# Definitions

## Convergence and Continuity

\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*}

A function $$f: {\mathbf{R}}\to {\mathbf{R}}$$ is continuous on $$X$$ iff for all $$x_0\in X$$, \begin{align*} &\forall \varepsilon \quad \exists \delta(\varepsilon, x_0) \text{ such that }\quad \forall y, {\left\lvert {x_0 - y} \right\rvert} < \delta &&\implies {\left\lvert {f(x_0) - f(y)} \right\rvert} < \varepsilon \\ \iff &\forall \varepsilon \quad \exists \delta(\varepsilon, x_0) \text{ such that }\quad \forall h, {\left\lvert {h} \right\rvert} < \delta &&\implies {\left\lvert {f(x_0) - f(x_0 \pm h)} \right\rvert} < \varepsilon .\end{align*}

$$f$$ is uniformly continuous on $$X$$ iff

\begin{align*} &\forall \varepsilon \quad \exists \delta(\varepsilon) \text{ such that }\quad \forall x, y, \in X \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) \text{ such that} \quad \, \forall x, h, \quad {\left\lvert {h} \right\rvert} < \delta &&\implies {\left\lvert {f(x) - f(x \pm h)} \right\rvert} < \varepsilon .\end{align*} These follow from the substitutions $$x_0-y = \mp h \implies y = x_0 \pm h$$.

The main difference is that $$\delta$$ may depend on $$x_0$$ and $${\varepsilon}$$ in continuity, but only depends on $${\varepsilon}$$ in the uniform version. I.e. once $$\delta$$ is fixed, for continuity one may only range over $$x$$, but in uniform continuity one can range over all pairs $$x,y$$.

Let $$X$$ be a metric space and $$A$$ a subset. Let $$A'$$ denote the limit points of $$A$$, and $$\overline{A} \coloneqq A\cup A'$$ 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 $$B_r(p) \coloneqq\left\{{q {~\mathrel{\Big\vert}~}d(p, q) < r}\right\}$$ for some $$r>0$$.

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

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

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

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

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

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

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

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

• Alternatively, $$\overline{A} = X$$, so the closure of $$A$$ is $$X$$.

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*}

\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: 1 \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*}

## Function Spaces

A metric space is complete if every Cauchy sequence converges.

If $$X$$ is complete, then absolutely convergent implies convergent.

Recall that a set $$S$$ in $$X$$ is dense $$\iff$$ every open $$U\subseteq X$$ intersects $$S$$. A set $$S$$ is nowhere dense in $$X$$ $$\iff$$ the closure of $$S$$ has empty interior $$\iff$$ every subset (or interval) contains an open set (or a subinterval) that does not intersect $$S$$. This just says $$S$$ is not dense in any subset $$S' \subseteq X$$, by negating what it means to be dense.

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

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

## Measure Theory

\begin{align*} \liminf_{n} E_{n} \coloneqq\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_{n} &= \left\{{x {~\mathrel{\Big\vert}~}x\in E_{n} \text{ for all but finitely many } n}\right\} \\ \limsup_{n} E_{n} \coloneqq\bigcap_{N=1}^\infty \bigcup_{n=N}^{\infty} E_{n} &= \left\{{x {~\mathrel{\Big\vert}~}x\in E_{n} \text{ for infinitely many } n}\right\} .\end{align*}

How to derive these definitions: use that $$\inf$$ corresponds to intersections/existence and $$\sup$$ corresponds to unions/forall.

• For $$\liminf E_n$$:
• $$x\in \liminf E_n \iff$$ there exists some $$N$$ such that $$x\in \cap_{n\geq N} E_n$$, i.e. $$x\in E_n$$ for all $$n\geq N$$. So $$x$$ is in all but finitely many $$n$$.
• How to remember: $$\liminf_{n} x_n = \sup_{n} \inf_{k\geq n} x_n$$ for sequences, where sups look like unions and infs look like intersections.
• Alternatively: there exists an $$n$$ (union) such that for all $$k\geq n$$ (intersection)…
• For $$\limsup E_n$$:
• $$x\in \limsup E_n \iff$$ for every $$N$$, there exists some $$n\geq N$$ such that $$x\in E_n$$. So $$x$$ is an infinitely many $$E_n$$.
• How to remember: $$\limsup_{n} x_n = \inf{n} \sup{k\geq n} x_n$$ for sequences, where sups look like unions and infs look like intersections.
• Alternatively: for all $$n$$ (intersection) there exists a $$k\geq n$$ (union)…

It’s also useful to note that $$\liminf E_n \subseteq \limsup E_n$$, since $$\liminf E_n$$ are elements that are eventually in all sets, and $$\limsup E_n$$ are elements in infinitely many sets.

Why these are useful: for finite measure spaces, \begin{align*} \mu\qty{\liminf_n E_n }\leq \liminf_n \mu(E_n) \leq \lim_n \mu(E_n) \leq \limsup_n \mu(E_n) \leq \mu\qty{\limsup_n E_n} .\end{align*} If the $$\limsup$$ and $$\liminf$$ sets are equal, then one can define the set $$\lim_n E_n \coloneqq\cup_n E_n$$ if $$E_n \nearrow E$$ or $$\lim_n E_n \coloneqq\cap_n E_n$$ if $$E_n\searrow E$$ in which case continuity of measure states \begin{align*} \mu\qty{\lim_n E_n} = \lim_n \mu(E_n) .\end{align*}

An $$F_\sigma$$ set is a union of closed sets, and a $$G_\delta$$ set is an intersection of opens. 2

The outer measure of a set is given by \begin{align*} m_*(E) \coloneqq\inf_{\substack{\left\{{Q_{i}}\right\} \rightrightarrows E \\ \text{closed cubes}}} \sum {\left\lvert {Q_{i}} \right\rvert} ,\end{align*} where $${\left\lvert {Q_i} \right\rvert}$$ is the standard Euclidean volume of a cube in $${\mathbf{R}}^n$$.

A subset $$E\subseteq {\mathbf{R}}^n$$ 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 $$m(E) \coloneqq m_*(E)$$.

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

## Integrals and $$L^p$$ Spaces

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

\begin{align*} {\left\lVert {f} \right\rVert}_\infty &\coloneqq\inf_{\alpha \geq 0} \left\{{\alpha {~\mathrel{\Big\vert}~}\mu\qty{\left\{{{\left\lvert {f} \right\rvert} \geq \alpha}\right\}} = 0}\right\} .\end{align*} In words, this is the smallest upper bound that holds almost everywhere, so $${\left\lvert {f(x)} \right\rvert} \leq {\left\lVert {f} \right\rVert}_\infty$$ holds for almost every $$x$$. A function $$f:X \to {\mathbf{C}}$$ 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) \coloneqq\left\{{f: X\to {\mathbf{C}}{~\mathrel{\Big\vert}~}f \text{ is essentially bounded }}\right\} \coloneqq\left\{{f: X\to {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lVert {f} \right\rVert}_{\infty }< \infty}\right\} .\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.

Some properties that approximate identities enjoy:

• $$\int \phi_t = {\left\lVert {\phi_t} \right\rVert}_1 = 1$$ for every $$t$$.
• $$\sup_t {\left\lVert {\phi_t} \right\rVert}_1 < \infty$$
• For every $$h>0$$, $$\lim_{t\to\infty} \int_{{\left\lvert {x} \right\rvert} \geq h} {\left\lvert {\phi_t(x)} \right\rvert}\,dx= 0$$.

## Functional Analysis

For $$X$$ a normed vector space and $$L \in X {}^{ \vee }$$, the dual norm or operator norm is defined by \begin{align*} {\left\lVert {L} \right\rVert}_{X {}^{ \vee }} \coloneqq\sup_{ \substack{x\in X \\ {\left\lVert {x} \right\rVert} = 1} } {\left\lvert {L(x)} \right\rvert} = \sup_{ \substack{x\in X \\ {\left\lVert {x} \right\rVert} \leq 1} } {\left\lvert {L(x)} \right\rvert} .\end{align*}

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

• $${\left\langle {u_i},~{u_j} \right\rangle} = 0$$ for all $$j \neq k$$ and
• $${\left\lVert {u_j} \right\rVert}^2 \coloneqq{\left\langle {u_j},~{u_j} \right\rangle} = 1$$ 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 }\coloneqq\left\{{L: H\to {\mathbf{C}}{~\mathrel{\Big\vert}~}L \text{ is continuous }}\right\} .\end{align*}

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

Footnotes
1.
Slogan: to negate, find a bad $$x$$ depending on $$n_0$$ that are larger than some $${\varepsilon}$$.
2.
Mnemonic: “F” stands for ferme, which is “closed” in French, and $$\sigma$$ corresponds to a “sum”, i.e. a union.