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
| 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 |
General advice: try swapping the orders of limits, sums, integrals, etc.
Limits:
Sequences and Series
Equalities
Continuity / differentiability: show it holds on \([-M, M]\) for all \(M\) to get it to hold on \({\mathbb{R}}\).
Simplifications:
Integrals
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:
\(f\) 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., \(X\subset A\cup A'\).
\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.
A countable collection of elements \(\left\{{ u_i }\right\}\) is orthonormal if and only if
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.
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.
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.
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).
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 \({\mathbb{Q}}\).
\(f(x) = \frac 1 n\) 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 \({\mathbb{R}}\setminus {\mathbb{Q}}\).
\(D_f\) 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.
\(f_n \to f\) 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).
:::{.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. :::
Let \(\left\{{f_k}\right\}\) be Cauchy in \(X\).
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 \(f(x) \mathrel{\vcenter{:}}=\lim_k f_k(x)\).
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 \(\varepsilon/2\).
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
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*}
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 \(\int f_n = \int f\).
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.
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
\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*}
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 \(x = 1/n^2\).
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).
\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*}
Break into disjoint annuli \(A_{2} = E_{2}\setminus E_{1}\), etc then apply countable disjoint additivity to \(E = {\coprod}A_{i}\).
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 \(\varepsilon>0\),
Lebesgue measure is translation and dilation invariant.
Obvious for cubes; if \(Q_{i} \rightrightarrows E\) then \(Q_{i} + k \rightrightarrows E + k\), etc.
There is a non-measurable set.
If \(E\) is Lebesgue measurable, then \(E = H {\coprod}N\) where \(H \in F_\sigma\) and \(N\) is null.
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.
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*}
Take the cone on \(f\) to get \(F(x, y) = f(x)\), then compose \(F\) with the linear transformation \(T = [1, -1; 1, 0]\).
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
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
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 \({\left\lvert {f_n} \right\rvert} < g_n \to g\in L^1\).
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*}
\(\int f \leq \liminf \int f_n\): \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*}
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*}
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*}
\(\implies\):
\(\impliedby\):
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 \(\sum \int f_n = \int \sum f_n\).
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*}
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 \(L^1\).
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: ?
\(\int {1\over 1 + x^2} = \arctan(x) \overset{x\to\infty}\to \pi/2 < \infty\)
Any bounded function (or continuous on a compact set, by EVT)
\(\int_0^1 {1 \over \sqrt{x}} < \infty\)
\(\int_0^1 {1\over x^{1-\varepsilon}} < \infty\)
\(\int_1^\infty {1\over x^{1+\varepsilon}} < \infty\)
For \(f\in L^+\), \begin{align*} \int f = 0 \quad\iff\quad f \equiv 0 \text{ almost everywhere} .\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*}
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 \(\int_A {\left\lvert {f} \right\rvert} < \varepsilon\).
\(m(E) \to 0 \implies \int_E f \to 0\).
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 \(m(\left\{{f(x) = \infty}\right\}) = 0\).
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*}
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*}
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*}
Fubini-Tonelli, and sketch region to change integration bounds, and continuity in \(L^1\).
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*}
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 \(1\leq p< \infty\), \((L^p)^\vee\cong L^q\).
?
Use Riesz Representation for Hilbert spaces.
\(L^1 \subset (L^\infty)^\vee\), since the isometric mapping is always injective, but never surjective.
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.
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!
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*}
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*}
\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*}
\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*}
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 \(f\ast g\in L^1\).
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*}
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*}
\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:
\(2 \implies 3\): 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.
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.
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
\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*}
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 \(\mathbf{x} = \mathbf{0}\).
\(\star\): Show that for \(E\subseteq {\mathbb{R}}^n\), TFAE:
\(\star\): 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*}
Uniform Convergence
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
Integrability
Convergence
Convolution
Fourier Analysis
Approximate Identities
\(L^p\) Spaces
Carefully state Tonelli’s theorem for a nonnegative function \(F(x, t)\) on \({\mathbb{R}}^n\times{\mathbb{R}}\).
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:
Let \(f, g\in L^1({\mathbb{R}}^n)\) and give a definition of \(f\ast g\).
Prove that if \(f, g\) are integrable and bounded, then \[\begin{align*} (f\ast g)(x) \overset{{\left\lvert {x} \right\rvert}\to\infty}\to 0 .\end{align*}\]
Let \(\left\{{u_n}\right\}_{n=1}^\infty\) be an orthonormal sequence in a Hilbert space \(H\).
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 \[\begin{align*} \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|_H^{2} .\end{align*}\]
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 \[\begin{align*} {\left\lVert {x} \right\rVert}_H = \qty{ \sum_{n\in {\mathbb{N}}} {\left\lvert {a_n} \right\rvert}^2}^{1\over 2} .\end{align*}\]
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, \[\begin{align*} {\left\lVert {x} \right\rVert}^2 = {\left\lVert {\sum_k a_k u_k} \right\rVert}^2 = \sum_k {\left\lVert {a_k u_k} \right\rVert}^2 = \sum_k {\left\lvert {a_k} \right\rvert}^2 .\end{align*}\]
Prove that if \(\left\{{u_n}\right\}\) is complete, Bessel’s inequality becomes an equality.
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 \[\begin{align*} x = \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle} u_k \implies {\left\lVert {x} \right\rVert}^2 = \sum_{k=1}^\infty {\left\lvert {{\left\langle {x},~{u_k} \right\rangle}} \right\rvert}^2. \hfill\blacksquare .\end{align*}\]
Prove Holder’s inequality: let \(f\in L^p, g\in L^q\) with \(p, q\) conjugate, and show that \[\begin{align*} {\left\lVert {fg} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}_{p} \cdot {\left\lVert {g} \right\rVert}_{q} .\end{align*}\]
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 \(f+g\).
Let \(X = [0, 1] \subset {\mathbb{R}}\).
Give a definition of the Banach space \(L^\infty(X)\) of essentially bounded functions of \(X\).
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
\[\begin{align*} \int f^p &= \int_{x < 1} f^p + \int_{x=1}f^p + \int_{x > 1} f^p\\ &= \int_{x < 1} f^p + \int_{x=1}1 + \int_{x > 1} f^p \\ &= \int_{x < 1} f^p + m(\left\{{f = 1}\right\}) + \int_{x > 1} f^p \\ &\overset{p\to\infty}\to 0 + m(\left\{{f = 1}\right\}) + \begin{cases} 0 & m(\left\{{x\geq 1}\right\}) = 0 \\ \infty & m(\left\{{x\geq 1}\right\}) > 0. \end{cases} \end{align*}\]
Let \(X\) be a normed vector space.
Give the definition of what it means for a map \(L:X\to {\mathbb{C}}\) to be a linear functional.
Define what it means for \(L\) to be bounded and show \(L\) is bounded \(\iff L\) is continuous.
Prove that \((X^\vee, {\left\lVert {{\,\cdot\,}} \right\rVert}_{^{\operatorname{op}}})\) is a Banach space.
Let \(f, g\in L^1([0, 1])\), define \(F(x) = \int_0^x f\) and \(G(x) = \int_0^x g\), and show \[\begin{align*} \int_0^1 F(x)g(x) \,dx = F(1)G(1) - \int_0^1 f(x) G(x) \, dx .\end{align*}\]
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.
Let \(g\in L^\infty([0, 1])\).
Prove \[\begin{align*} {\left\lVert {g} \right\rVert}_{L^p([0, 1])} \overset{p\to\infty}\to {\left\lVert {g} \right\rVert}_{L^\infty([0, 1])} .\end{align*}\]
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.
Note: (a) is a repeat.
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 \({\mathbb{R}}^n\times{\mathbb{R}}^n\).
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 \({\left\lvert {f\ast g} \right\rvert}^2 \leq {\left\lVert {g} \right\rVert}_1 \qty{ {\left\lvert {f} \right\rvert}^2 \ast {\left\lvert {g} \right\rvert}}\).
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 \({\left\lVert {f - P} \right\rVert}_{\infty} < \varepsilon\).
\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*}
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*}
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.
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 \(H\).
\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*}
\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*}
Mnemonic: “F” stands for ferme, which is “closed” in French, and \(\sigma\) corresponds to a “sum,” i.e. a union.↩︎
Slogan: to negate, find a bad \(x\) depending on \(n_0\) that are larger than some \(\varepsilon\).↩︎
Slogan: a uniform limit of continuous functions is continuous.↩︎
Note that this is only pointwise convergence of \(f\), whereas the full \(M{\hbox{-}}\)test gives uniform convergence.↩︎
It suffices to show \({\left\lvert {f_n(x)} \right\rvert} \leq M_n\) for some \(M_n\) not depending on \(x\).↩︎
So this implicitly holds if \(f\) is the pointwise limit of \(f_n\).↩︎
See Abbott theorem 6.4.3, pp 168.↩︎
This holds for outer measure iff \(\mathrm{dist}(A, B) > 0\).↩︎