LEC ANA2 16

Hilbert spaces

A Hilbert space is a complete inner product space. Examples include ${\mathbb{C}}^n$ with the usual inner product, and ${\ell }_2$ (Prp 161.3, Exr 164.6).

Definition 344.1(Lebesgue spaces).

On $C[a,b]$, Define

$$⟨f,g⟩\equiv \frac{1}{b-a}{\int }_a^{b}f\overline{g}.$$

Note that $C[a,b]$ is not complete under the norm induced by $⟨,⟩$, which may be denoted by $\Vert \cdot {\Vert }_2$ or $\Vert \cdot {\Vert }_{⟨,⟩}$. Define $L^2([a,b])$ to be the completion of $C[a,b]$ with respect to $⟨,⟩$. $L^2[a,b]$ is a Hilbert space.

2f39a2

Remark 344.2(Remark).

Alternatively, elements of $L^2[a,b]$ are equivalence classes of measurable functions $f:[a,b]\to \mathbb{C}$ for which

$${\int }_a^b|f(x){|}^{2}dx<\infty ,$$

where two functions belong to the same class iff they do not differ outside of a set of measure zero. Note that while $L^2[a,b]$ is a Hilbert space, it is not an algebra, since it is not closed under multiplication.

0109ea

Definition 344.3(Orthonormal basis).

Let $H$ be a Hilbert space. $S\subseteq H$ is an orthonormal basis if it is orthonormal and $\overline{\text{span}(S)}=H$.

5f2a03

Contrast with a Hamel basis.

Example 344.4(Example).

$\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ is an orthonormal basis for ${\ell }_2$, where ${\mathbf{e}}_n^k={\delta }_{n,k}$. Indeed, if $\mathbf{\lambda }\in {\ell }_2$,

$$\sum _{i=1}^n{\lambda }_i{\mathbf{e}}_i\to \mathbf{\lambda }.$$

Proposition 344.5(Proposition).

Any orthonormal subset of a separable Hilbert space is countable.

Proof.

Let $H$ be a separable Hilbert space, and $S\subseteq H$ be orthonormal. If $\mathbf{e},\mathbf{f}\in S$, $\Vert \mathbf{e}-\mathbf{f}{\Vert }^2=⟨\mathbf{e}-\mathbf{f},\mathbf{e}-\mathbf{f}⟩=2$. Let $\{x_n{\}}_{n=1}^{\infty }\subseteq H$ be dense in $H$. $B(x_n,\sqrt{2}/2)$ can contain at most one element of $S$. Thus, $S$ must be countable.□

Remark 344.6(Remark).

Let $\{{\mathbf{x}}_1,\dots ,{\mathbf{x}}_n\}$ be an orthogonal subset of a Hilbert space $H$. Then,

$${\Vert \sum _{i=1}^n{\mathbf{x}}_i\Vert }^2=\sum _{i=1}^n\Vert {\mathbf{x}}_i{\Vert }^2.$$

3a18a6

Proposition 344.7(Proposition).

Any orthonormal set is linearly independent.

Proof.

Let $S$ be an orthonormal set. Let $\{{\mathbf{x}}_1,\dots ,{\mathbf{x}}_n\}\subseteq S$. If ${\sum }_{i=1}^n{\alpha }_i{\mathbf{x}}_i=0$, ${\alpha }_j=⟨{\sum }_{i=1}^n{\alpha }_i{\mathbf{x}}_i,{\mathbf{x}}_j⟩=⟨0,{\mathbf{x}}_j⟩=0$.□

Proposition 344.8(Proposition).

Let $\{{\mathbf{e}}_1,\dots ,{\mathbf{e}}_n\}$ be an orthonormal set. Then

1. $\mathbf{x}-{\sum }_{i=1}^n⟨\mathbf{x},{\mathbf{e}}_i⟩{\mathbf{e}}_i$ is perpendicular to ${\mathbf{e}}_j$ for all $1⩽j⩽n$.

2. ${\sum }_{i=1}^n|⟨\mathbf{x},{\mathbf{e}}_i⟩{|}^2⩽\Vert \mathbf{x}{\Vert }^2$.

Proof.

$(1)$ is clear. Let $\mathbf{z}={\sum }_{i=1}^n⟨\mathbf{x},{\mathbf{e}}_i⟩{\mathbf{e}}_i$. By $(1)$, $\mathbf{x}-\mathbf{z}$ is orthogonal to $\mathbf{z}$. Using Rmk 6,

$$\begin{array}{rl}\Vert \mathbf{x}{\Vert }^2& ={\Vert \mathbf{x}-\mathbf{z}\Vert }^2+\Vert \mathbf{z}\Vert \\ & ⩾{\Vert \mathbf{z}\Vert }^2\\ & =\sum _{i=1}^n|⟨\mathbf{x},{\mathbf{e}}_i⟩{|}^2.\end{array}$$ □

Corollary 344.9(Corollary).

Let $H$ be a separable Hilbert space. Let $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ be an orthonormal set in $H$. Then

$$\sum _{i=1}^{\infty }|⟨\mathbf{x},{\mathbf{e}}_n⟩{|}^2<\infty .$$

9752cf

Lemma 344.10(Lemma).

The inner product in continuous map from $H\times H$ to $\mathbb{C}$, that is, if $\{{\mathbf{x}}_n\}\to \mathbf{x}$ and $\{{\mathbf{y}}_n\}\to \mathbf{y}$, then $⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩\to ⟨\mathbf{x},\mathbf{y}⟩$.

Proof.

$$\begin{array}{rl}\left|⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩-⟨\mathbf{x},\mathbf{y}⟩\right|& ⩽|⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩-⟨{\mathbf{x}}_n,\mathbf{y}⟩|+|⟨{\mathbf{x}}_n,\mathbf{y}⟩-⟨\mathbf{x},\mathbf{y}⟩|\\ & =|⟨{\mathbf{x}}_n,{\mathbf{y}}_n-\mathbf{y}⟩|+|⟨{\mathbf{x}}_n-\mathbf{x},\mathbf{y}⟩|\\ & ⩽\Vert {\mathbf{x}}_n\Vert \Vert {\mathbf{y}}_n-\mathbf{y}\Vert +\Vert {\mathbf{x}}_n-\mathbf{x}\Vert \Vert \mathbf{y}\Vert \\ & \to 0\text{ as }n\to \infty ,\end{array}$$

where the last inequality follows from Thm 87.8.□

181916

Proposition 344.11(Proposition).

Let $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ be an orthonormal set in a separable Hilbert space $H$. Let $\{{\alpha }_n{\}}_{n=1}^{\infty }\subseteq \mathbb{C}$. TFAE:

1. ${\sum }_{n=1}^{\infty }{\alpha }_n{\mathbf{e}}_n$ converges.

2. ${\sum }_{n=1}^{\infty }|{\alpha }_n{|}^2<\infty$.

3. There exists $\mathbf{x}\in \overline{\text{span}\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }}$ such that $⟨\mathbf{x},{\mathbf{e}}_n⟩={\alpha }_n$.

Proof.

$(1⟺2)$

$$\begin{array}{rl}& \sum _{n=1}^{\infty }{\alpha }_n{\mathbf{e}}_n<\infty \\ & ⟺\sum _{n=N}^M{\alpha }_n{\mathbf{e}}_n\to 0\text{ as }N,M\to \infty \\ & ⟺\Vert \sum _{n=N}^M{\alpha }_n{\mathbf{e}}_n\Vert \to 0\text{ as }N,M\to \infty \\ & ⟺\sum _{n=N}^M|{\alpha }_n{|}^2\to 0\text{ as }N,M\to \infty \\ & ⟺\sum _{n=1}^{\infty }|{\alpha }_n{|}^2<\infty .\end{array}$$

$(1,2⟹3)$ Take $\mathbf{x}={\sum }_{n=1}^{\infty }{\alpha }_n{\mathbf{e}}_n$. Using Lem 10,

$$\begin{array}{rl}⟨\mathbf{x},{\mathbf{e}}_k⟩& =⟨\sum _{n=1}^{\infty }{\alpha }_n{\mathbf{e}}_n,{\mathbf{e}}_k⟩\\ & =\sum _{n=1}^{\infty }⟨{\alpha }_n{\mathbf{e}}_n,{\mathbf{e}}_k⟩\\ & ={\alpha }_k.\end{array}$$

$(3⟹2)$ By Cor 9,

$$\begin{array}{r}\sum _{n=1}^{\infty }|{\alpha }_n{|}^2=\sum _{n=1}^{\infty }|⟨\mathbf{x},{\mathbf{e}}_n⟩{|}^2<\infty .\end{array}$$ □

84dbb8

Proposition 344.12(Proposition).

Let $H$ be a separable Hilbert space. Let $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ be an orthonormal set. TFAE:

1. $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ is an orthonormal basis.

2. $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ is a maximal orthonormal set.

3. $\mathbf{x}={\sum }_{n=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_n⟩{\mathbf{e}}_n$ for all $\mathbf{x}\in H$.

4. $⟨\mathbf{x},\mathbf{y}⟩={\sum }_{n=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_n⟩⟨{\mathbf{e}}_n,\mathbf{y}⟩$ for all $\mathbf{x},\mathbf{y}\in H$.

5. $\Vert \mathbf{x}{\Vert }^2={\sum }_{n=1}^{\infty }|⟨\mathbf{x},{\mathbf{e}}_n⟩{|}^2$ for all $\mathbf{x}\in H$.

Proof.

$(1⟺2)$ If $\{{\mathbf{e}}_n\}$ is a maximal orthonormal set, it follows that it is an orthonormal basis from $(2⟹3)$. Suppose $\{{\mathbf{e}}_n\}$ is an orthonormal basis, and $\mathbf{e}\perp {\mathbf{e}}_n$ for all $n$. Since $\overline{\text{span }\{{\mathbf{e}}_n\}}=H$, we must have ${\sum }_{n=1}^{\infty }{\alpha }_n{\mathbf{e}}_n=\mathbf{e}$ for some coefficients ${\alpha }_i$. Taking the inner product with $\mathbf{e}$ on both sides results in a contradiction by Lem 10.

$(2⟹3)$ By Cor 9 and Prp 11, ${\sum }_{n=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_n⟩{\mathbf{e}}_n$ converges. If $\mathbf{y}:=\mathbf{x}-{\sum }_{n=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_n⟩{\mathbf{e}}_n\ne 0$, $\mathbf{y}$ is perpendicular to ${\mathbf{e}}_n$ for all $n$, contradicting maximality.

$(3⟹4)$

$$\begin{array}{rl}⟨\mathbf{x},\mathbf{y}⟩& =⟨\sum _{i=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_i⟩{\mathbf{e}}_i,\sum _{j=1}^{\infty }⟨\mathbf{y},{\mathbf{e}}_j⟩{\mathbf{e}}_j⟩\\ & =\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_i⟩⟨{\mathbf{e}}_j,\mathbf{y}⟩⟨{\mathbf{e}}_i,{\mathbf{e}}_j⟩\\ & =\sum _{i=1}^{\infty }⟨\mathbf{x},{\mathbf{e}}_i⟩⟨{\mathbf{e}}_i,\mathbf{y}⟩.\end{array}$$

$(4⟹5)$ is immediate.

$(5⟹2)$ Suppose $\{{\mathbf{e}}_n{\}}_{n=1}^{\infty }$ is not maximal, that is, there exists nonzero $\mathbf{e}\in H$ such that $\mathbf{e}\perp {\mathbf{e}}_n$ for all $n$. Then, $\Vert \mathbf{e}{\Vert }^2={\sum }_{n=1}^{\infty }|⟨\mathbf{e},{\mathbf{e}}_n⟩{|}^2=0$, a contradiction.□

754891

Theorem 344.13(Theorem).

$S=\{e^{int}:n\in \mathbb{Z}\}\subseteq L^2[0,2\pi ]$ is an orthonormal basis.

Proof.

$S$ is orthonormal:

$$\begin{array}{rl}⟨e^{int},e^{imt}⟩& =\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{i(n-m)t}dt\\ & =\{\begin{array}{ll}1& n=m\\ {\left[\frac{e^{i(n-m)t}}{i(n-m)}\right]}_0^{2\pi }=0& n\ne m.\end{array}\end{array}$$

Let $\mathcal{C}:=C^0([0,2\pi ],\mathbb{C})$ be the space of all continuous periodic complex functions $f$ with period $2\pi$ - essentially, $C(S_1,\mathbb{C})$. Observe that $\mathcal{A}:=\text{span}\{e^{int}:n\in \mathbb{Z}\}$ is a unital self-adjoint subalgebra of $\mathcal{C}$ which separates points ¹. By Thm 315.11, $\mathcal{A}$ is dense in $(\mathcal{C},\Vert \cdot {\Vert }_{\infty })$. Thus, for every $f\in \mathcal{C}$, there exists $\{f_n\}\subseteq \mathcal{A}$ such that $\{f_n\}\rightrightarrows f$. It follows that $|f-f_n{|}^2\rightrightarrows 0$. By Thm 155.7,

$$\begin{array}{r}\underset{n\to \infty }{\lim }{\int }_0^{2\pi }|f-f_n{|}^{2}dx=0,\end{array}$$

so $\{f_n\}\to f$ under $\Vert \cdot {\Vert }_2$. It follows that $\mathcal{A}$ is dense in $(\mathcal{C},\Vert \cdot {\Vert }_2)$. Note that $(\mathcal{C},\Vert \cdot {\Vert }_2)$ is dense in $(C([0,2\pi ],\mathbb{C}),\Vert \cdot {\Vert }_2)$², which is dense in $L^2[0,2\pi ]$ by definition. $\mathcal{A}$ is dense in $L^2[0,2\pi ]$ by Lem 165.2.□

915903

https://www.desmos.com/3d/djkjo4hqid

Footnotes

1. Note that $0$ and $2\pi$ are considered to be the same point of $S_1$! ↩

2. It’s pretty obvious if you think about it. Remember that we’re claiming density under $\Vert \cdot {\Vert }_2$, not $\Vert \cdot {\Vert }_{\infty }$. ↩