§10.2 Orthonormal Systems of a Hilbert Space

10.2 Orthonormal Systems of a Hilbert Space

We first aim to provide a generalization of bases which will serve useful in generalizing vector spaces to infinite dimensions.

Definition 10.2.1 (Orthonormal System).

Let $𝐻$ be a Hilbert space. Let $𝐼$ be any set. A set ${𝑒_{𝑖}}_{𝑖 \in 𝐼} \subseteq 𝐻$ is an orthonormal system iff $\forall 𝑖, 𝑗 \in 𝐼$ ,

⟨ 𝑒_{𝑖}, 𝑒_{𝑗} ⟩ = 𝛿_{𝑖 𝑗} = {\begin{cases} 1 & if 𝑖 = 𝑗 \\ 0 & if 𝑖 \neq 𝑗 \end{cases} .

Generally one is concerned only with the cases where

𝐼

is finite or when

𝐼

is countably infinite (

𝐼 = ℕ

), although here we will consider the generalization to uncountable sets as it follows in the same way.

Definition 10.2.2 (Completeness of an Orthonormal System).

An orthonormal system in a Hilbert space is complete iff there exists no nonzero vector orthogonal to all vectors in the system.

The definition of completeness is intuitive: it means that there is no other nonzero vector that can be appended to the system while maintaining the orthogonality condition.

The orthonormal then generalizes the concept of a “basis” for finite-dimensional vector spaces. This will be made explicit in Corollary 10.2.4.1 $Corollary 10.2.4.1 Corollary 10.2.4.1$ .

Theorem 10.2.3.

Let ${𝑒_{𝑘}}_{𝑘 = 1}^{𝑛}$ be a finite orthonormal system in a Hilbert space $𝐻$ . Let $𝑈$ be the span of the system. The distance between a vector $𝑥 \in 𝐻$ and $𝑈$ , ${inf}_{𝑣 \in 𝑈} ‖ 𝑥 - 𝑣 ‖$ , is given by

‖ 𝑥 - 𝑢 ‖ = \sqrt{{‖ 𝑥 ‖}^{2} - \sum_{𝑘 = 1}^{𝑛} {| ⟨ 𝑥, 𝑒_{𝑘} ⟩ |}^{2}}

where

𝑢 \in 𝑈

is the unique value that attains this minimal distance.

Proof.

Let

𝑢 = \sum_{𝑘 = 1}^{𝑛} ⟨ 𝑥, 𝑒_{𝑘} ⟩ 𝑒_{𝑘} \in 𝑈 .

We aim to show this choice of $𝑢$ is optimized. Firstly, we show $𝑥 - 𝑢$ is orthogonal to $𝑈$ . For any $𝑗 \in ℕ_{\leq 𝑛}$ ,

\begin{aligned} ⟨ 𝑥 - 𝑢, 𝑒_{𝑗} ⟩ & = ⟨ 𝑥, 𝑒_{𝑗} ⟩ - ⟨ \sum_{𝑘 = 1}^{𝑛} ⟨ 𝑥, 𝑒_{𝑘} ⟩ 𝑒_{𝑘}, 𝑒_{𝑗} ⟩ \\ = ⟨ 𝑥, 𝑒_{𝑗} ⟩ - \sum_{𝑘 = 1}^{𝑛} ⟨ 𝑥, 𝑒_{𝑘} ⟩ ⟨ 𝑒_{𝑘}, 𝑒_{𝑗} ⟩ \\ = ⟨ 𝑥, 𝑒_{𝑗} ⟩ - \sum_{𝑘 = 1}^{𝑛} ⟨ 𝑥, 𝑒_{𝑘} ⟩ 𝛿_{𝑘 𝑗} = 0 . \end{aligned}

Moreover, for any $𝑣 \in 𝑈$ , by Theorem 10.1.28 $Theorem 10.1.28 Theorem 10.1.28$ , since $𝑥 - 𝑣 - (𝑥 - 𝑢) \in 𝑈$ , the orthogonal decomposition of $𝑥 - 𝑣 = 𝑎 + 𝑏$ (where $𝑎 \in 𝑈, 𝑏 \in 𝑈^{⟂}$ ) is uniquely determined by $𝑎 = 𝑢 - 𝑣$ , $𝑏 = 𝑥 - 𝑢$ .

By Theorem 10.1.22 $Theorem 10.1.22 Theorem 10.1.22$ ,

{‖ 𝑥 - 𝑣 ‖}^{2} = {‖ 𝑢 - 𝑣 ‖}^{2} + {‖ 𝑥 - 𝑢 ‖}^{2} \geq {‖ 𝑥 - 𝑢 ‖}^{2},

proving the minimality of $𝑢$ . Moreover, the point $𝑢$ is unique as equality is attained iff $‖ 𝑢 - 𝑣 ‖ = 0 ⟹ 𝑢 = 𝑣$ .

Then

\begin{aligned} ‖ 𝑥 - 𝑢 ‖ & = \sqrt{{‖ 𝑥 ‖}^{2} + {‖ 𝑢 ‖}^{2} - 2 ℜ𝔢 ⟨ 𝑥, 𝑢 ⟩} \\ = \sqrt{\begin{matrix} {‖ 𝑥 ‖}^{2} + \sum_{(𝑗, 𝑘) \in ℕ_{\leq 𝑛}^{2}} ⟨ ⟨ 𝑥, 𝑒_{𝑗} ⟩ 𝑒_{𝑗}, ⟨ 𝑥, 𝑒_{𝑘} ⟩ 𝑒_{𝑘} ⟩ \\ - 2 \sum_{𝑘 = 1}^{𝑛} ℜ𝔢 ⟨ 𝑥, ⟨ 𝑥, 𝑒_{𝑘} ⟩ 𝑒_{𝑘} ⟩ \end{matrix}} \\ = \sqrt{{‖ 𝑥 ‖}^{2} + \sum_{(𝑗, 𝑘) \in ℕ_{\leq 𝑛}^{2}} ⟨ 𝑥, 𝑒_{𝑗} ⟩ \overline{⟨ 𝑥, 𝑒_{𝑘} ⟩} 𝛿_{𝑗 𝑘} - 2 \sum_{𝑘 = 1}^{𝑛} {| ⟨ 𝑥, 𝑒_{𝑘} ⟩ |}^{2}} \\ = \sqrt{{‖ 𝑥 ‖}^{2} - \sum_{𝑘 = 1}^{𝑛} {| ⟨ 𝑥, 𝑒_{𝑘} ⟩ |}^{2}} . □ \end{aligned}

Therefore, one then derives

Corollary 10.2.3.1 (Bessel’s Inequality).

Let $𝐻$ be a Hilbert space and let ${𝑒_{𝑖}}_{𝑖 \in 𝐼}$ be an orthonormal system. Then for $𝑥 \in 𝐻$ ,

\sum_{𝑖 \in 𝐼} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} \leq {‖ 𝑥 ‖}^{2},

where the left-hand side converges unconditionally: $\exists 𝑆 \leq {‖ 𝑥 ‖}^{2}$ such that for any $𝜀 > 0$ , $\exists 𝐼^{'} \subseteq 𝐼$ countable such that for all $(𝐼 \supseteq) 𝐼^{″} \supseteq 𝐼^{'}$ countable,

| \sum_{𝑖 \in 𝐼^{″}} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} - 𝑆 | < 𝜀 .

Moreover,

𝑥

is orthogonal to all but countably many vectors in

{𝑒_{𝑖}}

Proof.

For each $𝑚 \in ℕ$ , let

𝐴_{𝑚} = {𝑒 \in {𝑒_{𝑖}}_{𝑖} : {| ⟨ 𝑥, 𝑒 ⟩ |}^{2} > \frac{1}{𝑚}} .

Then $𝐴_{𝑚}$ is a finite set, as otherwise

\sum_{𝑒 \in 𝐴_{𝑚}} {| ⟨ 𝑥, 𝑒 ⟩ |}^{2} = \infty,

contradicting Theorem 10.2.3 $Theorem 10.2.3 Theorem 10.2.3$ . Let $𝐴 = ⋃_{𝑚 \in ℕ} 𝐴_{𝑚}$ , which is a countable set. Moreover, for any ${𝑒_{𝑖}} ∋ 𝑒 \notin 𝐴$ , ${| ⟨ 𝑥, 𝑒 ⟩ |}^{2} \leq {inf}_{𝑚 \in ℕ} \frac{1}{𝑚} = 0$ .

Therefore only a countable number of vectors in ${𝑒_{𝑖}}_{𝑖 \in 𝐼}$ are not orthogonal to $𝑥$ . Let $𝐴$ be indexed by a sequence ${𝑣_{𝑛}}_{𝑛 \in ℕ} = {𝑒_{𝑖_{𝑛}}}_{𝑛 \in ℕ}$ . By Theorem 10.2.3 $Theorem 10.2.3 Theorem 10.2.3$ ,

\sum_{𝑛 = 1}^{\infty} {| ⟨ 𝑥, 𝑣_{𝑛} ⟩ |}^{2}

has all partial sums bounded by ${‖ 𝑥 ‖}^{2}$ ; therefore the series converges to a limit $𝑆 \leq {‖ 𝑥 ‖}^{2}$ by the Monotone Convergence Theorem.

Since $\sum_{𝑛 = 1}^{\infty} {| ⟨ 𝑥, 𝑣_{𝑛} ⟩ |}^{2} = 𝑆$ , for any $𝜀 > 0$ , $\exists 𝑁 \in ℕ$ such that for all $𝑘 \geq 𝑁$ , $| \sum_{𝑛 = 1}^{𝑘} {| ⟨ 𝑥, 𝑣_{𝑛} ⟩ |}^{2} - 𝑆 | < 𝜀$ . Then let $𝐼^{'} = {𝑖_{𝑛}}_{𝑛 = 1}^{𝑁}$ . Then for any $𝐼^{″} \supseteq 𝐼^{'}$ countable,

\sum_{𝑖 \in 𝐼^{'}} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} \leq \sum_{𝑖 \in 𝐼^{″}} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} \leq 𝑆,

which implies

𝜀 > | 𝑆 - \sum_{𝑖 \in 𝐼^{'}}^{𝑘} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} | \geq | 𝑆 - \sum_{𝑖 \in 𝐼^{″}} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} | \geq 0

and the assertion follows. ⁠ $□$

Then we may provide the following generalization of the Parseval’s Theorem from Fourier analysis:

Theorem 10.2.4 (Riesz–Fischer).

Let $𝐻$ be a Hilbert space and let ${𝑒_{𝑖}}_{𝑖 \in 𝐼}$ be a complete orthonormal system. Then:

For any $𝑥 \in 𝐻$ ,
${‖ 𝑥 ‖}^{2} = \sum_{𝑖 \in 𝐼} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2},$
and
$𝑥 = \sum_{𝑖 \in 𝐼} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖} .$
Conversely, for any complex collection ${𝛽_{𝑖}}_{𝑖 \in 𝐼}$ such that
$\sum_{𝑖 \in 𝐼} {| 𝛽_{𝑖} |}^{2} < \infty,$
then $\exists! 𝑥 \in 𝐻$ such that $\forall 𝑖 \in 𝐼$ , $⟨ 𝑥, 𝑒_{𝑖} ⟩ = 𝛽_{𝑖}$ . Then
${‖ 𝑥 ‖}^{2} = \sum_{𝑖 \in 𝐼} {| 𝛽_{𝑖} |}^{2}, 𝑥 = \sum_{𝑖 \in 𝐼} 𝛽_{𝑖} 𝑒_{𝑖}$

It is actually simpler to first prove Part 2 $Part 2 Part 2$ .

Proof of Part 2 $Part 2 Part 2$ of Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ .

Claim.

At most a countable number of

{𝛽_{𝑖}}_{𝑖 \in 𝐼}

are nonzero.

Proof of the claim.

This is essentially the same argument as used before with Corollary 10.2.3.1 $Corollary 10.2.3.1 Corollary 10.2.3.1$ . For each $𝑚 \in ℕ$ , let

𝐴_{𝑚} = {𝛽 \in {𝛽_{𝑖}}_{𝑖 \in 𝐼} : | 𝛽_{𝑖} | > \frac{1}{𝑚}} .

Then

𝐴_{𝑚}

is finite; otherwise

\sum_{𝑖 \in 𝐼} {| 𝛽_{𝑖} |}^{2} ≮ \infty

. Moreover the union of all

𝐴_{𝑚}

for

𝑚 \in ℕ

is a countable set and contains all nonzero

𝛽_{\cdot}

‘s. ⁠

□

If $⋃_{𝑚 \in ℕ} 𝐴_{𝑚}$ is finite ${𝛽_{𝑖_{𝑗}}}_{𝑗}$ , then letting

𝑥 = \sum_{𝑗} 𝛽_{𝑖_{𝑗}} 𝑒_{𝑖_{𝑗}},

we see that $⟨ 𝑥, 𝑒_{𝑖_{𝑗}} ⟩ = 𝛽_{𝑖_{𝑗}}$ and for all other $𝛽_{𝑖}$ ’s, $𝛽_{𝑖} = 0$ , which proves $⟨ 𝑥, 𝑒_{𝑖} ⟩ = 𝛽_{𝑖}$ from orthogonality. Then in this case, the assertion ${‖ 𝑥 ‖}^{2} = \sum_{𝑖 \in 𝐼} {| 𝛽_{𝑖} |}^{2}$ follows from Theorem 10.1.22 $Theorem 10.1.22 Theorem 10.1.22$ .

If $⋃_{𝑚} 𝐴_{𝑚}$ is countable and indexed by ${𝛽_{𝑖_{𝑗}}}_{𝑗 \in ℕ}$ , then for any $𝐾 \in ℕ$ , let

𝑥_{𝐾} = \sum_{𝑗 = 1}^{𝐾} 𝛽_{𝑖_{𝑗}} 𝑒_{𝑖_{𝑗}} .

Then for $𝑁 \geq 𝑀$ ,

{‖ 𝑥_{𝑁} - 𝑥_{𝑀} ‖}^{2} = {‖ \sum_{𝑀 + 1}^{𝑁} 𝛽_{𝑖_{𝑗}} 𝑒_{𝑖_{𝑗}} ‖}^{2} = \sum_{𝑀 + 1}^{𝑁} {| 𝛽_{𝑖_{𝑗}} |}^{2} \to 0 . (by Theorem 10.1.22 Theorem 10.1.22 Theorem 10.1.22)

Then ${𝑥_{𝐾}}_{𝐾}$ is a Cauchy sequence and hence converges to some $𝑥 \in 𝐻$ :

𝑥 = \sum_{𝑗 = 1}^{\infty} 𝛽_{𝑖_{𝑗}} 𝑒_{𝑖_{𝑗}} .

Since $| ‖ 𝑥 ‖ - ‖ 𝑥_{𝑁} ‖ | \leq ‖ 𝑥 - 𝑥_{𝑁} ‖ \to 0$ , ${‖ 𝑥_{𝑁} ‖}^{2} \to {‖ 𝑥 ‖}^{2}$ . Moreover, by Theorem 10.1.22 $Theorem 10.1.22 Theorem 10.1.22$ ,

{‖ 𝑥_{𝑁} ‖}^{2} = \sum_{𝑗 = 1}^{𝑁} {| 𝛽_{𝑖_{𝑗}} |}^{2} ⟹ {‖ 𝑥 ‖}^{2} = \sum_{𝑗 = 1}^{\infty} {| 𝛽_{𝑖_{𝑗}} |}^{2}

by letting $𝑁 \to \infty$ .

If there exists some other $\tilde{𝑥}$ satisfying $⟨ \tilde{𝑥}, 𝑒_{𝑖} ⟩ = 𝛽_{𝑖}$ , then $⟨ 𝑥 - \tilde{𝑥}, 𝑒_{𝑖} ⟩ = 0$ for each $𝑖 \in 𝐼$ . By the completeness of ${𝑒_{𝑖}}_{𝑖 \in 𝐼}$ , the only vector orthogonal to all vectors in the system is the zero vector, therefore $𝑥 - \tilde{𝑥} = 0 ⟹ 𝑥 = \tilde{𝑥}$ . ⁠ $□$

Proof of Part 1 $Part 1 Part 1$ of Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ .

By Bessel’s Inequality, we have

\sum_{𝑖 \in 𝐼} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} \leq {‖ 𝑥 ‖}^{2} .

By Part 2 $Part 2 Part 2$ applied to $𝛽_{𝑖} = ⟨ 𝑥, 𝑒_{𝑖} ⟩$ , $\exists! \tilde{𝑥}$ such that $\forall 𝑖 \in 𝐼$ ,

⟨ \tilde{𝑥}, 𝑒_{𝑖} ⟩ = ⟨ 𝑥, 𝑒_{𝑖} ⟩, (10.2 . 1)

given by $\tilde{𝑥} = \sum_{𝑖 = 𝐼} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖}$ . Then for this $\tilde{𝑥}$ ,

{‖ \tilde{𝑥} ‖}^{2} = \sum_{𝑖 \in 𝐼} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} . (10.2 . 2)

(10.2.1) $(10.2.1) (10.2.1)$ gives that for each $𝑖 \in 𝐼$ , $⟨ 𝑥 - \tilde{𝑥}, 𝑒_{𝑖} ⟩ = 0$ . Then $𝑥 - \tilde{𝑥}$ is orthogonal to all vectors in a complete orthonormal system, and is therefore zero by definition. Thus $𝑥 = \tilde{𝑥}$ , and by (10.2.2) $(10.2.2) (10.2.2)$ , we have

{‖ 𝑥 ‖}^{2} = \sum_{𝑖 \in 𝐼} {| ⟨ 𝑥, 𝑒_{𝑖} ⟩ |}^{2} . □

Corollary 10.2.4.1.

Let $𝐻$ be a Hilbert space such that there exists a complete orthonormal system ${𝑒_{𝑖}}_{𝑖 \in 𝐼}$ . Then ${span {𝑒_{𝑖}}}_{𝑖 \in 𝐼}$ is dense in $𝐻$ , or in other words:

\overline{{span {𝑒_{𝑖}}}_{𝑖}} = 𝐻 .

Proof.

By Riesz–Fischer (Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ ), any $𝑥 \in 𝐻$ ,

𝑥 = \sum_{𝑖} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖},

implying that for any $\forall 𝜀 > 0$ , $\exists 𝐼^{'} \subseteq 𝐼$ countable such that $\forall 𝐼^{″} \supseteq 𝐼^{'}$ countable,

‖ \sum_{𝑖 \in 𝐼^{″}} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖} - 𝑥 ‖ < \frac{𝜀}{2} .

Moreover, $\exists 𝐼^{‴} \subseteq 𝐼^{'}$ finite such that

‖ \sum_{𝑖 \in 𝐼^{‴}} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖} - \sum_{𝑖 \in 𝐼^{″}} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖} ‖ < \frac{𝜀}{2},

implying that

‖ \sum_{𝑖 \in 𝐼^{‴}} ⟨ 𝑥, 𝑒_{𝑖} ⟩ 𝑒_{𝑖} - 𝑥 ‖ < 𝜀 .

Therefore, there exists a (finite) linear combination of orthonormal vectors in

{𝑒_{𝑖}}

lying in any open ball centered at

𝑥

(by the arbitrariness of

𝜀

). ⁠

□

Remark.

The necessity of using closure or density comes from the observation that

span

appertains only to finite linear combinations, whereas in most cases we need a complete description of “countably infinite linear combinations.”

Example 10.2.5.

Let $ℓ^{2} = {{𝛼_{𝑗}}_{𝑗 \in ℕ} 𝐿 : \sum_{𝑗 = 1}^{\infty} {| 𝛼_{𝑗} |}^{2} < \infty}$ denote the set of all square-summable complex sequences. Then for $𝛼 = {𝛼_{𝑗}}_{𝑗}$ , $𝛽 = {𝛽_{𝑗}}_{𝑗}$ , define

⟨ 𝛼, 𝛽 ⟩ = \sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} \overline{𝛽_{𝑗}} .

Then

ℓ^{2}

is a vector space and a Hilbert space with the inner product

{⟨ 𝛼, 𝛽 ⟩}_{ℓ^{2}} = \sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} \overline{𝛽_{𝑗}}

and associated norm

{‖ \cdot ‖}_{ℓ^{2}}

We will not immediately state a full proof. In proving completeness, one could obviously write out the Cauchy sequences and prove that they converge. However, there is indeed an elegant method that uses the Riesz–Fischer Theorem (Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ ), under the assumption that there exists any Hilbert space $𝐻$ with an infinite countable complete orthonormal system, which we will first assume and justify later (in fact, there are many such spaces, such as the Bergman space $𝐴^{2} (𝔻)$ and the Hardy space $𝐻^{2}$ ).

Proof of Example 10.2.5 $Example 10.2.5 Example 10.2.5$ (with assumptions).

The space $ℓ^{2}$ is trivially closed under scalar multiplication. To show additivity, or that

{𝛼_{𝑗}}_{𝑗} + {𝛽_{𝑗}}_{𝑗} = {𝛼_{𝑗} + 𝛽_{𝑗}}_{𝑗} \in ℓ^{2}

holds, notice that

\sum_{𝑗 = 1}^{\infty} {| 𝛼_{𝑗} + 𝛽_{𝑗} |}^{2} \leq 2 \sum_{𝑗 = 1}^{\infty} [{| 𝛼_{𝑗} |}^{2} + {| 𝛽_{𝑗} |}^{2}] (by Theorem 10.1.18 Theorem 10.1.18 Theorem 10.1.18)

Observe that the product is well defined by the Cauchy–Schwarz Inequality:

\begin{aligned} \underset{𝑁 \to \infty}{lim sup} | \sum_{𝑗 = 1}^{𝑁} 𝛼_{𝑗} \overline{𝛽_{𝑗}} | & \leq \underset{𝑁 \to \infty}{lim sup} \sum_{𝑗 = 1}^{𝑁} | 𝛼_{𝑗} \overline{𝛽_{𝑗}} | \\ \leq \underset{𝑁 \to \infty}{lim sup} \sqrt{\sum_{𝑗 = 1}^{𝑁} {| 𝛼_{𝑗} |}^{2}} \sqrt{\sum_{𝑗 = 1}^{𝑁} {| 𝛽_{𝑗} |}^{2}} = ‖ 𝛼 ‖ ‖ 𝛽 ‖, \end{aligned}

and convergence follows from the Monotone Convergence Theorem. The verification of the other properties is rather simple. Observe that the associated norm is

{‖ 𝛼 ‖}_{ℓ^{2}} = \sqrt{\sum_{𝑗 = 1}^{\infty} {| 𝛼_{𝑗} |}^{2}} .

Let (using the assumption) $𝐻$ be a Hilbert space (with ${⟨ \cdot, \cdot ⟩}_{𝐻}$ , ${‖ \cdot ‖}_{𝐻}$ ) with an infinite countable complete orthonormal system ${𝑒_{𝑗}}_{𝑗 \in ℕ}$ . For any $𝛼 = {𝛼_{𝑗}}_{𝑗 \in ℕ} \in ℓ^{2}$ , by the Riesz–Fischer Theorem (Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ ), there is a single $𝑥 \in 𝐻$ such that

𝑥 = \sum_{𝑗} 𝛼_{𝑗} 𝑒_{𝑗}, and {⟨ 𝑥, 𝑒_{𝑗} ⟩}_{𝐻} = 𝛼_{𝑗} \forall 𝑗 \in ℕ .

Then the map

𝜑 : ℓ^{2} \to 𝐻 given by 𝜑 : 𝛼 \mapsto \sum_{𝑗} 𝛼_{𝑗} 𝑒_{𝑗}

is a bijection: the inverse is given by

𝜑^{- 1} : 𝑥 \mapsto {{⟨ 𝑥, 𝑒_{𝑗} ⟩}_{𝐻}}_{𝑗 \in ℕ} .

The vector structure of $𝐻$ is preserved under $𝜑$ . Moreover, $𝜑$ is trivially additive and homogeneous, thus linear.

Therefore, it remains to show that the space $ℓ^{2}$ is complete under the associated norm. Observe that $𝜑$ preserves norms, since

{‖ 𝑥 ‖}^{2} = \sum_{𝑗 = 1}^{\infty} {| {⟨ 𝑥, 𝑒_{𝑗} ⟩}_{𝐻} |}^{2} = {‖ {{⟨ 𝑥, 𝑒_{𝑗} ⟩}_{𝐻}}_{𝑗 \in ℕ} ‖}^{2}

by Riesz–Fischer (Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ ).

Therefore, since any Cauchy sequence ${𝛼_{𝑛}}_{𝑛}$ in $ℓ^{2}$ gives a Cauchy sequence ${𝜑 (𝛼_{𝑛})}_{𝑛} \subseteq 𝐻$ with a single accumulation point $𝑥 \in 𝐻$ ; from ${‖ 𝜑 (𝛼_{𝑛}) - 𝑥 ‖}_{𝐻} < 𝜀$ ( $𝑛 > 𝑁$ for some $𝑁$ ), we obtain $‖ 𝜑^{- 1} (𝜑 (𝛼_{𝑛}) - 𝑥) ‖ = ‖ 𝛼_{𝑛} - 𝜑^{- 1} (𝑥) ‖ < 𝜀$ . Therefore, ${𝛼_{𝑛}}_{𝑛}$ converges to $𝜑^{- 1} (𝑥)$ , and thus $ℓ^{2}$ is complete. ⁠ $□$

Example 10.2.6.

Let $𝐴^{2} (𝔻)$ be the Bergman space on $𝔻$ . With

⟨ 𝑓, 𝑔 ⟩ = \iint_{𝔻} 𝑓 (𝑧) \overline{𝑔 (𝑧)} d 𝑥 d 𝑦,

𝐴^{2} (𝔻)

is a Hilbert space by Example 10.1.15

Example 10.1.15 Example 10.1.15

. Then show that

{𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗}}_{𝑗 \in 0}^{\infty}

is a countable complete orthonormal system for

𝐴^{2} (𝔻)

Solution to Example 10.2.6 $Example 10.2.6 Example 10.2.6$ .

For any $𝑗$ , $𝑧 \mapsto 𝑧^{𝑗} \in 𝐴^{2} (𝔻)$ , so

\begin{aligned} \infty > ‖ 𝑧 \mapsto 𝑧^{𝑗} ‖ & = \sqrt{\iint_{𝔻} {| 𝑧 |}^{2 𝑗} d 𝑥 d 𝑦} = \sqrt{\int_{0}^{1} \int_{- π}^{π} 𝑟^{2 𝑗} 𝑟 d 𝜃 d 𝑟} \\ = \sqrt{2 π {\frac{𝑟^{2 𝑗 + 2}}{2 𝑗 + 2} |}_{0}^{1}} = \sqrt{\frac{π}{𝑗 + 1}} . \end{aligned}

Moreover, for $𝑗 \neq 𝑘$ ,

\begin{aligned} ⟨ 𝑧 \mapsto 𝑧^{𝑗}, 𝑧 \mapsto 𝑧^{𝑘} ⟩ & = \iint_{𝔻} 𝑧^{𝑗} {\overline{𝑧}}^{𝑘} d 𝑥 d 𝑦 = \int_{0}^{1} \int_{- π}^{π} 𝑟^{𝑗} e^{i 𝑗 𝜃} 𝑟^{𝑘} e^{- i 𝑘 𝜃} 𝑟 d 𝜃 d 𝑟 \\ = \int_{0}^{1} 𝑟^{𝑗 + 𝑘 + 1} d 𝑟 \int_{- π}^{π} e^{i 𝜃 (𝑗 - 𝑘)} d 𝜃 = 0 . \end{aligned}

Therefore, the set ${𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗}}_{𝑗 \in 0}^{\infty}$ satisfies the orthogonality condition, and

⟨ 𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗}, 𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗} ⟩ = 1 \forall 𝑗 \in ℤ_{\geq 0} .

Therefore it is an orthonormal system. Moreover, completeness follows naturally, because if there exists $𝑓 \in 𝐴^{2} (𝔻)$ such that $𝑓 ⟂ 𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗}$ for each $𝑗$ , then writing $𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}$ , we get

\begin{aligned} ⟨ 𝑓, 𝑧 \mapsto \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗} ⟩ & = 0 = \iint_{𝔻} (\sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}) \overline{(\frac{\sqrt{𝑗 + 1}}{\sqrt{π}} 𝑧^{𝑗})} d 𝑥 d 𝑦 \\ = \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} \int_{0}^{1} \int_{- π}^{π} \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑟^{𝑛 + 𝑗 + 1} e^{i (𝑛 - 𝑗) 𝜃} d 𝜃 d 𝑟 \\ (uniform convergence in 𝜃) \\ = \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} \int_{0}^{1} \sum_{𝑛 = 0}^{\infty} \int_{- π}^{π} 𝑎_{𝑛} 𝑟^{𝑛 + 𝑗 + 1} e^{i (𝑛 - 𝑗) 𝜃} d 𝜃 d 𝑟 \\ = \frac{\sqrt{𝑗 + 1}}{\sqrt{π}} \int_{0}^{1} \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑟^{𝑛 + 𝑗 + 1} {\begin{cases} {\frac{e^{i (𝑛 - 𝑗) 𝜃}}{i (𝑛 - 𝑗)} |}_{- π}^{π} & if 𝑛 \neq 𝑗 \\ 2 π & if 𝑛 = 𝑗 \end{cases} d 𝑟 \\ = 2 \sqrt{π (𝑗 + 1)} \int_{0}^{1} 𝑎_{𝑗} 𝑟^{2 𝑗 + 1} d 𝑟 = 𝑎_{𝑗} \sqrt{\frac{π}{𝑗 + 1}}, \end{aligned}

meaning that for each

𝑗 \in ℤ_{\geq 0}

𝑎_{𝑗} = 0

; then the only function orthogonal to every function in the system is the zero function. ⁠

□

(Then one could use this space in Example 10.2.5 $Example 10.2.5 Example 10.2.5$ .)

Definition 10.2.7 (Separability).

A metric space

(𝑋, 𝑑)

is said to be separable iff it has a countable dense subset.

Proposition 10.2.8.

A Hilbert space

𝐻

is separable iff it has a complete countable orthonormal system.

Proof.

(Forward implication)
Let $𝑆 \subseteq 𝐻$ be a countable dense subset. Then describing $𝑆$ with a sequence ${𝑠_{𝑗}}$ , we may then construct another sequence ${𝑢_{𝑗}}_{𝑗}$ recursively:
- Let $𝑢_{1}$ be the first nonzero element $𝑠_{𝑘_{1}}$ in ${𝑠_{𝑗}}$ .
- Assuming a linearly independent set ${𝑢_{𝑗}}_{𝑗 < 𝑘}$ , choose $𝑢_{𝑘}$ to be the first nonzero element in ${𝑠_{𝑗}}$ not in ${span {𝑢_{𝑗}}}_{𝑗 < 𝑘}$ .
- If such a $𝑢_{𝑘}$ ever does not exist, then end the process thereat. In this case ${span {𝑢_{𝑗}}}_{𝑗 < 𝑘}$ already densely spans $𝐻$ , is complete by Proposition 10.1.12 $Proposition 10.1.12 Proposition 10.1.12$ , and consequently hence spans $𝐻$ .
Then ${𝑢_{𝑗}}_{𝑗}$ densely spans $𝐻$ . Applying the Gram–Schmidt process then gives an orthonormal system which also densely spans $𝐻$ , namely ${𝑒_{𝑗}}_{𝑗}$ .
Suppose, for contradiction, that there exists a nonzero $𝑥 \in 𝐻$ such that $𝑥 ⟂ 𝑒_{𝑗}$ for every $𝑗$ . Then $𝑥 ⟂ {span {𝑒_{𝑗}}}_{𝑗}$ .
Since ${span {𝑒_{𝑗}}}_{𝑗}$ is dense in $𝐻$ , there exists a sequence ${𝑦_{𝑛}} \subseteq {span {𝑒_{𝑗}}}_{𝑗}$ such that $𝑦_{𝑛} \to 𝑥$ . By continuity of the inner product (Theorem 10.1.11 $Theorem 10.1.11 Theorem 10.1.11$ ),
${‖ 𝑥 ‖}^{2} = ⟨ 𝑥, 𝑥 ⟩ = lim_{𝑛 \to \infty} ⟨ 𝑥, 𝑦_{𝑛} ⟩ = 0,$
because $⟨ 𝑥, 𝑦_{𝑛} ⟩ = 0$ for every $𝑛$ . Hence $𝑥 = 0$ , contradicting our choice of $𝑥$ . Therefore ${𝑒_{𝑗}}_{𝑗}$ is complete.
(Converse)
Letting ${𝑒_{𝑗}}_{𝑗 \in ℕ_{< 𝐾}}$ ( $𝐾 \in ℕ \cup {\infty}$ ) be the complete orthonormal system, the set
$𝑆 = {\sum_{𝑗 = 1}^{𝑘} 𝛼_{𝑗} 𝑒_{𝑖_{𝑗}} : 𝑘 \in ℕ_{< 𝐾}, (\forall 𝑗 \in ℕ_{\leq 𝑘}) 𝑖_{𝑗} \in ℕ_{< 𝐾}, 𝛼_{𝑗} \in ℚ + i ℚ},$
is a countable set. Indeed, for each fixed $𝑘$ , the cardinality of
${\sum_{𝑗 = 1}^{𝑘} 𝛼_{𝑗} 𝑒_{𝑖_{𝑗}} : \forall 𝑗 \in ℕ_{\leq 𝑘}, 𝑖_{𝑗} \in ℕ_{< 𝐾}, 𝛼_{𝑗} \in ℚ + i ℚ}$
does not exceed that of $ℕ_{< 𝐾}^{𝑘} \times {(ℚ + i ℚ)}^{𝑘}$ , which is countable. Then the union of all such sets for $𝑘 \in ℕ_{< 𝐾}$ is a countable union of countable sets, hence countable.
Moreover, we can show that it is dense in $𝐻$ : $\forall 𝑥 \in 𝐻$ , we know from the proof Riesz–Fischer (Theorem 10.2.4 $Theorem 10.2.4 Theorem 10.2.4$ ) that it can be written as
$𝑥 = \sum_{𝑗 \in ℕ_{< 𝐾}} 𝛽_{𝑗} 𝑒_{𝑗}, (𝛽_{𝑗} = ⟨ 𝑥, 𝑒_{𝑗} ⟩),$
which may be approximated with an error under $\frac{𝜀}{2}$ by $\sum_{𝑗 = 1}^{𝑘} 𝛽_{𝑗} 𝑒_{𝑗}$ . By the density of rationals, each $𝛽_{𝑗}$ may be approximated by some $𝛼_{𝑗} \in ℚ + i ℚ$ , such that $| 𝛼_{𝑗} - 𝛽_{𝑗} | < \frac{𝜀}{2 𝑘} .$ Then
$‖ 𝑥 - \sum_{𝑗 = 1}^{𝑘} 𝛼_{𝑗} 𝑒_{𝑗} ‖ \leq ‖ 𝑥 - \sum_{𝑗 = 1}^{𝑘} 𝛽_{𝑗} 𝑒_{𝑗} ‖ + ‖ \sum_{𝑗 = 1}^{𝑘} (𝛽_{𝑗} - 𝛼_{𝑗}) 𝑒_{𝑗} ‖ < 𝜀 .$
Since $\sum_{𝑗 = 1}^{𝑘} 𝛼_{𝑗} 𝑒_{𝑗}$ lies in $𝑆$ , the density follows from the arbitrariness of $𝜀$ . $□$

We then conclude with the following technical result.

Theorem 10.2.9.

Let

𝐻

be a separable Hilbert space and let

𝐾 \subseteq 𝐻

be a Hilbert subspace. Then

𝐾

is separable.

Proof.

By Proposition 10.2.8 $Proposition 10.2.8 Proposition 10.2.8$ , $𝐻$ has a complete countable orthonormal basis ${𝑒_{𝑗}}_{𝑗}$ . Projecting each $𝑒_{𝑗}$ onto $𝐾$ by using the orthogonal decomposition (Theorem 10.1.28 $Theorem 10.1.28 Theorem 10.1.28$ ) gives a set ${𝑢_{𝑗}}$ in $𝐾$ .

In fact, the span of ${𝑢_{𝑗}}$ is dense in $𝐾$ : for any $𝜀 > 0, 𝑥 \in 𝐾$ , there exists $𝑦 \in {span {𝑒_{𝑗}}}_{𝑗}$ such that $‖ 𝑦 - 𝑥 ‖ < 𝜀$ . Projecting $𝑦 - 𝑥$ onto $𝐾$ gives a vector $𝑣$ satisfying $‖ 𝑣 ‖ \leq ‖ 𝑦 - 𝑥 ‖$ (by Theorem 10.1.22 $Theorem 10.1.22 Theorem 10.1.22$ , the Pythagorean theorem); and moreover, $𝑣 = 𝑦_{𝑝} - 𝑥$ (where $𝑦_{𝑝}$ is the projection of $𝑦$ onto $𝐾$ ). Moreover, $𝑦_{𝑝}$ is a linear combination of elements in ${𝑢_{𝑗}}$ , and is within $𝜀$ from $𝑥$ . Then the density follows from arbitrariness.

Then $𝐾$ is separable. ⁠ $□$