Functional Analysis

$L^p$ Spaces

remark:

For any measure space $(X, M, \mu)$ , one can define $L^2(\mu)$ with the inner product ${\left\langle {f},~{g} \right\rangle} \coloneqq\int_X f\overline{g} d\mu$ .

proposition (

$L^p$ spaces are Banach):

proposition (Duals of

$L^p$ spaces):

For $p^{-1}+ q^{-1}= 1$ , with $1<p<\infty$ , there is an isomorphism of Banach spaces $\begin{align*} \kappa: L^p(\mu) &\to L^q(\mu) \\ f &\mapsto (g \mapsto \int_X f g d\mu ) .\end{align*}$

This is surjective by Radon-Nikodym, and an isometry by Holder’s inequality, which is enough to be an isometric isomorphism.

General Facts

remark:

Working with inner products: $\begin{align*} {\left\langle {tx + sy},~{z} \right\rangle} &= t{\left\langle {x},~{z} \right\rangle} + s{\left\langle {y},~{z} \right\rangle} \\ {\left\langle {x},~{y} \right\rangle} &= \overline{{\left\langle {y},~{x} \right\rangle}} \\ x\neq 0 \implies {\left\langle {x},~{x} \right\rangle} > 0 .\end{align*}$ We define ${\left\lVert {x} \right\rVert} \coloneqq\sqrt{{\left\langle {x},~{x} \right\rangle}}$ .

fact (Cauchy-Schwarz):

For $x, y\in H$ , $\begin{align*} {\left\lvert { {\left\langle {x},~{y} \right\rangle}} \right\rvert} \leq {\left\lVert {x} \right\rVert} {\left\lVert {y} \right\rVert} ,\end{align*}$ with equality iff $x, y$ are linearly independent.

fact (Pythagoras):

$\begin{align*} {\left\langle {v},~{w} \right\rangle} = 0 \implies {\left\lVert {v+w} \right\rVert}^2 = {\left\lVert {v} \right\rVert}^2 + {\left\lVert {w} \right\rVert}^2 .\end{align*}$ More generally, if $x_i \perp x_j$ for all $i\neq j$ , then $\begin{align*} {\left\lVert {\sum_k x_k } \right\rVert}_H^2 = \sum_k {\left\lVert {x_k} \right\rVert}_H^2 .\end{align*}$

fact (Polarization):

For all $x, y\in H$ , $\begin{align*} 4 {\left\langle {x},~{y} \right\rangle} = {\left\lVert {x+y} \right\rVert}^2 - {\left\lVert {x-y} \right\rVert}^2 +i\qty{{\left\lVert {x+iy} \right\rVert}^2 - {\left\lVert {x-iy} \right\rVert}^2} .\end{align*}$

fact (Parallelogram law):

For all $x, y\in H$ , $\begin{align*} {\left\lVert {x+y} \right\rVert}^2 + {\left\lVert {x-y} \right\rVert}^2 = 2\qty{{\left\lVert {x} \right\rVert}^2 + {\left\lVert {y} \right\rVert}^2 } .\end{align*}$

proposition (Convergence implies convergence of inner products):

If $x_k\to x$ and $y_k\to y$ in $H$ , then ${\left\langle {x_k},~{y_k} \right\rangle} \to {\left\langle {x},~{y} \right\rangle}$ . Proof: $\begin{align*} {\left\lvert {{\left\langle {x_k},~{y_k} \right\rangle} - {\left\langle {x},~{y} \right\rangle} } \right\rvert} ={\left\lvert {{\left\langle {x_n - x},~{y_n} \right\rangle} + {\left\langle {x},~{y_n-y} \right\rangle} } \right\rvert} \leq {\left\lVert {x_n - x} \right\rVert}{\left\lVert {y_n} \right\rVert} + {\left\lVert {x} \right\rVert} {\left\lVert {y_n - y} \right\rVert} .\end{align*}$

Fourier Coefficients

theorem (Bessel's Inequality):

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 $\begin{align*} \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}$

Note that this generalizes to uncountable bases, and implies that only finitely many terms ${\left\langle {x},~{u_n} \right\rangle}$ can be nonzero.

proof (of Bessel's inequality):

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

theorem (Parseval):

Let $\left\{{u_n}\right\}_{n\in A}$ be an orthonormal set in a Hilbert space ${\mathcal{H}}$ . TFAE:

Completeness: $\left\{{u_n}\right\}$ is a complete basis, i.e. ${\left\langle {x},~{u_n} \right\rangle}=0$ for all $n$ implies $x=0$
Parseval’s identity: $\begin{align*} \sum_{n\in A} {\left\lvert { {\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 = {\left\lVert {x} \right\rVert}^2_{{\mathcal{H}}} .\end{align*}$
Every $x\in {\mathcal{H}}$ can be expressed uniquely as $\begin{align*} x = \sum_{n\in A} {\left\langle {x},~{u_n} \right\rangle}u_n ,\end{align*}$ where the sum has only countably many nonzero terms.

theorem (Riesz Representation for Hilbert Spaces):

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

proof (?):

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

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

Compute

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

Choose $y \coloneqq\overline{\Lambda(z)} z$ .
Check uniqueness:

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

theorem (Riesz-Fischer):

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

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 $\begin{align*} a_{n} = {\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle} \quad \forall n. \end{align*}$

$\mathbf{x}$ 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 $\mathbf{x} = \mathbf{0}$ .

proof (?):

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

Operator Norms

theorem (Functionals are continuous if and only if bounded):

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

$L$ is continuous
$L$ is continuous at zero
$L$ is bounded, i.e. $\exists c\geq 0$ such that ${\left\lvert {L(x)} \right\rvert} \leq c {\left\lVert {x} \right\rVert}$ for all $x\in H$

proof (?):

$2 \implies 3$ : 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 $\delta = \frac \varepsilon c$ .

theorem (The operator norm is a norm):

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

proof (?):

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

theorem (The operator norm on

$X\dual$ yields a Banach space):

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

proof (?):

Let $\left\{{L_{n}}\right\}$ be Cauchy in $X {}^{ \vee }$ .
Then for all $x\in C$ , $\left\{{L_{n}(x)}\right\} \subset {\mathbf{C}}$ 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 $\begin{align*} n, m \geq N \implies {\left\lVert {L_{n} - L_{m}} \right\rVert} < \varepsilon \implies {\left\lvert {L_{m}(x) - L_{n}(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} = 1 .\end{align*}$
Take $n\to \infty$ to obtain $\begin{align*}m \geq N &\implies {\left\lvert {L_{m}(x) - L(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} = 1\\ &\implies {\left\lVert {L_{m} - L} \right\rVert} < \varepsilon \to 0 .\end{align*}$
Continuity: $\begin{align*} {\left\lvert {L(x)} \right\rvert} &= {\left\lvert {L(x) - L_{n}(x) + L_{n}(x)} \right\rvert} \\ &\leq {\left\lvert {L(x) - L_{n}(x)} \right\rvert} + {\left\lvert {L_{n}(x)} \right\rvert} \\ &\leq \varepsilon {\left\lVert {x} \right\rVert} + c{\left\lVert {x} \right\rVert} \\ &= (\varepsilon + c){\left\lVert {x} \right\rVert} \hfill\blacksquare .\end{align*}$

Functional Analysis

LpL^p Spaces

General Facts

Fourier Coefficients

Operator Norms

$L^p$ Spaces