# 4. Hilbert Space Theory


Objective of this chapter is to completely characterize $L^2(\mu),$ the famous Hilbert space. To achieve our goal, we will use the fact that a Hilbert space can be seen as an infinite-dimensional vector space where there exists a “orthogonal basis”. i.e. any element in the space can be decomposed into an infinite linear combination of orthogonal components. In fact, the basis decomposition yields to the main result that $L^2$ is actually isomorphic to $\ell^2.$

## Counting measure

[TBD]

• countable $X$
• uncountable $X$

## Basic Hilbert space properties

### The element of the smallest norm

Let $E \sub H$ be a non-empty, closed, convex set. Then there uniquely exists $x \in E$ that satisfies $$\|x\| = \inf_{y \in E} \|y\|.$$

Let $M = \inf_{y \in E} \|y\|,$ then there exists $x_n \in E$ such that $\|x_n\| \to M.$ By the parallelogram law, \begin{aligned} \|x_n - x_m\|^2 &= 2(\|x_n\|^2 + \|x_m\|^2) - \|x_n + x_m\|^2 \\ &= 2(\|x_n\|^2 + \|x_m\|^2) - 4 \left\|\frac{x_n + x_m}{2}\right\|^2 \\ &\le 2(\|x_n\|^2 + \|x_m\|^2) - 4M^2 \\ &= 2(\|x_n\|^2 - m^2) + 2(\|x_m\|^2 - M^2) \end{aligned} Third equality follows from convexity of $E.$ Note that $(x_n+x_m)/2 \in E.$ Since $\|\cdot\|$ is continuous, $\|x_n\|^2 \to M^2$ as $n$ goes to infinity. Thus $(x_n)$ is a Cauchy sequence in $H.$
$H$ is a Hilbert space, which implies there exists $x \in H$ such that $x_n \to x.$ Again, by continuity of $\|\cdot\|,$ we get $\|x\|=M.$
Now we need to show the uniqueness. Suppose there are $x,y\in H$ with $\|x\| = \|y\| = M.$ Then again by parallelogram law, \begin{aligned} \|x-y\|^2 &= 2(\|x\|^2 + \|y\|^2) - \|x+y\|^2 \\ &\le 2(M^2 + M^2) - 4M^2 = 0. \end{aligned} Hence clearly $x=y.$

### Orthogonal decomposition

Let $M$ be a closed subspace of $H.$ For $x \in H,$ the followings hold.
(i) There uniquely exists $P_x \in M$ and $Q_x \in M^\perp$ such that $x = P_x + Q_x.$
(ii) $P_x, Q_x$ are the closeset point to $x$ from $M, M^\perp,$ respectively.
(iii) $P_x, Q_x$ are linear.
(iv) $\|x\|^2 = \|P_x\|^2 + \|Q_x\|^2.$

(i) Note that $M+x := \{y+x: y\in M\}$ is a non-empty, closed, convex set. Let $Q_x \in M+x$ be the element of the smallest norm in $M+x.$ Let $P_x = x - Q_x,$ then clearly $P_x \in M$ and $x = P_x + Q_x.$
Now we only need to prove that $Q_x \in M^\perp.$ Observe that given $t \in \mathbb{R}$ and $z\in M,$ $Q_x + tz \in M+x.$ Thus \begin{aligned} \|Qx\| &\le \|Q_x + tz\|, \\ \cancel{\|Q_x\|^2|} &\le \cancel{\|Q_x\|^2} + t^2\|z\|^2 + 2t(Q_x, z), \\ 0 &\le t^2\|z\|^2 + 2t(Q_x, z). \end{aligned} For the inequality to hold for all $t \in \mathbb{R},$ $(Q_x,z)$ should be zero for all $z\in M$, which implies $Q_x \in M^\perp.$

### Characterization of continuous linear functional

Let $L: H \to \mathbb{R}$ be a continuous linear functional. Then there uniquely exists $y\in H$ such that $Lx = (x, y).$

If $Lx = 0$ for all $x \in H,$ then trivially, let $y=0.$
If $Lx \ne 0$ for some $x \in H,$ let $M = \{x: Lx = 0\},$ then since $L$ is continuous, $M$ is a closed subspace of $H$ and $M \ne H.$ We now claim that there exists $w \in M^\perp$ such that $w\ne 0.$
For a given $x \in H \setminus M,$ there uniquely exists $P_x \in M$ and $Q_x \in M^\perp$ such that $x = P_x + Q_x.$ Let $u = Q_x \in M^\perp$ then $u \ne 0,$ otherwise contradiction.
Let $w = (Lx)u - (Lu)x,$ then $$Lw = (Lx)(Lu) - (Lu)(Lx) = 0,$$ thus $w \in M.$ $$(w, u) = (Lx)\|u\|^2 - (Lu)(x, u) = 0, \\ Lx = \frac{Lu}{\|u\|^2}(x, u) = \left(x, \frac{Lu}{\|u\|^2} u \right).$$ Let $y = \frac{Lu}{\|u\|^2} u.$

## Orthonormal basis of Hilbert space

### Existence of maximal orthonormal set

For any partially ordered set, there exists a maximal totally ordered subset.

Let $B \sub H$ be an orthonormal set in $H.$ Then there exists a maximal orthonormal set that contains $B.$

Let $\mathcal{P} = \{\text{orthonormal sets containing }B\}$ be a partially ordered set. By Hausdorff maximality theorem, there exists a maximal totally ordered subset $\mathcal{Q}.$ Define the candidate $$\mathcal{S} = \bigcup_{A \in \mathcal{Q}} A.$$ Our claim is that $\mathcal{S}$ is the maximal orthonormal set. Clearly $B \sub \mathcal{S}.$ Now pick $u_1, u_2 \in \mathcal{S},$ then there exists $A_1, A_2 \in \mathcal{Q}$ so that $u_1 \in A_1$ and $u_2 \in A_2.$ Since $\mathcal{Q}$ is totally ordered, without loss of generality, we can say $A_1 \sub A_2.$ This implies $u_1, u_2 \in A_2$ and thus $(u_1, u_2) = 0.$
Now, suppose $\mathcal{S}$ is not maximally orthonormal. Then there exists $u \in H$ such that $$(u, u_\alpha) \text{ for all } u_\alpha \in \mathcal{S}.$$ Then $\mathcal{Q} \cup \left\{\mathcal{S} \cup \{u_\alpha\}\right\}$ is also a totally ordered subset, which is a contradiction.

### Orthonormal basis of Hilbert space

$\{u_\alpha\}_{\alpha \in F}$ is an orthonormal set in $H,$ $F$ is finite. Define $$s_F(x) = \sum_{\alpha \in F} \hat x(\alpha) u_\alpha,\\ S=[\{u_\alpha\}_{\alpha \in F}].$$ Then the followings hold.
(i) $\|x - s_F(x)\| \le \|x - s\|$ for all $s \in S.$
(ii) The equality in (i) holds only if $s = s_F(x).$
(iii) $\sum_{\alpha \in F} \hat x(\alpha)^2 \le \|x\|^2.$
expand proof

(i) and (ii). Note that $s_F(x) \in S$ and $$(x-s_F(x), u_\alpha) = 0,~ \forall \alpha \in F.$$ This implies $$x - s_F(x) \in S^\perp.$$ Hence, for given $s \in S,$ \begin{aligned} \|x - s\|^2 &= \|x-s_F(x)\|^2 + \|s_F(x) - s\|^2. \end{aligned} (iii) Put $s=0,$ then \begin{aligned} \|x\|^2 &= \|x - s_F(x)\|^2 + \|s_F(x)\|^2 \\ &\ge \|s_F(x)\|^2 = \sum_{\alpha \in F} |\hat x(\alpha)|^2. \end{aligned}

Let $\{u_\alpha\}_{\alpha \in A}$ is an orthonormal set in $H,$ and define $$S=[\{u_\alpha\}_{\alpha \in A}].$$ Then the followings hold.
(i) $\sum_{\alpha \in A} |\hat x(\alpha)|^2 \le \|x\|^2.$
(ii) $f:H \to \ell^2(A),$ $x \mapsto \hat x$ is linear and continuous.
(iii) $\overline S \simeq \ell^2(A)$ with Hilbert space isomorphism $f.$
expand proof

(i) $$\sum_{\alpha \in F} \hat x(\alpha)^2 \le \|x\|^2,~ \forall \text{ finite } F.$$ Thus $$\sum_{\alpha \in A} \hat x(\alpha)^2 = \sup_{F \sub A:\text{ finite}} \sum_{\alpha \in F} \hat x(\alpha)^2 \le \|x\|^2.$$ (ii) Linearity of $f$ is trivial. Now, observe that for given $x,y \in H,$ \begin{aligned} \|f(x) - f(y)\| &= \|f(x-y)\| = \|\widehat{(x-y)}\| \\ &\le \|x-y\|, \end{aligned} thus $f$ is Lipschitz continuous.
(iii) For $f$ to be a Hilbert space isomophism, we need to prove that $f$ is isometric and surjective.
(isometry) Given $x \in S,$ observe that $x$ is a (finite) linear combination, thus if we let $x = \sum_{\alpha \in F} c_\alpha u_\alpha$ for some finite $F,$ \begin{aligned} \|x\|^2 &= \sum_{\alpha\in F} c_\alpha^2 \|u_\alpha\|^2 &&\text{(u_\alpha's are orthogonal)}\\ &= \sum_{\alpha \in F} c_\alpha^2, &&\text{(\|u_\alpha\|=1)}\\ \|\hat x\|^2 &= \sum_{\alpha \in F} \hat x(\alpha) ^2 \\ &= \sum_{\alpha \in F} c_\alpha^2. &&\text{(\hat x(\alpha) = c_\alpha)} \end{aligned} Hence the two values coincide.
If $x \in \overline S,$ note that $\overline S$ is a closed subspace in $H,$ thus is a Hilbert space itself. There exists $x_n \in S$ such that $x_n \to x$ as $n \to \infty.$ Since $\|\cdot\|$ and $(\cdot)^2$ and $f$ are continuous, $$\|x_n\|^2 \to \|x\|^2, \\ \|x_n\|^2 = \|\widehat x_n\|^2, \\ \|\widehat x_n\|^2 \to \|\widehat x\|^2$$ which implies $\|\widehat x\| = \|x\|.$
(surjectivity) Theorem 3.13 implies that $$f(S) = \{s: \text{is simple},~ c(s\ne0) < \infty\}$$ is dense in $\ell^2(A).$ Given $h \in \ell^2(A),$ there exists $h_n \in f(S)$ such that $h_n \to h.$ There also exists $x_n \in S$ such that $\widehat x_n = h_n.$ Note that since $\ell^2(A)$ is complete, this implies \begin{aligned} &(h_n = \widehat x_n) \text{ is Cauchy}\\ \iff & (x_n) \text{ is Cauchy} && \text{(isometry)} \\ \iff & x_n \to x \text{ for some } x \in \overline S &&\text{(Hilbert)} \end{aligned} Thus $f$ is surjective.

Let $\{u_\alpha\}_{\alpha \in A}$ is an orthonormal set in $H,$ and define $$S=[\{u_\alpha\}_{\alpha \in A}].$$ Then the followings are equivalent.
(i) $\{u_\alpha\}_{\alpha \in A}$ is a maximal orthonormal set.
(ii) $\overline S = H.$
(iii) $\|x\|^2 = \sum_{\alpha \in A}|\widehat x(\alpha)|^2$ for all $x \in H.$
(iv) $(x,y) = \sum_{\alpha \in A} \widehat x(\alpha) \widehat y(\alpha)$ for all $x,y\in H.$
expand proof

(i$\Rightarrow$ii) Suppose not. Then there exists $x \in H \setminus \overline S.$ Since $\overline S$ is a closed subspace, there is a unique decomposition $x = P_x + Q_x$ where $P_x \in \overline S$ and $Q_x \in \overline S ^\perp.$ Note that $Q_x \ne 0$ since otherwise $x \in \overline S$ which is a contradiction. Let $u = Q_x,$ then $$\{u_\alpha\}_{\alpha \in A} \cup \{u\}$$ is an orthonormal set, which is a contradiction.
(ii$\Rightarrow$iii) By the previous theorem, it is trivial.
(iii$\Rightarrow$iv) \begin{aligned} \|x+y\|^2 &= \|x\|^2 + \|y\|^2 + 2(x,y), \\ \|\widehat x + \widehat y\| &= \|\widehat x\|^2 + \|\widehat y\|^2 + 2(\widehat x, \widehat y), \\ (x,y) &= (\hat x, \hat y) = \sum_{\alpha \in A} \widehat x(\alpha) \widehat y(\alpha). && \text{(\|x\|=\|\hat x\|)} \end{aligned} (iv$\Rightarrow$i) Suppose not. Then there exists $u \in H$ that is orthogonal to $u_\alpha$ for all $\alpha \in A$ and $\|u\| = 1.$ i.e. $$\widehat u (\alpha) = (u, u_\alpha) = 0,~ \forall \alpha \in A.$$ Apply the result (iv), then \begin{aligned} 1 = (u,u) = \sum_{\alpha \in A} \widehat u(\alpha) \widehat u(\alpha) = 0. &&(\Rightarrow\mspace{-4mu}\Leftarrow) \end{aligned}

### Fourier transform in $\ell^2(A)$

[TBD]

References

• Rudin. 1986. Real and Complex Analysis. 3rd edition. McGraw-Hill.
• Real Analysis (Spring, 2021) @ Seoul National University, Republic of Korea (instructor: Prof. Insuk Seo).