Exercises from Folland:

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

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.

\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.

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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.

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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 \({\mathbb{Q}}\).

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\(f(x) = \frac 1 n\) 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 \({\mathbb{R}}\setminus {\mathbb{Q}}\).

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\(D_f\) 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. :::

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  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

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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

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. 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*}

1.6 Slightly Advanced Stuff

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*}

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  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 \(m(E\setminus K) < \varepsilon\).
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Lebesgue measure is translation and dilation invariant.

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

There is a non-measurable set.

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

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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.

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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*}

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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.


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 \({\left\lvert {f_n} \right\rvert} < g_n \to g\in L^1\).

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Proceed by showing \(\limsup \int f_n \leq \int f \leq \liminf \int f_n\):

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*}

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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 \(x\in {\mathbb{R}}^n\),

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

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

\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*}

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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 \(\sum \int f_n = \int \sum f_n\).

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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*}

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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 \(L^1\).

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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

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*}

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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*}

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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)

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

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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 \(m(\left\{{f(x) = \infty}\right\}) = 0\).

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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*}

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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*}

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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*}

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

3.4 Lp Facts

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

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

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

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

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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.

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\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!

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

\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*}

\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*}

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\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*}

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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*}

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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*}

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\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*}

\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*}

\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.

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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 \(U = \left\{{u_{n}}\right\}_{n=1}^\infty\) 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 \(\mathbf{x} = \mathbf{0}\).

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6 Extra Problems

6.1 Greatest Hits

6.2 By Topic

6.2.1 Topology

6.2.2 Continuity

6.2.3 Differentiation

6.2.4 Advanced Limitology

Uniform Convergence

Measure Theory





Fourier Analysis

Approximate Identities

\(L^p\) Spaces

7 Midterm Exam 2 (December 2014)

7.1 1

Note: (a) is a repeat.

7.2 2

  1. In parts:
  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*}

7.3 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 \({\left\lvert {f\ast g} \right\rvert}^2 \leq {\left\lVert {g} \right\rVert}_1 \qty{ {\left\lvert {f} \right\rvert}^2 \ast {\left\lvert {g} \right\rvert}}\).

7.4 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 \({\left\lVert {f - P} \right\rVert}_{\infty} < \varepsilon\).

8 Midterm Exam 1 (October 2018)

8.1 Problem 1

Let \(E \subseteq {\mathbb{R}}^n\) be bounded. Prove the following are equivalent:

  1. For any \(\epsilon>0\) there exists and open set \(G\) and a closed set \(F\) such that \begin{align*} F \subseteq E \subseteq G && m(G\setminus F) < \epsilon .\end{align*}

  2. There exists a \(G_ \delta\) set \(V\) and an \(F_ \sigma\) set \(H\) such that \begin{align*} m(V\setminus H) = 0 .\end{align*}

8.2 Problem 2

Let \(\left\{{ f_k }\right\} _{k=1}^{\infty }\) be a sequence of extended real-valued Lebesgue measurable functions.

  1. Prove that \(\sup_k f_k\) is a Lebesgue measurable function.

  2. Prove that if \(\lim_{k \to \infty } f_k(x)\) exists for every \(x \in {\mathbb{R}}^n\) then \(\lim_{k\to \infty } f_k\) is also a measurable function.

8.3 Problem 3

8.3.1 a

Prove that if \(E \subseteq {\mathbb{R}}^n\) is a Lebesgue measurable set, then for any \(h \in {\mathbb{R}}\) the set \begin{align*} E+h \mathrel{\vcenter{:}}=\left\{{x + h {~\mathrel{\Big|}~}x\in E }\right\} \end{align*} is also Lebesgue measurable and satisfies \(m(E + h) = m(E)\).

8.3.2 b

Prove that if \(f\) is a non-negative measurable function on \({\mathbb{R}}^n\) and \(h\in {\mathbb{R}}^n\) then the function \begin{align*} \tau_h d(x) \mathrel{\vcenter{:}}= f(x-h) \end{align*} is a non-negative measurable function and \begin{align*} \int f(x) \,dx= \int f(x-h) \,dx .\end{align*}

8.4 Problem 4

Let \(f: {\mathbb{R}}^n\to {\mathbb{R}}\) be a Lebesgue measurable function.

  1. Prove that for all \(\alpha> 0\) , \begin{align*} A_ \alpha \mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}{\left\lvert { f(x) } \right\rvert} > \alpha}\right\} \implies m(A_ \alpha) \leq {1\over \alpha} \int {\left\lvert {f (x)} \right\rvert} \,dx .\end{align*}

  2. Prove that \begin{align*} \int {\left\lvert { f(x) } \right\rvert} \,dx= 0 \iff f = 0 \text{ almost everywhere} .\end{align*}

8.5 Problem 5

Let \(\left\{{ f_k }\right\}_{k=1}^{\infty } \subseteq L^2([0, 1])\) be a sequence which converges in \(L^1\) to a function \(f\).

  1. Prove that \(f\in L^1([0, 1])\).

  2. Give an example illustrating that \(f_k\) may not converge to \(f\) almost everywhere.

  3. Prove that \(\left\{{f_k}\right\}\) must contain a subsequence that converges to \(f\) almost everywhere.

9 Midterm Exam 2 (November 2018)

9.1 Problem 1

Let \(f, g\in L^1([0, 1])\), define \(F(x) = \int_0^x f(y)\,dy\) and \(G(x) = \int_0^x g(y)\,dy\), 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*}

9.2 Problem 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 \(f\ast \phi_t \overset{t\to 0}\to f\) uniformly.

9.3 Problem 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 \({\left\lVert {\Lambda_g} \right\rVert}_{L^1([0, 1])^\vee}= {\left\lVert {g} \right\rVert}_{L^\infty([0, 1])}\).

9.4 Problem 4


10 Practice Exam (November 2014)

10.1 Problem 1

Let \(m_*(E)\) denote the Lebesgue outer measure of a set \(E \subseteq {\mathbb{R}}^n\).

  1. Prove using the definition of Lebesgue outer measure that \begin{align*} m \qty{ \bigcup_{j=1}^{\infty } E_j } \leq \sum_{j=1}^{\infty } m_*(E_j) .\end{align*}

  2. Prove that for any \(E \subseteq {\mathbb{R}}^n\) and any \(\epsilon> 0\) there exists an open set \(G\) with \(E \subseteq G\) and \begin{align*} m_*(E) \leq m_*(G) \leq m_*(E) + \epsilon .\end{align*}

10.2 Problem 2

  1. See

  2. Let \(f_k\) be a sequence of extended real-valued Lebesgue measurable function.

    1. Prove that \(\inf_k f_k, \sup_k f_k\) are both Lebesgue measurable function.

      Hint: argue that \begin{align*} \left\{{x {~\mathrel{\Big|}~}\inf_k f_k(x) < a}\right\} = \bigcup_k \left\{{x {~\mathrel{\Big|}~}f_k(x) < a}\right\} .\end{align*}

    2. Carefully state Fatou’s Lemma and deduce the Monotone Converge Theorem from it.

10.3 Problem 3

  1. Prove that if \(f, g\in L^+({\mathbb{R}})\) then \begin{align*} \int(f +g) = \int f + \int g .\end{align*} Extend this to establish that if \(\left\{{ f_k}\right\} \subseteq L^+({\mathbb{R}}^n)\) then \begin{align*} \int \sum_k f_k = \sum_k \int f_k .\end{align*}

  2. Let \(\left\{{E_j}\right\}_{j\in {\mathbb{N}}} \subseteq \mathcal{M}({\mathbb{R}}^n)\) with \(E_j \nearrow E\). Use the countable additivity of \(\mu_f\) on \(\mathcal{M}({\mathbb{R}}^n)\) established above to show that \begin{align*} \mu_f(E) = \lim_{j\to \infty } \mu_f(E_j) .\end{align*}

10.4 Problem 4

  1. Show that \(f\in L^1({\mathbb{R}}^n) \implies {\left\lvert {f(x)} \right\rvert} < \infty\) almost everywhere.

  2. Show that if \(\left\{{f_k}\right\} \subseteq L^1({\mathbb{R}}^n)\) with \(\sum {\left\lVert {f_k} \right\rVert}_1 < \infty\) then \(\sum f_k\) converges almost everywhere and in \(L^1\).

  3. Use the Dominated Convergence Theorem to evaluate \begin{align*} \lim_{t\to 0} \int_0^1 {e^{tx^2} - 1 \over t} \,dx .\end{align*}

11 Practice Exam (November 2014)

11.1 1: Fubini-Tonelli

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

  2. Let \(f:{\mathbb{R}}^n\to [0, \infty]\) and define \begin{align*} {\mathcal{A}}\mathrel{\vcenter{:}}=\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*}

11.2 2: Convolutions and the Fourier Transform

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

  2. 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*}

  3. In parts:

    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 \(L^1({\mathbb{R}}^n)\).

11.3 3: Hilbert Spaces

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

  1. 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*}

  2. 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*}

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

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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*}

11.4 4: \(L^p\) Spaces

  1. 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*}

  2. 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 \(f+g\).

  3. 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

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\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*}

11.5 5: Dual Spaces

Let \(X\) be a normed vector space.

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

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

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

12 Extra Problems from Problem Sets

12.1 2010 6.1

Show that \begin{align*} \int_{{\mathbb{B}}^n} {1 \over {\left\lvert {x} \right\rvert}^p } \,dx&< \infty \iff p < n \\ \int_{{\mathbb{R}}^n\setminus{\mathbb{B}}^n} {1 \over {\left\lvert {x} \right\rvert}^p } \,dx&< \infty \iff p > n .\end{align*}

12.2 2010 6.2

Show that \begin{align*} \int_{{\mathbb{R}}^n} {\left\lvert { f} \right\rvert} = \int_0^{\infty } m(A_t)\,dt&& A_t \mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} > t}\right\} .\end{align*}

12.3 2010 6.5

Suppose \(F \subseteq {\mathbb{R}}\) with \(m(F^c) < \infty\) and let \(\delta(x) \mathrel{\vcenter{:}}= d(x, F)\) and \begin{align*} I_F(x) \mathrel{\vcenter{:}}=\int_{\mathbb{R}}{ \delta(y) \over {\left\lvert {x-y} \right\rvert}^2 } \,dy .\end{align*}

  1. Show that \(\delta\) is continuous.

  2. Show that if \(x\in F^c\) then \(I_F(x) = \infty\).

  3. Show that \(I_F(x) < \infty\) for almost every \(x\)

12.4 2010 7.1

Let \((X, \mathcal{M}, \mu)\) be a measure space and prove the following properties of \(L^ \infty (X, \mathcal{M}, \mu)\):

12.5 2010 7.2

Show that for \(0 < p < q \leq \infty\), \({\left\lVert {a} \right\rVert}_{\ell^q} \leq {\left\lVert {a} \right\rVert}_{\ell^p}\) over \({\mathbb{C}}\), where \({\left\lVert {a} \right\rVert}_{\infty } \mathrel{\vcenter{:}}=\sup_j {\left\lvert {a_j} \right\rvert}\).

12.6 2010 7.3

Let \(f, g\) be non-negative measurable functions on \([0, \infty)\) with \begin{align*} A &\mathrel{\vcenter{:}}=\int_0^{\infty } f(y) y^{-1/2} \,dy< \infty \\ B &\mathrel{\vcenter{:}}=\qty{ \int_0^{\infty } {\left\lvert { g(y) } \right\rvert} }^2 \,dy< \infty .\end{align*}

Show that \begin{align*} \int_0^{\infty } \qty{ \int_0^{\infty } f(y) \,dy} {g(x) \over x} \,dx\leq AB .\end{align*}

12.7 2010 7.4

Let \((X, \mathcal{M}, \mu)\) be a measure space and \(0 < p < q< \infty\). Prove that if \(L^q(X) \subseteq L^p(X)\), then \(X\) does not contain sets of arbitrarily large finite measure.

12.8 2010 7.5

Suppose \(0 < a < b \leq \infty\), and find examples of functions \(f \in L^p((0, \infty ))\) if and only if:

Hint: consider functions of the following form: \begin{align*} f(x) \mathrel{\vcenter{:}}= x^{- \alpha} {\left\lvert { \log(x) } \right\rvert}^{ \beta} .\end{align*}

12.9 2010 7.6

Define \begin{align*} F(x) &\mathrel{\vcenter{:}}=\qty{ \sin(\pi x) \over \pi x}^2 \\ G(x) &\mathrel{\vcenter{:}}= \begin{cases} 1 - {\left\lvert {x} \right\rvert} & {\left\lvert {x} \right\rvert} \leq 1 \\ 0 & \text{else}. \end{cases} \end{align*}

  1. Show that \(\widehat{G}(\xi) = F(\xi)\)

  2. Compute \(\widehat{F}\).

  3. Give an example of a function \(g\not \in L^1({\mathbb{R}})\) which is the Fourier transform of an \(L^1\) function.

Hint: write \(\widehat{G}(\xi) = H(\xi) + H(-\xi)\) where \begin{align*} H(\xi) \mathrel{\vcenter{:}}= e^{2\pi i \xi} \int_0^1 y e^{2\pi i y \xi }\,dy .\end{align*}

12.10 2010 7.7

Show that for each \(\epsilon>0\) the following function is the Fourier transform of an \(L^1({\mathbb{R}}^n)\) function: \begin{align*} F(\xi) \mathrel{\vcenter{:}}=\qty{1 \over 1 + {\left\lvert {\xi} \right\rvert}^2}^{\epsilon} .\end{align*}

Hint: show that

\begin{align*} K_\delta(x) &\mathrel{\vcenter{:}}=\delta^{-n/2} e^{-\pi {\left\lvert {x} \right\rvert}^2 \over \delta} \\ f(x) &\mathrel{\vcenter{:}}=\int_0^{\infty } K_{\delta}(x) e^{-\pi \delta} \delta^{\epsilon - 1} \,d \delta \\ \Gamma(s) &\mathrel{\vcenter{:}}=\int_0^{\infty } e^{-t} t^{s-1} \,dt\\ \implies \widehat{f}(\xi) &= \int_0^{\infty } e^{- \pi \delta {\left\lvert {\xi} \right\rvert}^2} e^{ -\pi \delta} \delta^{\epsilon - 1} = \pi^{-s} \Gamma(\epsilon) F(\xi) .\end{align*}

12.11 2010 7 Challenge 1: Generalized Holder

Suppose that \begin{align*} 1\leq p_j \leq \infty, && \sum_{j=1}^n {1\over p_j} = {1\over r} \leq 1 .\end{align*}

Show that if \(f_j \in L^{p_j}\) for each \(1\leq j \leq n\), then \begin{align*} \prod f_j \in L^r, && {\left\lVert { \prod f_j } \right\rVert}_r \leq \prod {\left\lVert {f_j} \right\rVert}_{p_j} .\end{align*}

12.12 2010 7 Challenge 2: Young’s Inequality

Suppose \(1\leq p,q,r \leq \infty\) with \begin{align*} {1\over p } + {1 \over q} = 1 + {1 \over r} .\end{align*}

Prove that \begin{align*} f \in L^p, g\in L^q \implies f \ast g \in L^r \text{ and } {\left\lVert {f \ast g} \right\rVert}_r \leq {\left\lVert {f} \right\rVert}_p {\left\lVert {g} \right\rVert}_q .\end{align*}

12.13 2010 9.1

Show that the set \(\left\{{ u_k(j) \mathrel{\vcenter{:}}=\delta_{ij} }\right\} \subseteq \ell^2({\mathbb{Z}})\) and forms an orthonormal system.

12.14 2010 9.2

Consider \(L^2([0, 1])\) and define \begin{align*} e_0(x) &= 1 \\ e_1(x) &= \sqrt{3}(2x-1) .\end{align*}

  1. Show that \(\left\{{e_0, e_1}\right\}\) is an orthonormal system.

  2. Show that the polynomial \(p(x)\) where \(\deg(p) = 1\) which is closest to \(f(x) = x^2\) in \(L^2([0, 1])\) is given by \begin{align*} h(x) = x - {1\over 6} .\end{align*}

Compute \({\left\lVert {f - g} \right\rVert}_2\).

12.15 2010 9.3

Let \(E \subseteq H\) a Hilbert space.

  1. Show that \(E\perp \subseteq H\) is a closed subspace.

  2. Show that \((E^\perp)^\perp = \operatorname{cl}_H(E)\).

12.16 2010 9.5b

Let \(f\in L^1((0, 2\pi))\).

  1. Show that for an \(\epsilon>0\) one can write \(f = g+h\) where \(g\in L^2((0, 2\pi))\) and \({\left\lVert {H} \right\rVert}_1 < \epsilon\).

12.17 2010 9.6

Prove that every closed convex \(K \subset H\) a Hilbert space has a unique element of minimal norm.

12.18 2010 9 Challenge

Let \(U\) be a unitary operator on \(H\) a Hilbert space, let \(M \mathrel{\vcenter{:}}=\left\{{x\in H {~\mathrel{\Big|}~}Ux = x}\right\}\), let \(P\) be the orthogonal projection onto \(M\), and define \begin{align*} S_N \mathrel{\vcenter{:}}={1\over N} \sum_{n=0}^{N-1} U^n .\end{align*} Show that for all \(x\in H\), \begin{align*} {\left\lVert { S_N x - Px} \right\rVert}_H \overset{N\to \infty } \to 0 .\end{align*}

12.19 2010 10.1

Let \(\nu, \mu\) be signed measures, and show that \begin{align*} \nu \perp \mu \text{ and } \nu \ll {\left\lvert { \mu} \right\rvert} \implies \nu = 0 .\end{align*}

12.20 2010 10.2

Let \(f\in L^1({\mathbb{R}}^n)\) with \(f\neq 0\).

  1. Prove that there exists a \(c>0\) such that \begin{align*} Hf(x) \geq {c \over (1 + {\left\lvert {x} \right\rvert})^n } .\end{align*}

12.21 2010 10.3

Consider the function \begin{align*} f(x) \mathrel{\vcenter{:}}= \begin{cases} {1\over {\left\lvert {x} \right\rvert} \qty{ \log\qty{1\over x}}^2 } & {\left\lvert {x} \right\rvert} \leq {1\over 2} \\ 0 & \text{else}. \end{cases} \end{align*}

  1. Show that \(f \in L^1({\mathbb{R}})\).

  2. Show that there exists a \(c>0\) such that for all \({\left\lvert {x} \right\rvert} \leq 1/2\), \begin{align*} Hf(x) \geq {c \over {\left\lvert {x} \right\rvert} \log\qty{1\over {\left\lvert {x} \right\rvert}} } .\end{align*} Conclude that \(Hf\) is not locally integrable.

12.22 2010 10.4

Let \(f\in L^1({\mathbb{R}})\) and let \(\mathcal{U}\mathrel{\vcenter{:}}=\left\{{(x, y) \in {\mathbb{R}}^2 {~\mathrel{\Big|}~}y > 0}\right\}\) denote the upper half plane. For \((x, y) \in \mathcal{U}\) define \begin{align*} u(x, y) \mathrel{\vcenter{:}}= f \ast P_y(x) && \text{where } P_y(x) \mathrel{\vcenter{:}}={1\over \pi}\qty{y \over t^2 + y^2} .\end{align*}

  1. Prove that there exists a constant \(C\) independent of \(f\) such that for all \(x\in {\mathbb{R}}\), \begin{align*} \sup_{y > 0} {\left\lvert { u(x, y) } \right\rvert} \leq C\cdot Hf(x) .\end{align*}

    Hint: write the following and try to estimate each term: \begin{align*} u(x, y) = \int_{{\left\lvert {t} \right\rvert} < y} f(x - t) P_y(t) \,dt+ \sum_{k=0}^{\infty } \int_{A_k} f(x-t) P_y(t)\,dt&& A_k \mathrel{\vcenter{:}}=\left\{{2^ky \leq {\left\lvert {t} \right\rvert} < 2^{k+1}y}\right\} .\end{align*}

  2. Following the proof of the Lebesgue differentiation theorem, show that for \(f\in L^1({\mathbb{R}})\) and for almost every \(x\in {\mathbb{R}}\), \begin{align*} u(x, y) \overset{y\to 0} \to f(x) .\end{align*}

13 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*}

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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*}

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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.

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\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 \(H\).

14 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*}

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\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*}

15 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 \(\mathrm{dist}(A, B) > 0\).↩︎