{C}}{~\mathrel{\Big|}~}{\left\lVert {f} \right\rVert}_{\infty }< \infty}\right\} .\end{align*}

For $$X$$ a normed vector space and $$\Lambda \in X^\vee$$, \begin{align*} {\left\lVert {\Lambda} \right\rVert}_{X^\vee} \mathrel{\vcenter{:}}=\sup_{\left\{{x\in X {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert}_X \leq 1}\right\}} {\left\lvert {f(x)} \right\rvert} .\end{align*}

\begin{align*}f * g(x)=\int f(x-y) g(y) d y .\end{align*}

\begin{align*} \widehat{f}(\xi) = \int f(x) ~e^{2\pi i x \cdot \xi} ~dx .\end{align*}

\begin{align*} \phi_{t}(x) = t^{-n} \phi\left(t^{-1} x\right) .\end{align*}

For $$\phi\in L^1$$, the dilations satisfy $$\int \phi_{t} = \int \phi$$, and if $$\int \phi = 1$$ then $$\phi$$ is an approximate identity.

A space $$X$$ is a Baire space if and only if every countable intersections of open, dense sets is still dense.

### 1.2.1 Functional Analysis

A countable collection of elements $$\left\{{ u_i }\right\}$$ is orthonormal if and only if

1. $${\left\langle {u_i},~{u_j} \right\rangle} = 0$$ for all $$j \neq k$$ and
2. $${\left\lVert {u_j} \right\rVert}^2 \mathrel{\vcenter{:}}={\left\langle {u_j},~{u_j} \right\rangle} = 1$$ for all $$j$$.

A set $$\left\{{u_{n}}\right\}$$ is a basis for a Hilbert space $${\mathcal{H}}$$ iff it is dense in $${\mathcal{H}}$$.

A collection of vectors $$\left\{{u_{n}}\right\}\subset H$$ is complete iff $${\left\langle {x},~{u_{n}} \right\rangle} = 0$$ for all $$n \iff x = 0$$ in $$H$$.

The dual of a Hilbert space $$H$$ is defined as \begin{align*} H^\vee\mathrel{\vcenter{:}}=\left\{{L: H\to {\mathbb{C}}{~\mathrel{\Big|}~}L \text{ is continuous }}\right\} .\end{align*}

A map $$L: X \to {\mathbb{C}}$$ is a linear functional iff \begin{align*} L(\alpha\mathbf{x} + \mathbf{y}) = \alpha L(\mathbf{x}) + L(\mathbf{y}). .\end{align*}

The operator norm of an operator $$L$$ is defined as \begin{align*} {\left\lVert {L} \right\rVert}_{X^\vee} \mathrel{\vcenter{:}}=\sup_{ \substack{x\in X \\ {\left\lVert {x} \right\rVert} = 1} } {\left\lvert {L(x)} \right\rvert} .\end{align*}

A space is a Banach space if and only if it is a complete normed vector space.

A Hilbert space is an inner product space which is a Banach space under the induced norm.

## 1.3 Theorems

### 1.3.1 Topology / Sets

Every continuous function on a compact space is uniformly continuous.

Metric spaces are compact iff they are sequentially compact, (i.e. every sequence has a convergent subsequence).

The unit ball in $$C([0, 1])$$ with the sup norm is not compact.

(Click to expand)

Take $$f_k(x) = x^n$$, which converges to $$\chi(x=1)$$. The limit is not continuous, so no subsequence can converge.

A finite union of nowhere dense is again nowhere dense.

\begin{align*}\sum a_n < \infty \implies a_n \to 0 {\quad \operatorname{and} \quad} \sum_{k=N}^\infty a_n \overset{N\to\infty}\to 0\end{align*}

$$X\subseteq {\mathbb{R}}^n$$ is compact $$\iff X$$ is closed and bounded.

\begin{align*} \sum_{k=0}^\infty x^k = \frac 1 {1-x} \iff {\left\lvert {x} \right\rvert} < 1 .\end{align*}

\begin{align*} \sum_{k=0}^\infty \frac 1 {2^k} = 1 .\end{align*}

The Cantor set is closed with empty interior.

(Click to expand)

Its complement is a union of open intervals, and can’t contain an interval since intervals have positive measure and $$m(C_n)$$ tends to zero.

The Cantor set is nowhere dense.

Singleton sets in $${\mathbb{R}}$$ are closed, and thus $${\mathbb{Q}}$$ is an $$F_\sigma$$ set.

$${\mathbb{R}}$$ is a Baire space Thus $${\mathbb{R}}$$ can not be written as a countable union of nowhere dense sets.

Any nonempty set which is bounded from above (resp. below) has a well-defined supremum (resp. infimum).

### 1.3.2 Functions

There exist smooth compactly supported functions, e.g. take \begin{align*} f(x) = e^{-\frac{1}{x^2}} \chi_{(0, \infty)}(x) .\end{align*}

There is a function discontinuous precisely on $${\mathbb{Q}}$$.

(Click to expand)

$$f(x) = \frac 1 n$$ if $$x = r_n \in {\mathbb{Q}}$$ is an enumeration of the rationals, and zero otherwise. The limit at every point is 0.

There do not exist functions that are discontinuous precisely on $${\mathbb{R}}\setminus {\mathbb{Q}}$$.

(Click to expand)

$$D_f$$ is always an $$F_\sigma$$ set, which follows by considering the oscillation $$\omega_f$$. $$\omega_f(x) = 0 \iff f$$ is continuous at $$x$$, and $$D_f = \cup_n A_{\frac 1 n}$$ where $$A_\varepsilon = \left\{{\omega_f \geq \varepsilon}\right\}$$ is closed.

A function $$f: (a, b) \to {\mathbb{R}}$$ is Lipschitz $$\iff f$$ is differentiable and $$f'$$ is bounded. In this case, $${\left\lvert {f'(x)} \right\rvert} \leq C$$, the Lipschitz constant.

## 1.4 Uniform Convergence

$$f_n \to f$$ uniformly iff there exists an $$M_n$$ such that $${\left\lVert {f_n - f} \right\rVert}_\infty \leq M_n \to 0$$.

Negating: find an $$x$$ which depends on $$n$$ for which $${\left\lVert {f_n} \right\rVert}_\infty > \varepsilon$$ (negating small tails) or $${\left\lVert {f_n - f_m} \right\rVert} > \varepsilon$$ (negating the Cauchy criterion).

### 1.4.1 Example: Completeness of a Normed Function Space

:::{.proposition title=" $$C(I)$$ is complete"} The space $$X = C([0, 1])$$, continuous functions $$f: [0, 1] \to {\mathbb{R}}$$, equipped with the norm \begin{align*} {\left\lVert {f} \right\rVert}_\infty \mathrel{\vcenter{:}}=\sup_{x\in [0, 1]} {\left\lvert {f(x)} \right\rvert} \end{align*} is a complete metric space. :::

(Click to expand)
1. Let $$\left\{{f_k}\right\}$$ be Cauchy in $$X$$.

2. Define a candidate limit using pointwise convergence:

Fix an $$x$$; since \begin{align*} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} \leq {\left\lVert {f_k - f_k} \right\rVert} \to 0 \end{align*} the sequence $$\left\{{f_k(x)}\right\}$$ is Cauchy in $${\mathbb{R}}$$. So define $$f(x) \mathrel{\vcenter{:}}=\lim_k f_k(x)$$.

3. Show that $${\left\lVert {f_k - f} \right\rVert} \to 0$$: \begin{align*} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} < \varepsilon ~\forall x \implies \lim_{j} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} <\varepsilon ~\forall x \end{align*} Alternatively, $${\left\lVert {f_k-f} \right\rVert} \leq {\left\lVert {f_k - f_N} \right\rVert} + {\left\lVert {f_N - f_j} \right\rVert}$$, where $$N, j$$ can be chosen large enough to bound each term by $$\varepsilon/2$$.

4. Show that $$f\in X$$:

The uniform limit of continuous functions is continuous.

In other cases, you may need to show the limit is bounded, or has bounded derivative, or whatever other conditions define $$X$$.

If $$f_n\to f$$ pointwise and uniformly with each $$f_n$$ continuous, then $$f$$ is continuous.3

(Click to expand)
• Follows from an $$\varepsilon/3$$ argument: \begin{align*} {\left\lvert {F(x) - F(y} \right\rvert} \leq {\left\lvert {F(x) - F_N(x)} \right\rvert} + {\left\lvert {F_N(x) - F_N(y)} \right\rvert} + {\left\lvert {F_N(y) - F(y)} \right\rvert} \leq \varepsilon\to 0 .\end{align*}

• The first and last $$\varepsilon/3$$ come from uniform convergence of $$F_N\to F$$.
• The middle $$\varepsilon/3$$ comes from continuity of each $$F_N$$.
• So just need to choose $$N$$ large enough and $$\delta$$ small enough to make all 3 $$\varepsilon$$ bounds hold.

If $$f_n \to f$$ uniformly, then $$\int f_n = \int f$$.

### 1.4.2 Series

Let $$n$$ be a fixed dimension and set $$B = \left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} \leq 1}\right\}$$. \begin{align*} \sum \frac 1 {n^p} < \infty &\iff p>1 \\ \int_\varepsilon^\infty \frac 1 {x^p} < \infty &\iff p>1 \\ \int_0^1 \frac 1 {x^p} < \infty &\iff p<1 \\ \int_B \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p < n \\ \int_{B^c} \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p > n \\ .\end{align*}

If $$0\leq a_n \leq b_n$$, then

• $$\sum b_n < \infty \implies \sum a_n < \infty$$, and
• $$\sum a_n = \infty \implies \sum b_n = \infty$$.

If $$\sum f_n$$ converges then $$f_n \to 0$$ uniformly.

If $$f_n$$ are continuous and $$\sum f_n \to f$$ converges uniformly, then $$f$$ is continuous.

If $$f_n(x) \leq M_n$$ for a fixed $$x$$ where $$\sum M_n < \infty$$, then the series $$f(x) = \sum f_n(x)$$ converges.4

If $$\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \leq M_n$$ for each $$n$$ where $$\sum M_n < \infty$$, then $$\sum_{n=1}^\infty f_n(x)$$ converges uniformly and absolutely on $$A$$.5 Conversely, if $$\sum f_n$$ converges uniformly on $$A$$ then $$\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \to 0$$.

$$f_n$$ are uniformly Cauchy (so $${\left\lVert {f_n - f_m} \right\rVert}_\infty < \varepsilon$$) iff $$f_n$$ is uniformly convergent.

#### 1.4.2.1 Derivatives

If $$f_n$$ are differentiable, $$\sum f_n' \to g$$ uniformly, and there exists one point6 $$x_0$$ such that $$\sum f_n(x)$$ converges, then there exist an $$f$$ such that $$\sum f_n \to f$$ uniformly and $$f' = g$$.7

## 1.5 Commuting Limiting Operations

\begin{align*} \lim_{n\to \infty}\sup_{x\in X} {\left\lvert {f_n(x) } \right\rvert} \neq \sup_{x\in X} {\left\lvert {\lim_{n\to\infty} f_n(x) } \right\rvert} .\end{align*}

\begin{align*} \lim_{k\to \infty} \lim_{n\to\infty} f_n(x_k) \neq \lim_{n\to \infty} \lim_{k\to\infty} f_n(x_k) .\end{align*}

\begin{align*} \lim_{n\to \infty} {\frac{\partial }{\partial x}\,} f_n \neq {\frac{\partial }{\partial n}\,} \qty{\lim_{n\to \infty} f_n} .\end{align*}

\begin{align*} \lim_{n\to \infty} \int_a^b f_n(x) \,dx \neq \int_a^b \lim_{n\to \infty} \qty{ f_n(x) } \,dx .\end{align*}

If $$[a, b] \subset {\mathbb{R}}$$ is a closed interval and $$f$$ is continuous, then for every $$\varepsilon> 0$$ there exists a polynomial $$p_\varepsilon$$ such that $${\left\lVert {f- p_\varepsilon} \right\rVert}_{L^\infty([a, b])} \overset{\varepsilon\to 0}\to 0$$.

Equivalently, polynomials are dense in the Banach space $$C([0, 1], {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)$$.

Let $$E \subseteq {\mathbb{R}}^n$$ be measurable with $$m(E) > 0$$ and $$\left\{{f_k: E \to {\mathbb{R}}}\right\}$$ be measurable functions such that \begin{align*} f(x) \mathrel{\vcenter{:}}=\lim_{k\to\infty} f_k(x) < \infty \end{align*} exists almost everywhere.

Then $$f_k \to f$$ almost uniformly, i.e. \begin{align*} \forall\varepsilon > 0, ~\exists F \subseteq E ~\text{closed such that } & m(E\setminus F) < \varepsilon ~\text{ and }~ f_k\to f ~\text{uniformly on}~ F .\end{align*}

## 1.7 Examples

A series of continuous functions that does not converge uniformly but is still continuous: \begin{align*} g(x) \mathrel{\vcenter{:}}=\sum {1 \over 1 + n^2 x} .\end{align*}

Take $$x = 1/n^2$$.

# 2 Measure Theory

## 2.1 Theorems

Every open subset of $${\mathbb{R}}$$ (resp $${\mathbb{R}}^n$$) can be written as a unique countable union of disjoint (resp. almost disjoint) intervals (resp. cubes).

1. Monotonicity: $$E\subseteq F \implies m_*(E) \leq m_*(F)$$.
2. Countable Subadditivity: $$m_*(\cup E_{i}) \leq \sum m_*(E_{i})$$.
3. Approximation: For all $$E$$ there exists a $$G \supseteq E$$ such that $$m_*(G) \leq m_*(E) + \varepsilon$$.
4. Disjoint8 Additivity: $$m_*(A {\coprod}B) = m_*(A) + m_*(B)$$.

\begin{align*}m(A) = m(B) + m(C) {\quad \operatorname{and} \quad} m(C) < \infty \implies m(A) - m(C) = m(B).\end{align*}

\begin{align*} E_{i} \nearrow E &\implies m(E_{i}) \to m(E) \\ m(E_{1}) < \infty \text{ and } E_{i} \searrow E &\implies m(E_{i}) \to m(E) .\end{align*}

(Click to expand)
1. Break into disjoint annuli $$A_{2} = E_{2}\setminus E_{1}$$, etc then apply countable disjoint additivity to $$E = {\coprod}A_{i}$$.

2. Use $$E_{1} = ({\coprod}E_{j}\setminus E_{j+1}) {\coprod}(\cap E_{j})$$, taking measures yields a telescoping sum,and use countable disjoint additivity.

Suppose $$E$$ is measurable; then for every $$\varepsilon>0$$,

1. There exists an open $$O\supset E$$ with $$m(O\setminus E) < \varepsilon$$
2. There exists a closed $$F\subset E$$ with $$m(E\setminus F) < \varepsilon$$
3. There exists a compact $$K\subset E$$ with $$m(E\setminus K) < \varepsilon$$.
(Click to expand)
• (1): Take $$\left\{{Q_{i}}\right\} \rightrightarrows E$$ and set $$O = \cup Q_{i}$$.
• (2): Since $$E^c$$ is measurable, produce $$O\supset E^c$$ with $$m(O\setminus E^c) < \varepsilon$$.
• Set $$F = O^c$$, so $$F$$ is closed.
• Then $$F\subset E$$ by taking complements of $$O\supset E^c$$
• $$E\setminus F = O\setminus E^c$$ and taking measures yields $$m(E\setminus F) < \varepsilon$$
• (3): Pick $$F\subset E$$ with $$m(E\setminus F) < \varepsilon/2$$.
• Set $$K_{n} = F\cap{\mathbb{D}}_{n}$$, a ball of radius $$n$$ about $$0$$.
• Then $$E\setminus K_{n} \searrow E\setminus F$$
• Since $$m(E) < \infty$$, there is an $$N$$ such that $$n\geq N \implies m(E\setminus K_{n}) < \varepsilon$$.

Lebesgue measure is translation and dilation invariant.

(Click to expand)

Obvious for cubes; if $$Q_{i} \rightrightarrows E$$ then $$Q_{i} + k \rightrightarrows E + k$$, etc.

There is a non-measurable set.

(Click to expand)
• Use AOC to choose one representative from every coset of $${\mathbb{R}}/{\mathbb{Q}}$$ on $$[0, 1)$$, which is countable, and assemble them into a set $$N$$
• Enumerate the rationals in $$[0, 1]$$ as $$q_{j}$$, and define $$N_{j} = N + q_{j}$$. These intersect trivially.
• Define $$M \mathrel{\vcenter{:}}={\coprod}N_{j}$$, then $$[0, 1) \subseteq M \subseteq [-1, 2)$$, so the measure must be between 1 and 3.
• By translation invariance, $$m(N_{j}) = m(N)$$, and disjoint additivity forces $$m(M) = 0$$, a contradiction.

If $$E$$ is Lebesgue measurable, then $$E = H {\coprod}N$$ where $$H \in F_\sigma$$ and $$N$$ is null.

(Click to expand)

For every $$\frac 1 n$$ there exists a closed set $$K_{n} \subset E$$ such that $$m(E\setminus K_{n}) \leq \frac 1 n$$. Take $$K = \cup K_{n}$$, wlog $$K_{n} \nearrow K$$ so $$m(K) = \lim m(K_{n}) = m(E)$$. Take $$N\mathrel{\vcenter{:}}= E\setminus K$$, then $$m(N) = 0$$.

If $$A_{n}$$ are all measurable, $$\limsup A_{n}$$ and $$\liminf A_{n}$$ are measurable.

(Click to expand)

Measurable sets form a sigma algebra, and these are expressed as countable unions/intersections of measurable sets.

Let $$\{E_{k}\}$$ be a countable collection of measurable sets. Then \begin{align*} \sum_{k} m(E_{k}) < \infty \implies \text{ almost every } x\in {\mathbb{R}}\text{ is in at most finitely many } E_{k} .\end{align*}

(Click to expand)
• If $$E = \limsup_{j} E_{j}$$ with $$\sum m(E_{j}) < \infty$$ then $$m(E) = 0$$.
• If $$E_{j}$$ are measurable, then $$\limsup_{j} E_{j}$$ is measurable.
• If $$\sum_{j} m(E_{j}) < \infty$$, then $$\sum_{j=N}^\infty m(E_{j}) \overset{N\to\infty}\to 0$$ as the tail of a convergent sequence.
• $$E = \limsup_{j} E_{j} = \cap_{k=1}^\infty \cup_{j=k}^\infty E_{j} \implies E \subseteq \cup_{j=k}^\infty$$ for all $$k$$
• $$E \subset \cup_{j=k}^\infty \implies m(E) \leq \sum_{j=k}^\infty m(E_{j}) \overset{k\to\infty}\to 0$$.
• Characteristic functions are measurable
• If $$f_{n}$$ are measurable, so are $${\left\lvert {f_{n}} \right\rvert}, \limsup f_{n}, \liminf f_{n}, \lim f_{n}$$,
• Sums and differences of measurable functions are measurable,
• Cones $$F(x,y) = f(x)$$ are measurable,
• Compositions $$f\circ T$$ for $$T$$ a linear transformation are measurable,
• “Convolution-ish” transformations $$(x,y) \mapsto f(x-y)$$ are measurable
(Click to expand)

Take the cone on $$f$$ to get $$F(x, y) = f(x)$$, then compose $$F$$ with the linear transformation $$T = [1, -1; 1, 0]$$.

# 3 Integration

## 3.1 Theorems

If $$f\in L^\infty(X)$$, then $$f$$ is equal to some bounded function $$g$$ almost everywhere.

$$f(x) = x\chi_{\mathbb{Q}}(x)$$ is essentially bounded but not bounded.

\begin{align*} \int_0^1 {1\over x^p} < \infty \iff p < 1 \\ \int_1^\infty {1\over x^p} < \infty \iff p > 1 .\end{align*}

Large powers of $$x$$ help us in neighborhoods of infinity and hurt around zero

### 3.1.1 Convergence Theorems

If $$f_n \in L^+$$ and $$f_n \nearrow f$$ almost everywhere, then \begin{align*} \lim \int f_n = \int \lim f_n = \int f \quad \text{i.e.}~~ \int f_n \to \int f .\end{align*}

Needs to be positive and increasing.

If $$f_n \in L^1$$ and $$f_n \to f$$ almost everywhere with $${\left\lvert {f_n} \right\rvert} \leq g$$ for some $$g\in L^1$$, then $$f\in L^1$$ and \begin{align*} \lim \int f_n = \int \lim f_n = \int f \quad \text{i.e.}~~ \int f_n \to \int f < \infty ,\end{align*}

and more generally, \begin{align*} \int {\left\lvert {f_n - f} \right\rvert} \to 0 .\end{align*}

Positivity not needed.

If

• $$f_n \in L^1$$ with $$f_n \to f$$ almost everywhere,
• There exist $$g_n \in L^1$$ with $${\left\lvert {f_n} \right\rvert} \leq g_n$$, $$g_n \geq 0$$.
• $$g_n\to g$$ almost everywhere with $$g\in L^1$$, and
• $$\lim \int g_n = \int g$$,

then $$f\in L^1$$ and $$\lim \int f_n = \int f < \infty$$.

Note that this is the DCT with $${\left\lvert {f_n} \right\rvert} < {\left\lvert {g} \right\rvert}$$ relaxed to $${\left\lvert {f_n} \right\rvert} < g_n \to g\in L^1$$.

(Click to expand)

Proceed by showing $$\limsup \int f_n \leq \int f \leq \liminf \int f_n$$:

• $$\int f \geq \limsup \int f_n$$: \begin{align*} \int g - \int f &= \int \qty{g-f} \\ &\leq \liminf \int \qty{g_n - f_n} \quad \text{Fatou} \\ &= \lim \int g_n + \liminf \int (-f_n) \\ &= \lim \int g_n - \limsup \int f_n \\ &= \int g - \limsup \int f_n \\ \\ \implies \int f &\geq \limsup \int f_n .\end{align*}

• Here we use $$g_n - f_n \overset{n\to\infty} g-f$$ with $$0 \leq {\left\lvert {f_n} \right\rvert} - f_n \leq g_n - f_n$$, so $$g_n - f_n$$ are nonnegative (and measurable) and Fatou’s lemma applies.
• $$\int f \leq \liminf \int f_n$$: \begin{align*} \int g + \int f &= \int(g+f) \\ &\leq \liminf \int \qty{g_n + f_n} \\ &= \lim \int g_n + \liminf \int f_n \\ &= \int g + \liminf f_n \\ \\ \int f &\leq \liminf \int f_n .\end{align*}

• Here we use that $$g_n + f_n \to g+f$$ with $$0 \leq {\left\lvert {f_n} \right\rvert} + f_n \leq g_n + f_n$$ so Fatou’s lemma again applies.

If $$f\in L^1$$, then \begin{align*} \int{\left\lvert {f_n - f} \right\rvert} \to 0 \iff \int {\left\lvert {f_n} \right\rvert} \to \int {\left\lvert {f} \right\rvert} .\end{align*}

(Click to expand)

Let $$g_n = {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert}$$, then $$g_n \to {\left\lvert {f} \right\rvert}$$ and \begin{align*} {\left\lvert {g_n} \right\rvert} = {\left\lvert { {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert} } \right\rvert} \leq {\left\lvert {f_n - (f_n - f)} \right\rvert} = {\left\lvert {f} \right\rvert} \in L^1 ,\end{align*} so the DCT applies to $$g_n$$ and \begin{align*} {\left\lVert {f_n - f} \right\rVert}_1 = \int {\left\lvert {f_n - f} \right\rvert} + {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n} \right\rvert} = \int {\left\lvert {f_n} \right\rvert} - g_n\\ \to_{DCT} \lim \int {\left\lvert {f_n} \right\rvert} - \int {\left\lvert {f} \right\rvert} .\end{align*}

If $$f_n$$ is a sequence of nonnegative measurable functions, then \begin{align*} \int \liminf_n f_n &\leq \liminf_n \int f_n \\ \limsup_n \int f_n &\leq \int \limsup_n f_n .\end{align*}

For $$f(x, y)$$ non-negative and measurable, for almost every $$x\in {\mathbb{R}}^n$$,

• $$f_x(y)$$ is a measurable function
• $$F(x) = \int f(x, y) ~dy$$ is a measurable function,
• For $$E$$ measurable, the slices $$E_x \mathrel{\vcenter{:}}=\left\{{y {~\mathrel{\Big|}~}(x, y) \in E}\right\}$$ are measurable.
• $$\int f = \int \int F$$, i.e. any iterated integral is equal to the original.

For $$f(x, y)$$ integrable, for almost every $$x\in {\mathbb{R}}^n$$,

• $$f_x(y)$$ is an integrable function
• $$F(x) \mathrel{\vcenter{:}}=\int f(x, y) ~dy$$ is an integrable function,
• For $$E$$ measurable, the slices $$E_x \mathrel{\vcenter{:}}=\left\{{y {~\mathrel{\Big|}~}(x, y) \in E}\right\}$$ are measurable.
• $$\int f = \int \int f(x,y)$$, i.e. any iterated integral is equal to the original

If any iterated integral is absolutely integrable, i.e. $$\int \int {\left\lvert {f(x, y)} \right\rvert} < \infty$$, then $$f$$ is integrable and $$\int f$$ equals any iterated integral.

Let $$E$$ be a measurable subset of $${\mathbb{R}}^n$$. Then

• For almost every $$x\in {\mathbb{R}}^{n_1}$$, the slice $$E_x \mathrel{\vcenter{:}}=\left\{{y \in {\mathbb{R}}^{n_2} \mathrel{\Big|}(x,y) \in E}\right\}$$ is measurable in $${\mathbb{R}}^{n_2}$$.
• The function

\begin{align*} F: {\mathbb{R}}^{n_1} &\to {\mathbb{R}}\\ x &\mapsto m(E_x) = \int_{{\mathbb{R}}^{n_2}} \chi_{E_x} ~dy \end{align*} is measurable and \begin{align*} m(E) = \int_{{\mathbb{R}}^{n_1}} m(E_x) ~dx = \int_{{\mathbb{R}}^{n_1}} \int_{{\mathbb{R}}^{n_2}} \chi_{E_x} ~dy ~dx .\end{align*}

(Click to expand)

$$\implies$$:

• Let $$f$$ be measurable on $${\mathbb{R}}^n$$.
• Then the cylinders $$F(x, y) = f(x)$$ and $$G(x, y) = f(y)$$ are both measurable on $${\mathbb{R}}^{n+1}$$.
• Write $$\mathcal{A} = \left\{{G \leq F}\right\} \cap\left\{{G \geq 0}\right\}$$; both are measurable.

$$\impliedby$$:

• Let $$A$$ be measurable in $${\mathbb{R}}^{n+1}$$.
• Define $$A_x = \left\{{y\in {\mathbb{R}}\mathrel{\Big|}(x, y) \in \mathcal{A}}\right\}$$, then $$m(A_x) = f(x)$$.
• By the corollary, $$A_x$$ is measurable set, $$x \mapsto A_x$$ is a measurable function, and $$m(A) = \int f(x) ~dx$$.
• Then explicitly, $$f(x) = \chi_{A}$$, which makes $$f$$ a measurable function.

If $${\left\lvert {{\frac{\partial }{\partial t}\,}f(x, t)} \right\rvert} \leq g(x) \in L^1$$, then letting $$F(t) = \int f(x, t) ~dt$$, \begin{align*} {\frac{\partial }{\partial t}\,} F(t) &\mathrel{\vcenter{:}}=\lim_{h \rightarrow 0} \int \frac{f(x, t+h)-f(x, t)}{h} d x \\ &\mathop{\mathrm{=}}^{\scriptstyle\text{DCT}} \int {\frac{\partial }{\partial t}\,} f(x, t) ~dx .\end{align*}

To justify passing the limit, let $$h_k \to 0$$ be any sequence and define \begin{align*} f_k(x, t) = \frac{f(x, t+h_k)-f(x, t)}{h_k} ,\end{align*} so $$f_k \overset{\text{pointwise}}\to {\frac{\partial }{\partial t}\,}f$$.

Apply the MVT to $$f_k$$ to get $$f_k(x, t) = f_k(\xi, t)$$ for some $$\xi \in [0, h_k]$$, and show that $$f_k(\xi, t) \in L_1$$.

If $$f_n$$ are non-negative and $$\sum \int {\left\lvert {f} \right\rvert}_n < \infty$$, then $$\sum \int f_n = \int \sum f_n$$.

(Click to expand)
• Idea: MCT.
• Let $$F_N = \sum^N f_n$$ be a finite partial sum;
• Then there are simple functions $$\phi_n \nearrow f_n$$
• So $$\sum^N \phi_n \nearrow F_N$$ and MCT applies

If $$\left\{{f_n}\right\}$$ integrable with either $$\sum \int {\left\lvert {f_n} \right\rvert} < \infty$$ or $$\int\sum {\left\lvert {f_n} \right\rvert} < \infty$$, then \begin{align*} \int\sum f_n = \sum \int f_n .\end{align*}

(Click to expand)
• By Tonelli, if $$f_n(x) \geq 0$$ for all $$n$$, taking the counting measure allows interchanging the order of “integration.”
• By Fubini on $${\left\lvert {f_n} \right\rvert}$$, if either “iterated integral” is finite then the result follows.

If $$f_k \in L^1$$ and $$\sum {\left\lVert {f_k} \right\rVert}_1 < \infty$$ then $$\sum f_k$$ converges almost everywhere and in $$L^1$$.

(Click to expand)

Define $$F_N = \sum^N f_k$$ and $$F = \lim_N F_N$$, then $${\left\lVert {F_N} \right\rVert}_1 \leq \sum^N {\left\lVert {f_k} \right\rVert} < \infty$$ so $$F\in L^1$$ and $${\left\lVert {F_N - F} \right\rVert}_1 \to 0$$ so the sum converges in $$L^1$$. Almost everywhere convergence: ?

## 3.2 Examples of (Non)Integrable Functions

• $$\int {1\over 1 + x^2} = \arctan(x) \overset{x\to\infty}\to \pi/2 < \infty$$

• Any bounded function (or continuous on a compact set, by EVT)

• $$\int_0^1 {1 \over \sqrt{x}} < \infty$$

• $$\int_0^1 {1\over x^{1-\varepsilon}} < \infty$$

• $$\int_1^\infty {1\over x^{1+\varepsilon}} < \infty$$

• $$\int_0^1 {1\over x} = \infty$$.
• $$\int_1^\infty {1\over x} = \infty$$.
• $$\int_1^\infty {1 \over \sqrt{x}} = \infty$$
• $$\int_1^\infty {1\over x^{1-\varepsilon}} = \infty$$
• $$\int_0^1 {1\over x^{1+\varepsilon}} = \infty$$

## 3.3$$L^1$$ Facts

For $$f\in L^+$$, \begin{align*} \int f = 0 \quad\iff\quad f \equiv 0 \text{ almost everywhere} .\end{align*}

(Click to expand)
• Obvious for simple functions:
• If $$f(x) = \sum_{j=1}^n c_j \chi_{E_j}$$, then $$\int f = 0$$ iff for each $$j$$, either $$c_j=0$$ or $$m(E_j) = 0$$.
• Since nonzero $$c_j$$ correspond to sets where $$f\neq 0$$, this says $$m\qty{\left\{{f\neq 0}\right\}} = 0$$.
• $$\impliedby$$:
• If $$f= 0$$ almost everywhere and $$\phi \nearrow f$$, then $$\phi = 0$$ almost everywhere since $$\phi(x) \leq f(x)$$ -Then \begin{align*} \int f = \sup_{\phi \leq f} \int \phi = \sup_{\phi \leq f} 0 = 0 .\end{align*}
• $$\implies$$:
• Instead show negating “$$f=0$$ almost everywhere” implies $$\int f \neq 0$$.
• Write $$\left\{{f\neq 0}\right\} = \cup_{n\in {\mathbb{N}}} S_n$$ where $$S_n \mathrel{\vcenter{:}}=\left\{{x{~\mathrel{\Big|}~}f(x) \geq {1\over n}}\right\}$$.
• Since “not $$f=0$$ almost everywhere,” there exists an $$n$$ such that $$m(S_n) > 0$$.
• Then \begin{align*} 0 < {1\over n} \chi_{E_n} \leq f \implies 0 < \int {1\over n} \chi_{E_n} \leq \int f .\end{align*}

The Lebesgue integral is translation invariant, i.e. \begin{align*} \int f(x) ~dx = \int f(x + h) ~dx &&\text{for any} h .\end{align*}

(Click to expand)
• Let $$E\subseteq X$$; for characteristic functions, \begin{align*} \int_X \chi_E(x+h) = \int_{X} \chi_{E+h}(x) = m(E+h) = m(E) = \int_X \chi_E(x) \end{align*} by translation invariance of measure.
• So this also holds for simple functions by linearity.
• For $$f\in L^+$$, choose $$\phi_n \nearrow f$$ so $$\int \phi_n \to \int f$$.
• Similarly, $$\tau_h \phi_n \nearrow \tau_h f$$ so $$\int \tau_h f \to \int f$$
• Finally $$\left\{{\int \tau_h \phi}\right\} = \left\{{\int \phi}\right\}$$ by step 1, and the suprema are equal by uniqueness of limits.

If $$X \subseteq A \cup B$$, then $$\int_X f \leq \int_A f + \int_{A^c} f$$ with equality iff $$X = A{\coprod}B$$.

If $$f \in L^1$$ and $$f$$ is uniformly continuous, then $$f(x) \overset{{\left\lvert {x} \right\rvert}\to\infty}\to 0$$.

This doesn’t hold for general $$L^1$$ functions, take any train of triangles with height 1 and summable areas.

If $$f\in L^1$$, then for every $$\varepsilon$$ there exists a radius $$R$$ such that if $$A = B_R(0)^c$$, then $$\int_A {\left\lvert {f} \right\rvert} < \varepsilon$$.

(Click to expand)
• Approximate with compactly supported functions.
• Take $$g\overset{L_1}\to f$$ with $$g\in C_c$$
• Then choose $$N$$ large enough so that $$g=0$$ on $$E\mathrel{\vcenter{:}}= B_N(0)$$
• Then \begin{align*} \int_E {\left\lvert {f} \right\rvert} \leq \int_E{\left\lvert {f-g} \right\rvert} + \int_E {\left\lvert {g} \right\rvert}.\end{align*}

$$m(E) \to 0 \implies \int_E f \to 0$$.

(Click to expand)

Approximate with compactly supported functions. Take $$g\overset{L_1}\to f$$, then $$g \leq M$$ so $$\int_E{f} \leq \int_E{f-g} + \int_E g \to 0 + M \cdot m(E) \to 0$$.

If $$f\in L^1$$, then $$m(\left\{{f(x) = \infty}\right\}) = 0$$.

(Click to expand)

Idea: Split up domain Let $$A = \left\{{f(x) = \infty}\right\}$$, then $$\infty > \int f = \int_A f + \int_{A^c} f = \infty \cdot m(A) + \int_{A^c} f \implies m(X) =0$$.

\begin{align*} {\left\lVert {\tau_h f - f} \right\rVert}_1 \overset{h\to 0}\to 0 \end{align*}

(Click to expand)

Approximate with compactly supported functions. Take $$g\overset{L_1}\to f$$ with $$g\in C_c$$. \begin{align*} \int f(x+h) - f(x) &\leq \int f(x+h) - g(x+h) + \int g(x+h) - g(x) + \int g(x) - f(x) \\ &\overset{?\to?}\to 2 \varepsilon + \int g(x+h) - g(x) \\ &= \int_K g(x+h) - g(x) + \int_{K^c} g(x+h) - g(x)\\ &\overset{??}\to 0 ,\end{align*} which follows because we can enlarge the support of $$g$$ to $$K$$ where the integrand is zero on $$K^c$$, then apply uniform continuity on $$K$$.

\begin{align*} F(x):=\int_{0}^{x} f(y) d y \quad \text { and } \quad G(x):=\int_{0}^{x} g(y) d y \\ \implies \int_{0}^{1} F(x) g(x) d x=F(1) G(1)-\int_{0}^{1} f(x) G(x) d x .\end{align*}

(Click to expand)

Fubini-Tonelli, and sketch region to change integration bounds.

\begin{align*} A_{h}(f)(x):=\frac{1}{2 h} \int_{x-h}^{x+h} f(y) d y \implies {\left\lVert {A_h(f) - f} \right\rVert} \overset{h\to 0}\to 0 .\end{align*}

(Click to expand)

Fubini-Tonelli, and sketch region to change integration bounds, and continuity in $$L^1$$.

## 3.4 Lp Facts

The following are dense subspaces of $$L^2([0, 1])$$:

• Simple functions
• Step functions
• $$C_0([0, 1])$$
• Smoothly differentiable functions $$C_0^\infty([0, 1])$$
• Smooth compactly supported functions $$C_c^\infty$$

\begin{align*} m(X) < \infty \implies \lim_{p\to\infty} {\left\lVert {f} \right\rVert}_p = {\left\lVert {f} \right\rVert}_\infty .\end{align*}

(Click to expand)

Let $$M = {\left\lVert {f} \right\rVert}_\infty$$.

• For any $$L < M$$, let $$S = \left\{{{\left\lvert {f} \right\rvert} \geq L}\right\}$$.
• Then $$m(S) > 0$$ and

\begin{align*} {\left\lVert {f} \right\rVert}_{p} &= \left( \int_X {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\geq \left( \int_S {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\geq L ~m(S)^{\frac 1 p} \overset{p\to\infty}\to L \\ &\implies \liminf_p {\left\lVert {f} \right\rVert}_{p} \geq M .\end{align*}

We also have \begin{align*} {\left\lVert {f} \right\rVert}_{p} &= \left( \int_X {\left\lvert {f} \right\rvert}^p \right)^{\frac 1 p} \\ &\leq \left( \int_X M^p \right)^{\frac 1 p} \\ &= M ~m(X)^{\frac 1 p} \xrightarrow{p\to\infty} M \\ &\implies \limsup_p {\left\lVert {f} \right\rVert}_{p} \leq M \hfill\blacksquare .\end{align*}

For $$1\leq p< \infty$$, $$(L^p)^\vee\cong L^q$$.

(Click to expand)

?

(Click to expand)

Use Riesz Representation for Hilbert spaces.

$$L^1 \subset (L^\infty)^\vee$$, since the isometric mapping is always injective, but never surjective.

# 4 Fourier Transform and Convolution

## 4.1 The Fourier Transform

If $$\widehat{f} = \widehat{g}$$ then $$f=g$$ almost everywhere.

\begin{align*} f\in L^1 \implies \widehat{f}(\xi) \rightarrow 0 \text { as }|\xi| \rightarrow \infty ,\end{align*}

if $$f \in L^1$$, then $$\widehat{f}$$ is continuous and bounded.

(Click to expand)
• Boundedness: \begin{align*} {\left\lvert {\widehat{f}(\xi)} \right\rvert} \leq \int {\left\lvert {f} \right\rvert}\cdot {\left\lvert {e^{2\pi i x\cdot \xi }} \right\rvert} = {\left\lVert {f} \right\rVert}_{1} .\end{align*}

• Continuity:

• $${\left\lvert {\widehat{f}(\xi_{n}) - \widehat{f} (\xi) } \right\rvert}$$

• Apply DCT to show $$a\overset{n\to\infty}\to 0$$.

\begin{align*} f(x)=\int_{\mathbb{R}^{n}} \widehat{f}(x) e^{2 \pi i x \cdot \xi} d \xi .\end{align*}

Fubini-Tonelli does not work here!

(Click to expand)

Idea: Fubini-Tonelli doesn’t work directly, so introduce a convergence factor, take limits, and use uniqueness of limits.

• Take the modified integral:

\begin{align*} I_{t}(x) &= \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~e^{-\pi t^2 {\left\lvert {\xi} \right\rvert}^2} \\ &= \int \widehat{f}(\xi) \phi(\xi) \\ &= \int f(\xi) \widehat{\phi}(\xi) \\ &= \int f(\xi) \widehat{\widehat{g}}(\xi - x) \\ &= \int f(\xi) g_{t}(x - \xi) ~d\xi \\ &= \int f(y-x) g_{t}(y) ~dy \quad (\xi = y-x)\\ &= (f \ast g_{t}) \\ &\to f \text{ in $L^1$ as }t \to 0 .\end{align*}

• We also have \begin{align*} \lim_{t\to 0} I_{t}(x) &= \lim_{t\to 0} \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~e^{-\pi t^2 {\left\lvert {\xi} \right\rvert}^2} \\ &= \lim_{t\to 0} \int \widehat{f}(\xi) \phi(\xi) \\ &=_{DCT} \int \widehat{f}(\xi) \lim_{t\to 0} \phi(\xi) \\ &= \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} \\ .\end{align*}

• So \begin{align*} I_{t}(x) \to \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~\text{ pointwise and }~{\left\lVert {I_{t}(x) - f(x)} \right\rVert}_{1} \to 0 .\end{align*}

• So there is a subsequence $$I_{t_{n}}$$ such that $$I_{t_{n}}(x) \to f(x)$$ almost everywhere

• Thus $$f(x) = \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi}$$ almost everywhere by uniqueness of limits.

\begin{align*} g(x) \mathrel{\vcenter{:}}= e^{-\pi {\left\lvert {t} \right\rvert}^2} \implies \widehat{g}(\xi) = g(\xi) {\quad \operatorname{and} \quad} \widehat{g}_{t}(x) = g(tx) = e^{-\pi t^2 {\left\lvert {x} \right\rvert}^2} .\end{align*}

## 4.2 Approximate Identities

\begin{align*} \phi(x) \mathrel{\vcenter{:}}= e^{-\pi x^2} .\end{align*}

\begin{align*} {\left\lVert {f \ast \phi_{t} - f} \right\rVert}_{1} \overset{t\to 0}\to 0 .\end{align*}

(Click to expand)

\begin{align*} {\left\lVert {f - f\ast \phi_{t}} \right\rVert}_1 &= \int f(x) - \int f(x-y)\phi_{t}(y) ~dy dx \\ &= \int f(x)\int \phi_{t}(y) ~dy - \int f(x-y)\phi_{t}(y) ~dy dx \\ &= \int \int \phi_{t}(y)[f(x) - f(x-y)] ~dy dx \\ &=_{FT} \int \int \phi_{t}(y)[f(x) - f(x-y)] ~dx dy \\ &= \int \phi_{t}(y) \int f(x) - f(x-y) ~dx dy \\ &= \int \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy \\ &= \int_{y < \delta} \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy + \int_{y \geq \delta} \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy \\ &\leq \int_{y < \delta} \phi_{t}(y) \varepsilon + \int_{y \geq \delta} \phi_{t}(y) \left( {\left\lVert {f} \right\rVert}_1 + {\left\lVert {\tau_{y} f} \right\rVert}_1 \right) dy \quad\text{by continuity in } L^1 \\ &\leq \varepsilon + 2{\left\lVert {f} \right\rVert}_1 \int_{y \geq \delta} \phi_{t}(y) dy \\ &\leq \varepsilon + 2{\left\lVert {f} \right\rVert}_1 \cdot \varepsilon \quad\text{since $\phi_{t}$ has small tails} \\ &\overset{\varepsilon\to 0}\to 0 .\end{align*}

\begin{align*} f,g \in L^1 \text{ and bounded} \implies \lim_{|x| \rightarrow \infty} (f * g)(x) = 0 .\end{align*}

(Click to expand)
• Choose $$M \geq f,g$$.

• By small tails, choose $$N$$ such that $$\int_{B_{N}^c} {\left\lvert {f} \right\rvert}, \int_{B_{n}^c} {\left\lvert {g} \right\rvert} < \varepsilon$$

• Note \begin{align*} {\left\lvert {f \ast g} \right\rvert} \leq \displaystyle\int {\left\lvert {f(x-y)} \right\rvert} ~{\left\lvert {g(y)} \right\rvert} ~dy \mathrel{\vcenter{:}}= I .\end{align*}

• Use $${\left\lvert {x} \right\rvert} \leq {\left\lvert {x-y} \right\rvert} + {\left\lvert {y} \right\rvert}$$, take $${\left\lvert {x} \right\rvert}\geq 2N$$ so either \begin{align*} {\left\lvert {x-y} \right\rvert} \geq N \implies I \leq \int_{\left\{{x-y \geq N}\right\}} {\left\lvert {f(x-y)} \right\rvert}M ~dy\leq \varepsilon M \to 0 \end{align*} then \begin{align*} {\left\lvert {y} \right\rvert} \geq N \implies I \leq \int_{\left\{{y \geq N}\right\}} M{\left\lvert {g(y)} \right\rvert} ~dy\leq M \varepsilon \to 0 .\end{align*}

Take $$q = 1$$ in Young’s inequality to obtain \begin{align*} {\left\lVert {f \ast g} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}p {\left\lVert {g} \right\rVert}1 .\end{align*}

If $$f, g \in L^1$$ then $$f\ast g\in L^1$$.

# 5 Functional Analysis

## 5.1 Theorems

For any orthonormal set $$\left\{{u_{n}}\right\} \subseteq {\mathcal{H}}$$ a Hilbert space (not necessarily a basis), \begin{align*} \left\|x-\sum_{n=1}^{N}\left\langle x, u_{n}\right\rangle u_{n}\right\|^{2}=\|x\|^{2}-\sum_{n=1}^{N}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \end{align*} and thus \begin{align*} \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}

(Click to expand)
• Let $$S_{N} = \sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle} u_{n}$$ \begin{align*} {\left\lVert {x - S_{N}} \right\rVert}^2 &= {\left\langle {x - S_{n}},~{x - S_{N}} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re{\left\langle {x},~{S_{N}} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re {\left\langle {x},~{\sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle}u_{n}} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re \sum_{n=1}^N {\left\langle {x},~{ {\left\langle {x},~{u_{n}} \right\rangle}u_{n}} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re \sum_{n=1}^N \overline{{\left\langle {x},~{u_{n}} \right\rangle}}{\left\langle {x},~{u_{n}} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 + \left\|\sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle} u_{n}\right\|^2 - 2 \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 - 2 \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\ &= {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 .\end{align*}

• By continuity of the norm and inner product, we have \begin{align*} \lim_{N\to\infty} {\left\lVert {x - S_{N}} \right\rVert}^2 &= \lim_{N\to\infty} {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\ \implies {\left\lVert {x - \lim_{N\to\infty} S_{N}} \right\rVert}^2 &= {\left\lVert {x} \right\rVert}^2 - \lim_{N\to\infty}\sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2\\ \implies {\left\lVert {x - \sum_{n=1}^\infty {\left\langle {x},~{u_{n}} \right\rangle} u_{n}} \right\rVert}^2 &= {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 .\end{align*}

• Then noting that $$0 \leq {\left\lVert {x - S_{N}} \right\rVert}^2$$, \begin{align*} 0 &\leq {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\ \implies \sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 &\leq {\left\lVert {x} \right\rVert}^2 \hfill\blacksquare .\end{align*}

If $$\Lambda$$ is a continuous linear functional on a Hilbert space $$H$$, then there exists a unique $$y \in H$$ such that \begin{align*} \forall x\in H,\quad \Lambda(x) = {\left\langle {x},~{y} \right\rangle}. .\end{align*}

(Click to expand)
• Define $$M \mathrel{\vcenter{:}}=\ker \Lambda$$.
• Then $$M$$ is a closed subspace and so $$H = M \oplus M^\perp$$
• There is some $$z\in M^\perp$$ such that $${\left\lVert {z} \right\rVert} = 1$$.
• Set $$u \mathrel{\vcenter{:}}=\Lambda(x) z - \Lambda(z) x$$
• Check

\begin{align*}\Lambda(u) = \Lambda(\Lambda(x) z - \Lambda(z) x) = \Lambda(x) \Lambda(z) - \Lambda(z) \Lambda(x) = 0 \implies u\in M\end{align*}

• Compute

\begin{align*} 0 &= {\left\langle {u},~{z} \right\rangle} \\ &= {\left\langle {\Lambda(x) z - \Lambda(z) x},~{z} \right\rangle} \\ &= {\left\langle {\Lambda(x) z},~{z} \right\rangle} - {\left\langle {\Lambda(z) x},~{z} \right\rangle} \\ &= \Lambda(x) {\left\langle {z},~{z} \right\rangle} - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\ &= \Lambda(x) {\left\lVert {z} \right\rVert}^2 - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\ &= \Lambda(x) - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\ &= \Lambda(x) - {\left\langle {x},~{\overline{\Lambda(z)} z} \right\rangle} ,\end{align*}

• Choose $$y \mathrel{\vcenter{:}}=\overline{\Lambda(z)} z$$.
• Check uniqueness:

\begin{align*} {\left\langle {x},~{y} \right\rangle} &= {\left\langle {x},~{y'} \right\rangle} \quad\forall x \\ \implies {\left\langle {x},~{y-y'} \right\rangle} &= 0 \quad\forall x \\ \implies {\left\langle {y-y'},~{y-y'} \right\rangle} &= 0 \\ \implies {\left\lVert {y-y'} \right\rVert} &= 0 \\ \implies y-y' &= \mathbf{0} \implies y = y' .\end{align*}

Let $$L:X \to {\mathbb{C}}$$ be a linear functional, then the following are equivalent:

1. $$L$$ is continuous
2. $$L$$ is continuous at zero
3. $$L$$ is bounded, i.e. $$\exists c\geq 0 {~\mathrel{\Big|}~}{\left\lvert {L(x)} \right\rvert} \leq c {\left\lVert {x} \right\rVert}$$ for all $$x\in H$$
(Click to expand)

$$2 \implies 3$$: Choose $$\delta < 1$$ such that \begin{align*} {\left\lVert {x} \right\rVert} \leq \delta \implies {\left\lvert {L(x)} \right\rvert} < 1. \end{align*} Then \begin{align*} {\left\lvert {L(x)} \right\rvert} &= {\left\lvert {L\left( \frac{{\left\lVert {x} \right\rVert}}{\delta} \frac{\delta }{{\left\lVert {x} \right\rVert}} x \right)} \right\rvert} \\ &= \frac{{\left\lVert {x} \right\rVert}}{\delta} ~{\left\lvert {L\left( \delta \frac{x }{{\left\lVert {x} \right\rVert}} \right)} \right\rvert} \\ &\leq \frac{{\left\lVert {x} \right\rVert}}{\delta} 1 ,\end{align*} so we can take $$c = \frac 1 \delta$$. $$\hfill\blacksquare$$

$$3 \implies 1$$:

We have $${\left\lvert {L(x-y)} \right\rvert} \leq c{\left\lVert {x-y} \right\rVert}$$, so given $$\varepsilon \geq 0$$ simply choose $$\delta = \frac \varepsilon c$$.

If $$H$$ is a Hilbert space, then $$(H^\vee, {\left\lVert {{\,\cdot\,}} \right\rVert}_{\text{op}})$$ is a normed space.

(Click to expand)

The only nontrivial property is the triangle inequality, but \begin{align*} {\left\lVert {L_{1} + L_{2}} \right\rVert}_{^{\operatorname{op}}} = \sup {\left\lvert {L_{1}(x) + L_{2}(x)} \right\rvert} \leq \sup {\left\lvert {L_{1}(x)} \right\rvert} + {\left\lvert {\sup L_{2}(x)} \right\rvert} = {\left\lVert {L_{1}} \right\rVert}_{^{\operatorname{op}}}+ {\left\lVert {L_{2}} \right\rVert}_{^{\operatorname{op}}} .\end{align*}

If $$X$$ is a normed vector space, then $$(X^\vee, {\left\lVert {{\,\cdot\,}} \right\rVert}_{\text{op}})$$ is a Banach space.

(Click to expand)
• Let $$\left\{{L_{n}}\right\}$$ be Cauchy in $$X^\vee$$.

• Then for all $$x\in C$$, $$\left\{{L_{n}(x)}\right\} \subset {\mathbb{C}}$$ is Cauchy and converges to something denoted $$L(x)$$.

• Need to show $$L$$ is continuous and $${\left\lVert {L_{n} - L} \right\rVert} \to 0$$.

• Since $$\left\{{L_{n}}\right\}$$ is Cauchy in $$X^\vee$$, choose $$N$$ large enough so that \begin{align*} n, m \geq N \implies {\left\lVert {L_{n} - L_{m}} \right\rVert} < \varepsilon \implies {\left\lvert {L_{m}(x) - L_{n}(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} = 1 .\end{align*}

• Take $$n\to \infty$$ to obtain \begin{align*}m \geq N &\implies {\left\lvert {L_{m}(x) - L(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} = 1\\ &\implies {\left\lVert {L_{m} - L} \right\rVert} < \varepsilon \to 0 .\end{align*}

• Continuity: \begin{align*} {\left\lvert {L(x)} \right\rvert} &= {\left\lvert {L(x) - L_{n}(x) + L_{n}(x)} \right\rvert} \\ &\leq {\left\lvert {L(x) - L_{n}(x)} \right\rvert} + {\left\lvert {L_{n}(x)} \right\rvert} \\ &\leq \varepsilon {\left\lVert {x} \right\rVert} + c{\left\lVert {x} \right\rVert} \\ &= (\varepsilon + c){\left\lVert {x} \right\rVert} \hfill\blacksquare .\end{align*}

Let $$U = \left\{{u_{n}}\right\}_{n=1}^\infty$$ be an orthonormal set (not necessarily a basis), then

1. There is an isometric surjection

\begin{align*} \mathcal{H} &\to \ell^2({\mathbb{N}}) \\ \mathbf{x} &\mapsto \left\{{{\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle}}\right\}_{n=1}^\infty \end{align*}

i.e. if $$\left\{{a_{n}}\right\} \in \ell^2({\mathbb{N}})$$, so $$\sum {\left\lvert {a_{n}} \right\rvert}^2 < \infty$$, then there exists a $$\mathbf{x} \in \mathcal{H}$$ such that \begin{align*} a_{n} = {\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle} \quad \forall n. \end{align*}

1. $$\mathbf{x}$$ can be chosen such that \begin{align*} {\left\lVert {\mathbf{x}} \right\rVert}^2 = \sum {\left\lvert {a_{n}} \right\rvert}^2 \end{align*}

Note: the choice of $$\mathbf{x}$$ is unique $$\iff$$ $$\left\{{u_{n}}\right\}$$ is complete, i.e. $${\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle} = 0$$ for all $$n$$ implies $$\mathbf{x} = \mathbf{0}$$.

(Click to expand)
• Given $$\left\{{a_{n}}\right\}$$, define $$S_{N} = \sum^N a_{n} \mathbf{u}_{n}$$.
• $$S_{N}$$ is Cauchy in $$\mathcal{H}$$ and so $$S_{N} \to \mathbf{x}$$ for some $$\mathbf{x} \in \mathcal{H}$$.
• $${\left\langle {x},~{u_{n}} \right\rangle} = {\left\langle {x - S_{N}},~{u_{n}} \right\rangle} + {\left\langle {S_{N}},~{u_{n}} \right\rangle} \to a_{n}$$
• By construction, $${\left\lVert {x-S_{N}} \right\rVert}^2 = {\left\lVert {x} \right\rVert}^2 - \sum^N {\left\lvert {a_{n}} \right\rvert}^2 \to 0$$, so $${\left\lVert {x} \right\rVert}^2 = \sum^\infty {\left\lvert {a_{n}} \right\rvert}^2$$.

# 6 Extra Problems

## 6.1 Greatest Hits

• $$\star$$: Show that for $$E\subseteq {\mathbb{R}}^n$$, TFAE:

1. $$E$$ is measurable
2. $$E = H\cup Z$$ here $$H$$ is $$F_\sigma$$ and $$Z$$ is null
3. $$E = V\setminus Z'$$ where $$V\in G_\delta$$ and $$Z'$$ is null.
• $$\star$$: Show that if $$E\subseteq {\mathbb{R}}^n$$ is measurable then $$m(E) = \sup \left\{{ m(K) {~\mathrel{\Big|}~}K\subset E\text{ compact}}\right\}$$ iff for all $$\varepsilon> 0$$ there exists a compact $$K\subseteq E$$ such that $$m(K) \geq m(E) - \varepsilon$$.

• $$\star$$: Show that cylinder functions are measurable, i.e. if $$f$$ is measurable on $${\mathbb{R}}^s$$, then $$F(x, y) \mathrel{\vcenter{:}}= f(x)$$ is measurable on $${\mathbb{R}}^s\times{\mathbb{R}}^t$$ for any $$t$$.

• $$\star$$: Prove that the Lebesgue integral is translation invariant, i.e. if $$\tau_h(x) = x+h$$ then $$\int \tau_h f = \int f$$.

• $$\star$$: Prove that the Lebesgue integral is dilation invariant, i.e. if $$f_\delta(x) = {f({x\over \delta}) \over \delta^n}$$ then $$\int f_\delta = \int f$$.

• $$\star$$: Prove continuity in $$L^1$$, i.e. \begin{align*} f \in L^{1} \Longrightarrow \lim _{h \rightarrow 0} \int|f(x+h)-f(x)|=0 .\end{align*}

• $$\star$$: Show that \begin{align*}f,g \in L^1 \implies f\ast g \in L^1 {\quad \operatorname{and} \quad} {\left\lVert {f\ast g} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1.\end{align*}

• $$\star$$: Show that if $$X\subseteq {\mathbb{R}}$$ with $$\mu(X) < \infty$$ then \begin{align*} {\left\lVert {f} \right\rVert}_p \overset{p\to\infty}\to {\left\lVert {f} \right\rVert}_\infty .\end{align*}

## 6.2 By Topic

### 6.2.1 Topology

• Show that every compact set is closed and bounded.
• Show that if a subset of a metric space is complete and totally bounded, then it is compact.
• Show that if $$K$$ is compact and $$F$$ is closed with $$K, F$$ disjoint then $$\operatorname{dist}(K, F) > 0$$.

### 6.2.2 Continuity

• Show that a continuous function on a compact set is uniformly continuous.

### 6.2.3 Differentiation

• Show that if $$f\in C^1({\mathbb{R}})$$ and both $$\lim_{x\to \infty} f(x)$$ and $$\lim_{x\to \infty} f'(x)$$ exist, then $$\lim_{x\to\infty} f'(x)$$ must be zero.

• If $$f$$ is continuous, is it necessarily the case that $$f'$$ is continuous?
• If $$f_n \to f$$, is it necessarily the case that $$f_n'$$ converges to $$f'$$ (or at all)?
• Is it true that the sum of differentiable functions is differentiable?
• Is it true that the limit of integrals equals the integral of the limit?
• Is it true that a limit of continuous functions is continuous?
• Show that a subset of a metric space is closed iff it is complete.
• Show that if $$m(E) < \infty$$ and $$f_n\to f$$ uniformly, then $$\lim \int_E f_n = \int_E f$$.

Uniform Convergence

• Show that a uniform limit of bounded functions is bounded.
• Show that a uniform limit of continuous function is continuous.
• I.e. if $$f_n\to f$$ uniformly with each $$f_n$$ continuous then $$f$$ is continuous.
• Show that if $$f_n\to f$$ pointwise, $$f_n' \to g$$ uniformly for some $$f, g$$, then $$f$$ is differentiable and $$g = f'$$.
• Prove that uniform convergence implies pointwise convergence implies a.e. convergence, but none of the implications may be reversed.
• Show that $$\sum {x^n \over n!}$$ converges uniformly on any compact subset of $${\mathbb{R}}$$.

Measure Theory

• Show that continuity of measure from above/below holds for outer measures.

• Show that a countable union of null sets is null.

Measurability

• Show that $$f=0$$ a.e. iff $$\int_E f = 0$$ for every measurable set $$E$$.

Integrability

• Show that if $$f$$ is a measurable function, then $$f=0$$ a.e. iff $$\int f = 0$$.
• Show that a bounded function is Lebesgue integrable iff it is measurable.
• Show that simple functions are dense in $$L^1$$.
• Show that step functions are dense in $$L^1$$.
• Show that smooth compactly supported functions are dense in $$L^1$$.

Convergence

• Prove Fatou’s lemma using the Monotone Convergence Theorem.
• Show that if $$\left\{{f_n}\right\}$$ is in $$L^1$$ and $$\sum \int {\left\lvert {f_n} \right\rvert} < \infty$$ then $$\sum f_n$$ converges to an $$L^1$$ function and \begin{align*}\int \sum f_n = \sum \int f_n.\end{align*}

Convolution

• Show that if $$f\in L^1$$ and $$g$$ is bounded, then $$f\ast g$$ is bounded and uniformly continuous.
• If $$f, g$$ are compactly supported, is it necessarily the case that $$f\ast g$$ is compactly supported?
• Show that under any of the following assumptions, $$f\ast g$$ vanishes at infinity:
• $$f, g\in L^1$$ are both bounded.
• $$f, g\in L^1$$ with just $$g$$ bounded.
• $$f, g$$ smooth and compactly supported (and in fact $$f\ast g$$ is smooth)
• $$f\in L^1$$ and $$g$$ smooth and compactly supported (and in fact $$f\ast g$$ is smooth)
• Show that if $$f\in L^1$$ and $$g'$$ exists with $${\frac{\partial g}{\partial x_i}\,}$$ all bounded, then \begin{align*}{\frac{\partial }{\partial x_i}\,}(f\ast g) = f \ast {\frac{\partial g}{\partial x_i}\,}\end{align*}

Fourier Analysis

• Show that if $$f\in L^1$$ then $$\widehat{f}$$ is bounded and uniformly continuous.
• Is it the case that $$f\in L^1$$ implies $$\widehat{f}\in L^1$$?
• Show that if $$f, \widehat{f} \in L^1$$ then $$f$$ is bounded, uniformly continuous, and vanishes at infinity.
• Show that this is not true for arbitrary $$L^1$$ functions.
• Show that if $$f\in L^1$$ and $$\widehat{f} = 0$$ almost everywhere then $$f = 0$$ almost everywhere.
• Prove that $$\widehat{f} = \widehat{g}$$ implies that $$f=g$$ a.e.
• Show that if $$f, g \in L^1$$ then \begin{align*}\int \widehat{f} g = \int f\widehat{g}.\end{align*}
• Give an example showing that this fails if $$g$$ is not bounded.
• Show that if $$f\in C^1$$ then $$f$$ is equal to its Fourier series.

Approximate Identities

• Show that if $$\phi$$ is an approximate identity, then \begin{align*}{\left\lVert {f\ast \phi_t - f} \right\rVert}_1 \overset{t\to 0}\to 0.\end{align*}
• Show that if additionally $${\left\lvert {\phi(x)} \right\rvert} \leq c(1 + {\left\lvert {x} \right\rvert})^{-n-\varepsilon}$$ for some $$c,\varepsilon>0$$, then this converges is almost everywhere.
• Show that is $$f$$ is bounded and uniformly continuous and $$\phi_t$$ is an approximation to the identity, then $$f\ast \phi_t$$ uniformly converges to $$f$$.

$$L^p$$ Spaces

• Show that if $$E\subseteq {\mathbb{R}}^n$$ is measurable with $$\mu(E) < \infty$$ and $$f\in L^p(X)$$ then \begin{align*}{\left\lVert {f} \right\rVert}_{L^p(X)} \overset{p\to\infty}\to {\left\lVert {f} \right\rVert}_\infty.\end{align*}
• Is it true that the converse to the DCT holds? I.e. if $$\int f_n \to \int f$$, is there a $$g\in L^p$$ such that $$f_n < g$$ a.e. for every $$n$$?
• Prove continuity in $$L^p$$: If $$f$$ is uniformly continuous then for all $$p$$, \begin{align*}{\left\lVert {\tau_h f - f} \right\rVert}_p \overset{h\to 0}\to 0.\end{align*}
• Prove the following inclusions of $$L^p$$ spaces for $$m(X) < \infty$$: \begin{align*} L^\infty(X) &\subset L^2(X) \subset L^1(X) \\ \ell^2({\mathbb{Z}}) &\subset \ell^1({\mathbb{Z}}) \subset \ell^\infty({\mathbb{Z}}) .\end{align*}

# 7 Practice Exam (November 2014)

## 7.1 1: Fubini-Tonelli

### 7.1.1 a

Carefully state Tonelli’s theorem for a nonnegative function $$F(x, t)$$ on $${\mathbb{R}}^n\times{\mathbb{R}}$$.

### 7.1.2 b

Let $$f:{\mathbb{R}}^n\to [0, \infty]$$ and define \begin{align*} {\mathcal{A}}\coloneqq\left\{{(x, t) \in {\mathbb{R}}^n\times{\mathbb{R}}{~\mathrel{\Big|}~}0\leq t \leq f(x)}\right\} .\end{align*}

Prove the validity of the following two statements:

1. $$f$$ is Lebesgue measurable on $${\mathbb{R}}^{n} \iff {\mathcal{A}}$$ is a Lebesgue measurable subset of $${\mathbb{R}}^{n+1}$$.
2. If $$f$$ is Lebesgue measurable on $${\mathbb{R}}^n$$ then \begin{align*} m(\mathcal{A})=\int_{\mathbb{R}^{n}} f(x) d x=\int_{0}^{\infty} m\left(\left\{x \in \mathbb{R}^{n}{~\mathrel{\Big|}~}f(x) \geq t\right\}\right) d t .\end{align*}

## 7.2 2: Convolutions and the Fourier Transform

### 7.2.1 a

Let $$f, g\in L^1({\mathbb{R}}^n)$$ and give a definition of $$f\ast g$$.

### 7.2.2 b

Prove that if $$f, g$$ are integrable and bounded, then \begin{align*} (f\ast g)(x) \overset{{\left\lvert {x} \right\rvert}\to\infty}\to 0 .\end{align*}

### 7.2.3 c

1. Define the Fourier transform of an integrable function $$f$$ on $${\mathbb{R}}^n$$.
2. Give an outline of the proof of the Fourier inversion formula.
3. Give an example of a function $$f\in L^1({\mathbb{R}}^n)$$ such that $$\widehat{f}$$ is not in $$L^1({\mathbb{R}}^n)$$.

## 7.3 3: Hilbert Spaces

Let $$\left\{{u_n}\right\}_{n=1}^\infty$$ be an orthonormal sequence in a Hilbert space $$H$$.

### 7.3.1 a

Let $$x\in H$$ and verify that \begin{align*} \left\|x-\sum_{n=1}^{N}\left\langle x, u_{n}\right\rangle u_{n}\right\|_H^{2} = \|x\|_H^{2}-\sum_{n=1}^{N}\left|\left\langle x, u_{n}\right\rangle\right|^{2} .\end{align*} for any $$N\in {\mathbb{N}}$$ and deduce that \begin{align*} \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|_H^{2} .\end{align*}

### 7.3.2 b

Let $$\left\{{a_n}\right\}_{n\in {\mathbb{N}}} \in \ell^2({\mathbb{N}})$$ and prove that there exists an $$x\in H$$ such that $$a_n = {\left\langle {x},~{u_n} \right\rangle}$$ for all $$n\in {\mathbb{N}}$$, and moreover $$x$$ may be chosen such that \begin{align*} {\left\lVert {x} \right\rVert}_H = \qty{ \sum_{n\in {\mathbb{N}}} {\left\lvert {a_n} \right\rvert}^2}^{1\over 2} .\end{align*}

Proof
• Take $$\left\{{a_n}\right\} \in \ell^2$$, then note that $$\sum {\left\lvert {a_n} \right\rvert}^2 < \infty \implies$$ the tails vanish.

• Define $$x \mathrel{\vcenter{:}}=\displaystyle\lim_{N\to\infty} S_N$$ where $$S_N = \sum_{k=1}^N a_k u_k$$

• $$\left\{{S_N}\right\}$$ is Cauchy and $$H$$ is complete, so $$x\in H$$.

• By construction, \begin{align*} {\left\langle {x},~{u_n} \right\rangle} = {\left\langle {\sum_k a_k u_k},~{u_n} \right\rangle} = \sum_k a_k {\left\langle {u_k},~{u_n} \right\rangle} = a_n \end{align*} since the $$u_k$$ are all orthogonal.

• By Pythagoras since the $$u_k$$ are normal, \begin{align*} {\left\lVert {x} \right\rVert}^2 = {\left\lVert {\sum_k a_k u_k} \right\rVert}^2 = \sum_k {\left\lVert {a_k u_k} \right\rVert}^2 = \sum_k {\left\lvert {a_k} \right\rvert}^2 .\end{align*}

### 7.3.3 c

Prove that if $$\left\{{u_n}\right\}$$ is complete, Bessel’s inequality becomes an equality.

Proof

Let $$x$$ and $$u_n$$ be arbitrary.

\begin{align*} {\left\langle {x - \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} &= {\left\langle {x},~{u_n} \right\rangle} - {\left\langle {\sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} \\ &= {\left\langle {x},~{u_n} \right\rangle} - \sum_{k=1}^\infty {\left\langle {{\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} \\ &= {\left\langle {x},~{u_n} \right\rangle} - \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle} {\left\langle {u_k },~{u_n} \right\rangle} \\ &= {\left\langle {x},~{u_n} \right\rangle} - {\left\langle {x},~{u_n} \right\rangle} = 0 \\ \implies x - \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle}u_k &= 0 \quad\text{by completeness} .\end{align*}

So \begin{align*} x = \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle} u_k \implies {\left\lVert {x} \right\rVert}^2 = \sum_{k=1}^\infty {\left\lvert {{\left\langle {x},~{u_k} \right\rangle}} \right\rvert}^2. \hfill\blacksquare .\end{align*}

## 7.4 4: $$L^p$$ Spaces

### 7.4.1 a

Prove Holder’s inequality: let $$f\in L^p, g\in L^q$$ with $$p, q$$ conjugate, and show that \begin{align*} {\left\lVert {fg} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}_{p} \cdot {\left\lVert {g} \right\rVert}_{q} .\end{align*}

### 7.4.2 b

Prove Minkowski’s Inequality: \begin{align*} 1\leq p < \infty \implies {\left\lVert {f+g} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}_{p}+ {\left\lVert {g} \right\rVert}_{p} .\end{align*} Conclude that if $$f, g\in L^p({\mathbb{R}}^n)$$ then so is $$f+g$$.

### 7.4.3 c

Let $$X = [0, 1] \subset {\mathbb{R}}$$.

1. Give a definition of the Banach space $$L^\infty(X)$$ of essentially bounded functions of $$X$$.

2. Let $$f$$ be non-negative and measurable on $$X$$, prove that \begin{align*} \int_X f(x)^p \,dx \overset{p\to\infty}\to \begin{dcases} \infty \quad\text{or} \\ m\qty{\left\{{f^{-1}(1)}\right\}} \end{dcases} ,\end{align*} and characterize the functions of each type

(Click to expand)

\begin{align*} \int f^p &= \int_{x < 1} f^p + \int_{x=1}f^p + \int_{x > 1} f^p\\ &= \int_{x < 1} f^p + \int_{x=1}1 + \int_{x > 1} f^p \\ &= \int_{x < 1} f^p + m(\left\{{f = 1}\right\}) + \int_{x > 1} f^p \\ &\overset{p\to\infty}\to 0 + m(\left\{{f = 1}\right\}) + \begin{cases} 0 & m(\left\{{x\geq 1}\right\}) = 0 \\ \infty & m(\left\{{x\geq 1}\right\}) > 0. \end{cases} \end{align*}

## 7.5 5: Dual Spaces

Let $$X$$ be a normed vector space.

### 7.5.1 a

Give the definition of what it means for a map $$L:X\to {\mathbb{C}}$$ to be a linear functional.

### 7.5.2 b

Define what it means for $$L$$ to be bounded and show $$L$$ is bounded $$\iff L$$ is continuous.

### 7.5.3 c

Prove that $$(X^\vee, {\left\lVert {{\,\cdot\,}} \right\rVert}_{^{\operatorname{op}}})$$ is a Banach space.

# 8 Midterm Exam 2 (November 2018)

## 8.1 1 (Integration by Parts)

Let $$f, g\in L^1([0, 1])$$, define $$F(x) = \int_0^x f$$ and $$G(x) = \int_0^x g$$, and show \begin{align*} \int_0^1 F(x)g(x) \,dx = F(1)G(1) - \int_0^1 f(x) G(x) \, dx .\end{align*}

## 8.2 2

Let $$\phi\in L^1({\mathbb{R}}^n)$$ such that $$\int \phi = 1$$ and define $$\phi_t(x) = t^{-n}\phi(t^{-1}x)$$.

Show that if $$f$$ is bounded and uniformly continuous then $$f\ast \phi_t \overset{t\to 0} f$$ uniformly.

## 8.3 3

Let $$g\in L^\infty([0, 1])$$.

1. Prove \begin{align*} {\left\lVert {g} \right\rVert}_{L^p([0, 1])} \overset{p\to\infty}\to {\left\lVert {g} \right\rVert}_{L^\infty([0, 1])} .\end{align*}

2. Prove that the map \begin{align*} \Lambda_g: L^1([0, 1]) &\to {\mathbb{C}}\\ f &\mapsto \int_0^1 fg \end{align*} defines an element of $$L^1([0, 1])^\vee$$ with $${\left\lVert {\Lambda_g} \right\rVert}_{L^1([0, 1])^\vee}= {\left\lVert {g} \right\rVert}_{L^\infty([0, 1])}$$.

Note: 4 is a repeat.

# 9 Midterm Exam 2 (December 2014)

## 9.1 1

Note: (a) is a repeat.

• Let $$\Lambda\in L^2(X)^\vee$$.
• Show that $$M\mathrel{\vcenter{:}}=\left\{{f\in L^2(X) {~\mathrel{\Big|}~}\Lambda(f) = 0}\right\} \subseteq L^2(X)$$ is a closed subspace, and $$L^2(X) = M \oplus M\perp$$.
• Prove that there exists a unique $$g\in L^2(X)$$ such that $$\Lambda(f) = \int_X g \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu$$.

## 9.2 2

1. In parts:
• Given a definition of $$L^\infty({\mathbb{R}}^n)$$.
• Verify that $${\left\lVert {{\,\cdot\,}} \right\rVert}_\infty$$ defines a norm on $$L^\infty({\mathbb{R}}^n)$$.
• Carefully proved that $$(L^\infty({\mathbb{R}}^n), {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)$$ is a Banach space.
1. Prove that for any measurable $$f:{\mathbb{R}}^n \to {\mathbb{C}}$$, \begin{align*} L^1({\mathbb{R}}^n) \cap L^\infty({\mathbb{R}}^n) \subset L^2({\mathbb{R}}^n) {\quad \operatorname{and} \quad} {\left\lVert {f} \right\rVert}_2 \leq {\left\lVert {f} \right\rVert}_1^{1\over 2} \cdot {\left\lVert {f} \right\rVert}_\infty^{1\over 2} .\end{align*}

## 9.3 3

1. Prove that if $$f, g: {\mathbb{R}}^n\to {\mathbb{C}}$$ is both measurable then $$F(x, y) \mathrel{\vcenter{:}}= f(x)$$ and $$h(x, y)\mathrel{\vcenter{:}}= f(x-y) g(y)$$ is measurable on $${\mathbb{R}}^n\times{\mathbb{R}}^n$$.

2. Show that if $$f\in L^1({\mathbb{R}}^n) \cap L^\infty({\mathbb{R}}^n)$$ and $$g\in L^1({\mathbb{R}}^n)$$, then $$f\ast g \in L^1({\mathbb{R}}^n) \cap L^\infty({\mathbb{R}}^n)$$ is well defined, and carefully show that it satisfies the following properties: \begin{align*} {\left\lVert {f\ast g} \right\rVert}_\infty &\leq {\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_\infty {\left\lVert {f\ast g} \right\rVert}_1 &\leq {\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 {\left\lVert {f\ast g} \right\rVert}_2 &\leq {\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_2 .\end{align*}

Hint: first show $${\left\lvert {f\ast g} \right\rvert}^2 \leq {\left\lVert {g} \right\rVert}_1 \qty{ {\left\lvert {f} \right\rvert}^2 \ast {\left\lvert {g} \right\rvert}}$$.

## 9.4 4 (Weierstrass Approximation Theorem)

Note: (a) is a repeat.

Let $$f: [0, 1]\to {\mathbb{R}}$$ be continuous, and prove the Weierstrass approximation theorem: for any $$\varepsilon> 0$$ there exists a polynomial $$P$$ such that $${\left\lVert {f - P} \right\rVert}_{\infty} < \varepsilon$$.

# 10 Common Inequalities

\begin{align*} {\left\lvert {{\left\lVert {x} \right\rVert} - {\left\lVert {y} \right\rVert}} \right\rvert} \leq {\left\lVert {x - y} \right\rVert} .\end{align*}

\begin{align*} \mu(\{x:|f(x)|>\alpha\}) \leq\left(\frac{{\left\lVert {f} \right\rVert}_{p}}{\alpha}\right)^{p} .\end{align*}

\begin{align*} \frac 1 p + \frac 1 q = 1 \implies {\left\lVert {f g} \right\rVert}_{1} \leq {\left\lVert {f} \right\rVert}_{p} {\left\lVert {g} \right\rVert}_{q} .\end{align*}

(Click to expand)

It suffices to show this when $${\left\lVert {f} \right\rVert}_p = {\left\lVert {g} \right\rVert}_q = 1$$, since \begin{align*} \|f g\|_{1} \leq\|f\|_{p}\|f\|_{q} \Longleftrightarrow \int \frac{|f|}{\|f\|_{p}} \frac{|g|}{\|g\|_{q}} \leq 1 .\end{align*}

Using $$AB \leq \frac 1 p A^p + \frac 1 q B^q$$, we have \begin{align*} \int|f \| g| \leq \int \frac{|f|^{p}}{p} \frac{|g|^{q}}{q}=\frac{1}{p}+\frac{1}{q}=1 .\end{align*}

For finite measure spaces, \begin{align*} 1 \leq p < q \leq \infty \implies L^q \subset L^p \quad (\text{ and } \ell^p \subset \ell^q) .\end{align*}

(Click to expand)

Fix $$p, q$$, let $$r = \frac q p$$ and $$s = \frac{r}{r-1}$$ so $$r^{-1}+ s^{-1}= 1$$. Then let $$h = {\left\lvert {f} \right\rvert}^p$$:

\begin{align*} {\left\lVert {f} \right\rVert}_{p}^p = {\left\lVert {h\cdot 1} \right\rVert}_{1} \leq {\left\lVert {1} \right\rVert}_{s} {\left\lVert {h} \right\rVert}_{r} = \mu(X)^{\frac 1 s} {\left\lVert {f} \right\rVert}_{q}^{\frac q r} \implies {\left\lVert {f} \right\rVert}_{p} \leq \mu(X)^{\frac 1 p - \frac 1 q} {\left\lVert {f} \right\rVert}_{q} .\end{align*}

Note: doesn’t work for $$\ell_p$$ spaces, but just note that $$\sum {\left\lvert {x_n} \right\rvert} < \infty \implies x_n < 1$$ for large enough $$n$$, and thus $$p<q \implies {\left\lvert {x_n} \right\rvert}^q \leq {\left\lvert {x_n} \right\rvert}^q$$.

\begin{align*} {\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert} = {\left\lVert {fg} \right\rVert}_{1} \leq {\left\lVert {f} \right\rVert}_{2} {\left\lVert {g} \right\rVert}_{2} && \text{with equality} \iff f = \lambda g .\end{align*}

In general, Cauchy-Schwarz relates inner product to norm, and only happens to relate norms in $$L^1$$.

\begin{align*} 1\leq p < \infty \implies {\left\lVert {f+g} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}_{p}+ {\left\lVert {g} \right\rVert}_{p} .\end{align*}

This does not handle $$p=\infty$$ case. Use to prove $$L^p$$ is a normed space.

(Click to expand)
• We first note \begin{align*} {\left\lvert {f+g} \right\rvert}^p = {\left\lvert {f+g} \right\rvert}{\left\lvert {f+g} \right\rvert}^{p-1} \leq \left( {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\right) {\left\lvert {f+g} \right\rvert}^{p-1} .\end{align*}

• Note that if $$p,q$$ are conjugate exponents then \begin{align*} \frac 1 q &= 1 - \frac 1 p = \frac{p-1} p \\ q &= \frac p {p-1} .\end{align*}

• Then taking integrals yields \begin{align*} {\left\lVert {f+g} \right\rVert}_p^p &= \int {\left\lvert {f+g} \right\rvert}^p \\ &\leq \int \left( {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\right) {\left\lvert {f+g} \right\rvert}^{p-1} \\ &= \int {\left\lvert {f} \right\rvert} {\left\lvert {f+g} \right\rvert}^{p-1} + \int {\left\lvert {g} \right\rvert} {\left\lvert {f+g} \right\rvert}^{p-1} \\ &= {\left\lVert {f(f+g)^{p-1}} \right\rVert}_1 + {\left\lVert {g(f+g)^{p-1}} \right\rVert}_1 \\ &\leq {\left\lVert {f} \right\rVert}_p ~{\left\lVert {(f+g)^{p-1})} \right\rVert}_q + {\left\lVert {g} \right\rVert}_p ~{\left\lVert {(f+g)^{p-1})} \right\rVert}_q \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) {\left\lVert { (f+g)^{p-1})} \right\rVert}_q \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \left( \int {\left\lvert {f+g} \right\rvert}^{(p-1)q} \right)^{\frac 1 q} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \left( \int {\left\lvert {f+g} \right\rvert}^{p} \right)^{1 - \frac 1 p} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{\int {\left\lvert {f+g} \right\rvert}^{p} }{\left( \int {\left\lvert {f+g} \right\rvert}^{p} \right)^{\frac 1 p}} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{{\left\lVert {f+g} \right\rVert}_p^p}{{\left\lVert {f+g} \right\rVert}_p} .\end{align*}

• Cancelling common terms yields \begin{align*} 1 &\leq \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{1}{{\left\lVert {f+g} \right\rVert}_p} \\ &\implies {\left\lVert {f+g} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p .\end{align*}

\begin{align*} \frac 1 p + \frac 1 q = \frac 1 r + 1 \implies \|f \ast g\|_{r} \leq\|f\|_{p}\|g\|_{q} \end{align*}

\begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 & \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1 \\ \|f * g\|_{p} & \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}p, \\ {\left\lVert {f\ast g} \right\rVert}_\infty & \leq {\left\lVert {f} \right\rVert}_2 {\left\lVert {g} \right\rVert}_2 \\ {\left\lVert {f\ast g} \right\rVert}_\infty & \leq {\left\lVert {f} \right\rVert}_p {\left\lVert {g} \right\rVert}_q .\end{align*}

For $$x\in H$$ a Hilbert space and $$\left\{{e_k}\right\}$$ an orthonormal sequence, \begin{align*} \sum_{k=1}^{\infty}\| {\left\langle {x},~{e_{k} } \right\rangle} \|^{2} \leq \|x\|^{2} .\end{align*}

Note that this does not need to be a basis.

Equality in Bessel’s inequality, attained when $$\left\{{e_k}\right\}$$ is a basis, i.e. it is complete, i.e. the span of its closure is all of $$H$$.

# 11 Less Explicitly Used Inequalities

\begin{align*} \sqrt{ab} \leq \frac{a+b}{2} .\end{align*}

\begin{align*} f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) .\end{align*}

\begin{align*} AB \leq {A^p \over p} + {B^q \over q} .\end{align*}

\begin{align*} (a+b)^p \leq 2^{p-1} (a^p + b^p) .\end{align*}

\begin{align*} (1 + x)^n \geq 1 +nx \quad x\geq -1, \text{ or } n\in 2{\mathbb{Z}}\text{ and } \forall x .\end{align*}

\begin{align*} \forall t\in {\mathbb{R}},\quad 1 + t \leq e^t .\end{align*}

(Click to expand)
• It’s an equality when $$t=0$$.
• $${\frac{\partial }{\partial t}\,} 1+ t < {\frac{\partial t}{\partial e}\,}^t \iff t<0$$

\begin{align*} {1\over r} \mathrel{\vcenter{:}}={1\over p} + {1\over q} - 1 \implies {\left\lVert {f \ast g} \right\rVert}_{r} \leq {\left\lVert {f} \right\rVert}_{p} {\left\lVert {g} \right\rVert}{q} .\end{align*}

# 12 Bibliography

1. Mnemonic: “F” stands for ferme, which is “closed” in French, and $$\sigma$$ corresponds to a “sum,” i.e. a union.↩︎

2. Slogan: to negate, find a bad $$x$$ depending on $$n_0$$ that are larger than some $$\varepsilon$$.↩︎

3. Slogan: a uniform limit of continuous functions is continuous.↩︎

4. Note that this is only pointwise convergence of $$f$$, whereas the full $$M{\hbox{-}}$$test gives uniform convergence.↩︎

5. It suffices to show $${\left\lvert {f_n(x)} \right\rvert} \leq M_n$$ for some $$M_n$$ not depending on $$x$$.↩︎

6. So this implicitly holds if $$f$$ is the pointwise limit of $$f_n$$.↩︎

7. See Abbott theorem 6.4.3, pp 168.↩︎

8. This holds for outer measure iff $$\mathrm{dist}(A, B) > 0$$.↩︎