§10.1 Geometry of a Hilbert Space

10.1 Geometry of a Hilbert Space

The concept of a vector space is an important one in linear algebra. However, its definition itself does not actually imply that the “vectors” are tuples of values of finite dimension. As a matter of fact the term “vector” is a very general one. Developments in the 20th century have rigorously defined vectors with infinite dimensions.

Definition 10.1.1 (Inner Product).

Let $𝑉$ be a vector space over $ℂ$ . Then a function

⟨ \cdot, \cdot ⟩ : 𝑉^{2} \to ℂ

is said to be a positive-definite nondegenerate Hermitian inner product iff

For any $𝑣_{1}, 𝑣_{2}, 𝑤 \in 𝑉$ , $⟨ 𝑣_{1} + 𝑣_{2}, 𝑤 ⟩ = ⟨ 𝑣_{1}, 𝑤 ⟩ + ⟨ 𝑤_{2}, 𝑤 ⟩$ (additivity in the first argument).
For any $𝛼 \in ℂ$ , $𝑣, 𝑤 \in 𝑉$ , $⟨ 𝛼 𝑣, 𝑤 ⟩ = 𝛼 ⟨ 𝑣, 𝑤 ⟩$ (homogeneity in the first argument).
For any $𝑣, 𝑤 \in 𝑉$ , $⟨ 𝑣, 𝑤 ⟩ = \overline{⟨ 𝑤, 𝑣 ⟩}$ (conjugate-symmetry/Hermetian).
For any $𝑣 \in 𝑉$ , $⟨ 𝑣, 𝑣 ⟩ \geq 0$ . (positive semi-definiteness)
For $𝑣 \in 𝑉$ , $⟨ 𝑣, 𝑣 ⟩ = 0$ iff $𝑣 = 0$ (nondegeneracy).

Then the space $𝑉$ together with the inner product is known as an inner product space.

Note that Part 5 $Part 5 Part 5$ and Part 4 $Part 4 Part 4$ , imply positive-definitness: for $𝑣 \neq 0$ , $⟨ 𝑣, 𝑣 ⟩ > 0$ .

To avoid pedantry, for the proceeding passages, the use of “inner product” will assume positive-definiteness, nondegeneracy, and conjugate-symmetry.

Proposition 10.1.2.

Let $𝑉$ be a vector space over $ℂ$ and let $⟨ \cdot, \cdot ⟩$ be a positive-definite nondegenerate Hermetian inner product. Then for $𝑣, 𝑤, 𝑤_{1}, 𝑤_{2} \in 𝑉$ and $𝛼 \in ℂ$ ,

$⟨ 𝑣, 𝑤_{1} + 𝑤_{2} ⟩ = ⟨ 𝑣, 𝑤_{1} ⟩ + ⟨ 𝑣, 𝑤_{2} ⟩$ (additivity in the second argument).
$⟨ 𝑣, 𝛼 𝑤 ⟩ = \overline{𝛼} ⟨ 𝑣, 𝑤 ⟩$ (conjugate-homogeneity in the second argument).
$⟨ 𝑣, 0 ⟩ = ⟨ 0, 𝑣 ⟩ = 0$ .
$⟨ 𝑣 + 𝑤, 𝑣 + 𝑤 ⟩ = ⟨ 𝑣, 𝑣 ⟩ + ⟨ 𝑤, 𝑤 ⟩ + 2 ℜ𝔢 ⟨ 𝑣, 𝑤 ⟩$ .

Proof.

Observe that Part 1 $Part 1 Part 1$ holds since

\begin{aligned} ⟨ 𝑣, 𝑤_{1} + 𝑤_{2} ⟩ & = \overline{⟨ 𝑤_{1} + 𝑤_{2}, 𝑣 ⟩} \\ = \overline{⟨ 𝑤_{1}, 𝑣 ⟩} + \overline{⟨ 𝑤_{2}, 𝑣 ⟩} = ⟨ 𝑣, 𝑤_{1} ⟩ + ⟨ 𝑣, 𝑤_{2} ⟩ \end{aligned}

and conjugate-homogeneity (Part 2 $Part 2 Part 2$ ) holds in the second argument since

⟨ 𝑣, 𝛼 𝑤 ⟩ = \overline{⟨ 𝛼 𝑤, 𝑣 ⟩} = \overline{𝛼 ⟨ 𝑤, 𝑣 ⟩} = \overline{𝛼} ⟨ 𝑣, 𝑤 ⟩ .

Letting

𝛼 = 0

⟨ 𝛼 𝑣, 𝑣 ⟩

gives Part 3

Part 3 Part 3

, while Part 4

Part 4 Part 4

is a consequence of additivity. ⁠

□

We provide examples of vector spaces and their corresponding nondegenerate Hermetian inner products:

Example 10.1.3.

Let $𝑉 = 𝐶^{𝑛}$ ( $𝑛 \in ℕ$ ) be a vector space under the usual operations, and define

⟨ 𝐯, 𝐰 ⟩ = \sum_{𝑘 = 1}^{𝑛} 𝑣_{𝑘} \overline{𝑤_{𝑘}},

where

𝐯 = (𝑣_{1}, \dots, 𝑣_{𝑛})

and

𝐰 = (𝑤_{1}, \dots, 𝑤_{𝑛})

are vectors in

𝑉

. Then

⟨ \cdot, \cdot ⟩

is a nondegenerate Hermetian inner product.

The verification of the necessary properties is trivial.

Example 10.1.4.

Let $𝑉 = 𝐶 ([0, 1], ℂ)$ , with addition and multiplication defined as in the usual sense. Then if $𝑓, 𝑔 \in 𝑉$ , define

⟨ 𝑓, 𝑔 ⟩ = \int_{0}^{1} 𝑓 (𝑥) \overline{𝑔 (𝑥)} d 𝑥 .

Proof.

The verification of properties Part 1 $Part 1 Part 1$ , Part 2 $Part 2 Part 2$ , Part 3 $Part 3 Part 3$ , Part 4 $Part 4 Part 4$ (in Definition 10.1.1 $Definition 10.1.1 Definition 10.1.1$ ) are somewhat trivial. The reverse implication of Part 4 $Part 4 Part 4$ is obvious. For $𝑓 \in 𝑉$ such that

⟨ 𝑓, 𝑓 ⟩ = 0 ⟹ \int_{0}^{1} {| 𝑓 (𝑥) |}^{2} d 𝑥 = 0,

if ${| 𝑓 (𝑥^{'}) |}^{2} > 0$ at some $𝑥^{'} \in (0, 1)$ , then for $𝑥$ in some $𝛿$ -neighborhood of $𝑥^{'}$ , ${| 𝑓 (𝑥) |}^{2} > \frac{1}{2} {| 𝑓 (𝑥^{'}) |}^{2}$ by continuity, implying that

\int_{0}^{1} {| 𝑓 (𝑥) |}^{2} d 𝑥 \geq \int_{𝑥^{'} - 𝛿}^{𝑥^{'} + 𝛿} {| 𝑓 (𝑥) |}^{2} d 𝑥 > 𝛿 {| 𝑓 (𝑥^{'}) |}^{2} > 0,

which is impossible; thus

𝑓 \equiv 0

forcibly. ⁠

□

Definition 10.1.5 (Associated Norm).

If $𝑉$ is a vector space and $⟨ \cdot, \cdot ⟩$ is a nondegenerate positive-definite Hermitian inner product on $𝑉$ , then the associated norm of some $𝑣 \in 𝑉$ is given by

‖ 𝑣 ‖ = \sqrt{⟨ 𝑣, 𝑣 ⟩} .

Definition 10.1.6 (Isometry).

Let $𝑉$ and $𝑊$ be inner product spaces over the same scalar field, with inner products ${⟨ \cdot, \cdot ⟩}_{𝑉}$ and ${⟨ \cdot, \cdot ⟩}_{𝑊}$ , respectively. A linear map $𝑇 : 𝑉 \to 𝑊$ is an isometry iff

{⟨ 𝑇 (𝑣), 𝑇 (𝑤) ⟩}_{𝑊} = {⟨ 𝑣, 𝑤 ⟩}_{𝑉} for all 𝑣, 𝑤 \in 𝑉 .

In particular,

𝑇

preserves norms and distances.

Definition 10.1.7 (Isomorphism).

Let

𝑉

and

𝑊

be vector spaces over the same scalar field. A linear bijection

𝑇 : 𝑉 \to 𝑊

is an isomorphism.

Example 10.1.8.

Show that every $𝑛$ -dimensional (by linear independence) complex inner product space is isometrically isomorphic to $ℂ^{𝑛}$ equipped with the Euclidean (Hermitian) inner product

{⟨ 𝐳, 𝐰 ⟩}_{ℂ^{𝑛}} = \sum_{𝑘 = 1}^{𝑛} 𝑧_{𝑘} \overline{𝑤_{𝑘}} .

Solution to Example 10.1.8 $Example 10.1.8 Example 10.1.8$ .

Let $𝑉$ be the $𝑛$ -dimensional complex inner product space. By Corollary 10.1.1.5.1 $Corollary 10.1.1.5.1 Corollary 10.1.1.5.1$ , $𝑉$ has an orthonormal basis ${𝑒_{1}, \dots, 𝑒_{𝑛}}$ . Define

𝑇 : 𝑉 \to ℂ^{𝑛}, 𝑇 (𝑣) = ({⟨ 𝑣, 𝑒_{1} ⟩}_{𝑉}, \dots, {⟨ 𝑣, 𝑒_{𝑛} ⟩}_{𝑉}) .

The map $𝑇$ is linear and bijective, since

𝑣 = \sum_{𝑘 = 1}^{𝑛} {⟨ 𝑣, 𝑒_{𝑘} ⟩}_{𝑉} 𝑒_{𝑘} .

Moreover, for any $𝑣, 𝑤 \in 𝑉$ ,

{⟨ 𝑇 (𝑣), 𝑇 (𝑤) ⟩}_{ℂ^{𝑛}} = \sum_{𝑘 = 1}^{𝑛} {⟨ 𝑣, 𝑒_{𝑘} ⟩}_{𝑉} \overline{{⟨ 𝑤, 𝑒_{𝑘} ⟩}_{𝑉}} = {⟨ 𝑣, 𝑤 ⟩}_{𝑉} .

Thus

𝑇

is an isometric isomorphism. ⁠

□

(The same construction shows that every $𝑛$ -dimensional real inner product space is isometrically isomorphic to $ℝ^{𝑛}$ with its usual Euclidean dot product.)

Theorem 10.1.9 (Cauchy–Schwarz Inequality).

If $𝑉$ is a vector space and $⟨ \cdot, \cdot ⟩$ is a nondegenerate positive-definite Hermitian inner product on $𝑉$ , then for $𝑣, 𝑤 \in 𝑉$ ,

| ⟨ 𝑣, 𝑤 ⟩ | \leq ‖ 𝑣 ‖ \cdot ‖ 𝑤 ‖ .

Proof.

We only need to consider cases when $𝑣 \neq 0$ and $𝑤 \neq 0$ (as these cases are already satisfied by Part 3 $Part 3 Part 3$ of Proposition 10.1.2 $Proposition 10.1.2 Proposition 10.1.2$ ).

By the positive-definiteness of the product,

\begin{aligned} 0 & \leq ⟨ 𝑣 - \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} 𝑤, 𝑣 - \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} 𝑤 ⟩ \\ = ⟨ 𝑣, 𝑣 ⟩ + ⟨ \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} 𝑤, \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} 𝑤 ⟩ + 2 ℜ𝔢 ⟨ 𝑣, - \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} 𝑤 ⟩ \\ = {‖ 𝑣 ‖}^{2} + {| \frac{⟨ 𝑣, 𝑤 ⟩}{{‖ 𝑤 ‖}^{2}} |}^{2} {‖ 𝑤 ‖}^{2} - 2 ℜ𝔢 \frac{\overline{⟨ 𝑣, 𝑤 ⟩}}{{‖ 𝑤 ‖}^{2}} ⟨ 𝑣, 𝑤 ⟩ \\ = {‖ 𝑣 ‖}^{2} - \frac{{| ⟨ 𝑣, 𝑤 ⟩ |}^{2}}{{‖ 𝑤 ‖}^{2}}, \end{aligned}

implying that

{‖ 𝑣 ‖}^{2} {‖ 𝑤 ‖}^{2} \geq {| ⟨ 𝑣, 𝑤 ⟩ |}^{2} ⟹ | ⟨ 𝑣, 𝑤 ⟩ | \leq ‖ 𝑣 ‖ \cdot ‖ 𝑤 ‖ . □

Then Theorem 10.1.9 $Theorem 10.1.9 Theorem 10.1.9$ and Part 4 $Part 4 Part 4$ of Proposition 10.1.2 $Proposition 10.1.2 Proposition 10.1.2$ are sufficient to imply the triangle inequality of the associated norm:

Theorem 10.1.10 (Triangle Inequality of the Associated Norm).

If $𝑉$ is a vector space and $⟨ \cdot, \cdot ⟩$ is a nondegenerate positive-definite Hermitian inner product on $𝑉$ , then for $𝑣, 𝑤 \in 𝑉$ ,

‖ 𝑣 ‖ + ‖ 𝑤 ‖ \geq ‖ 𝑣 + 𝑤 ‖ .

Proof.

By Part 4 $Part 4 Part 4$ of Proposition 10.1.2 $Proposition 10.1.2 Proposition 10.1.2$ and the Cauchy–Schwarz Inequality (Theorem 10.1.9 $Theorem 10.1.9 Theorem 10.1.9$ ),

\begin{aligned} {‖ 𝑣 + 𝑤 ‖}^{2} & = {‖ 𝑣 ‖}^{2} + {‖ 𝑤 ‖}^{2} + 2 ℜ𝔢 ⟨ 𝑣, 𝑤 ⟩ \\ \leq {‖ 𝑣 ‖}^{2} + {‖ 𝑤 ‖}^{2} + 2 | ⟨ 𝑣, 𝑤 ⟩ | \\ \leq {‖ 𝑣 ‖}^{2} + {‖ 𝑤 ‖}^{2} + 2 ‖ 𝑣 ‖ \cdot ‖ 𝑤 ‖ (by Cauchy - Schwarz) \\ = {(‖ 𝑣 ‖ + ‖ 𝑤 ‖)}^{2} . \end{aligned}

Then taking the square root of both sides completes the proof. ⁠

□

Then it follows that $𝑑 (𝑣, 𝑤) = ‖ 𝑣 - 𝑤 ‖$ can be used to define a metric as it satisfies all the properties of Definition 3.2.1.12 $Definition 3.2.1.12 Definition 3.2.1.12$ . Along with $𝑉$ , the associated norm then defines a metric space. Then we can define the notion of convergence of sequences in the vector space.

Theorem 10.1.11 (Continuity of the Inner Product).

Let $𝑉$ be an inner product space. If $𝑥_{𝑛} \to 𝑥$ and $𝑦_{𝑛} \to 𝑦$ with respect to the associated norm, then

⟨ 𝑥_{𝑛}, 𝑦_{𝑛} ⟩ \to ⟨ 𝑥, 𝑦 ⟩ .

Equivalently, the map

⟨ \cdot, \cdot ⟩ : 𝑉 \times 𝑉 \to ℂ

is jointly continuous.

Proof.

Since every convergent sequence is bounded, there is some $𝑀 > 0$ such that $‖ 𝑦_{𝑛} ‖ \leq 𝑀$ for every $𝑛$ . By the Cauchy–Schwarz Inequality (Theorem 10.1.9 $Theorem 10.1.9 Theorem 10.1.9$ ), for sufficiently large $𝑁$

\begin{aligned} | ⟨ 𝑥_{𝑛}, 𝑦_{𝑛} ⟩ - ⟨ 𝑥, 𝑦 ⟩ | & = | ⟨ 𝑥_{𝑛} - 𝑥, 𝑦_{𝑛} ⟩ + ⟨ 𝑥, 𝑦_{𝑛} - 𝑦 ⟩ | \\ \leq ‖ 𝑥_{𝑛} - 𝑥 ‖ ‖ 𝑦_{𝑛} ‖ + ‖ 𝑥 ‖ ‖ 𝑦_{𝑛} - 𝑦 ‖ \\ \leq (‖ 𝑦 ‖ + 𝜀) ‖ 𝑥_{𝑛} - 𝑥 ‖ + ‖ 𝑥 ‖ ‖ 𝑦_{𝑛} - 𝑦 ‖ \to 0 . □ \end{aligned}

For a general metric space $(𝑋, 𝑑)$ , if for a sequence ${𝑥_{𝑛}}_{𝑛 \in ℕ} \subseteq 𝑋$ such that $\forall 𝜀 > 0$ , $\exists 𝑁 \in ℕ : \forall 𝑚, 𝑛 > 𝑁$ , $𝑑 (𝑥_{𝑛}, 𝑥_{𝑚}) < 𝜀$ ; then ${𝑥_{𝑛}}$ is known as a Cauchy sequence.

If the metric space is $ℝ$ under the standard Euclidean metric, then any Cauchy sequence is convergent. However, note that on other sets this is not necessarily true; if we consider the set of rationals $ℚ$ , we may choose a sequence ${𝑥_{𝑛}}$ where the $𝑛$ -th element is the first $𝑛$ digits of the decimal expansion of $π$ : $3, 3.1, 3.14, 3.141, \dots$ ; it follows trivially that ${𝑥_{𝑛}}$ is a Cauchy sequence, although it is not convergent in $ℚ$ as the accumulation point $π \in ℝ$ is not in $ℚ$ .

A metric space is said to be complete iff all Cauchy sequences are complete. Intuitively, the set of rationals cannot be complete as there are accumulation points in $ℝ ∖ ℚ$ to which Cauchy sequences tend, so it is incomplete in the sense that potential accumulation points are not included. In most analytical settings, we are concerned only with vector space over complete scalar fields such as $ℝ$ or $ℂ$ .

Proposition 10.1.12.

Any finite-dimensional vector space

𝑉

over a complete field

𝔽

is complete with respect the metric space derived from any (nondegenerate positive-definite Hermitian) inner-product norm.

Proof.

Let ${𝑒_{𝑗}}_{𝑗 = 1}^{𝑛}$ be an orthonormal basis for $𝑉$ , obtained by the Gram–Schmidt process (see Section 10.1.1 $Section 10.1.1 Section 10.1.1$ ), and fix a particular inner product $⟨ \cdot, \cdot ⟩$ and associated norm $‖ \cdot ‖$ . Then for any Cauchy sequence ${𝑣_{𝑘}}_{𝑘 \in ℕ}$ , writing for each $𝑘$ :

𝑣_{𝑘} = \sum_{𝑗 = 1}^{𝑛} 𝑣_{𝑘}^{(𝑗)} 𝑒_{𝑗},

‖ 𝑣_{𝑘} - 𝑣_{𝑙} ‖ = ‖ \sum_{𝑗 = 1}^{𝑛} (𝑣_{𝑘}^{(𝑗)} - 𝑣_{𝑙}^{(𝑗)}) 𝑒_{𝑗} ‖ < 𝜀,

it follows that for each $𝑗$ , each ${𝑣_{𝑘}^{(𝑗)}}_{𝑘}$ is a Cauchy sequence, and converges by the completeness of $𝔽$ to a value $𝑣^{(𝑗)}$ .

Moreover, $\forall 𝜀$ , exists $𝑁 \in ℕ$ such that $\forall 𝑘 > 𝑁, 𝑗 \in ℕ_{\leq 𝑛}$ , $| 𝑣_{𝑘}^{(𝑗)} - 𝑣^{(𝑗)} | < \frac{𝜀}{𝑛}$ .

Let $𝑣 = \sum_{𝑗 = 1}^{𝑛} 𝑣^{(𝑗)} 𝑒_{𝑗}$ . For any $𝑘 > 𝑁$ ,

‖ 𝑣_{𝑘} - 𝑣 ‖ \leq \sum_{𝑗 = 1}^{𝑛} ‖ 𝑣_{𝑘}^{(𝑗)} - 𝑣_{𝑙}^{(𝑗)} ‖ < 𝜀 .

Thus, under any inner product, any Cauchy sequence in $𝑉$ converges. ⁠ $□$

It is often convenient in analysis to take limits, and thus in most cases we are concerned only with complete spaces. However, infinite-dimensional vector spaces do not need to be complete under any inner-product norm, and thus completeness must be then stated as a precursory condition before further analysis can be done. These conditions now motivate for the following definition.

Definition 10.1.13 (Hilbert Space).

A complex vector space

𝑉

with a (nondegenerate) positive-definite Hermetian inner product

⟨ \cdot, \cdot ⟩

is said to be a Hilbert space iff it is complete under the associated norm

‖ \cdot ‖

Example 10.1.14.

The vector space $ℂ^{𝑛}$ with the inner product

⟨ 𝐯, 𝐰 ⟩ = \sum_{𝑘 = 1}^{𝑛} 𝑣_{𝑘} \overline{𝑤_{𝑘}}

is a Hilbert space, and the metric derived from the associated norm provides the standard sense of Euclidean distance:

‖ 𝐯 - 𝐰 ‖ = \sqrt{\sum_{𝑘 = 1}^{𝑛} {| 𝑣_{𝑘} - 𝑤_{𝑘} |}^{2}} .

Example 10.1.15 (Bergman Space).

Let $Ω \subseteq ℂ$ be an open region. Let

𝐴^{2} (Ω) = {𝑓 holomorphic on Ω : \iint_{Ω} {| 𝑓 (𝑧) |}^{2} d 𝑥 d 𝑦 < \infty} .

Then $𝐴^{2} (Ω)$ is a vector space over $ℂ$ with the standard definition of addition and scalar multiplication. Then the inner product defined by

⟨ 𝑓, 𝑔 ⟩ = \iint_{Ω} 𝑓 (𝑧) \overline{𝑔 (𝑧)} d 𝑥 d 𝑦 where 𝑧 = 𝑥 + i 𝑦 (10.1 . 1)

for any two

𝑓, 𝑔 \in 𝐴^{2} (Ω)

defines a Hilbert space with

𝐴^{2} (Ω)

Proof.

Firstly, the inner product is well-defined as

\begin{aligned} | 𝑓 (𝑧) \overline{𝑔 (𝑧)} | & \leq 2 | 𝑓 (𝑧) | | 𝑔 (𝑧) | \\ = {| 𝑓 (𝑧) |}^{2} + {| 𝑔 (𝑧) |}^{2} - {(| 𝑓 (𝑧) | - | 𝑔 (𝑧) |)}^{2} \\ \leq {| 𝑓 (𝑧) |}^{2} + {| 𝑔 (𝑧) |}^{2}, \end{aligned}

which provides the absolute convergence of (10.1.1) $(10.1.1)$ , and Part 5 $Part 5 Part 5$ of Definition 10.1.1 $Definition 10.1.1 Definition 10.1.1$ follows from the same logic as in Example 10.1.4 $Example 10.1.4 Example 10.1.4$ .

Proposition 10.1.16.

For any $ℎ \in 𝐴^{2} (Ω)$ and any $\forall 𝑧 \in Ω$ , then whenever $𝑟 > 0$ such that $\overline{𝐷 (𝑧, 𝑟)} ⋐ Ω$ ,

| ℎ (𝑧) | \leq \frac{1}{𝑟 \sqrt{π}} ‖ ℎ ‖ .

Proof of Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ .

By the mean-value property,

ℎ (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} ℎ (𝑧 + 𝜌 e^{i 𝜃}) d 𝜃 .

Therefore,

\begin{aligned} \frac{1}{𝑟^{2}} \int_{0}^{𝑟} 2 𝜌 ℎ (𝑧) d 𝜌 & = \frac{1}{𝑟^{2}} \int_{0}^{𝑟} 2 𝜌 [\frac{1}{2 π} \int_{0}^{2 π} ℎ (𝑧 + 𝜌 e^{i 𝜃}) d 𝜃] d 𝜌 \\ ℎ (𝑧) & = \frac{1}{π 𝑟^{2}} \int_{0}^{𝑟} \int_{0}^{2 π} ℎ (𝑧 + 𝜌 e^{i 𝜃}) 𝜌 d 𝜃 d 𝜌 \\ = \frac{1}{π 𝑟^{2}} \int_{𝐷 (𝑧, 𝑟)} ℎ (𝜁) d 𝜉 \land d 𝜂 . (where 𝜁 = 𝜉 + i 𝜂) \end{aligned}

Thus,

| ℎ (𝑧) | \leq \frac{1}{π 𝑟^{2}} \int_{𝐷 (𝑧, 𝑟)} | ℎ (𝜁) | \cdot 1 d 𝜉 \land d 𝜂 .

By the Cauchy–Schwarz Inequality ( $\int | 𝑓 𝑔 | \leq \sqrt{\int {| 𝑓 |}^{2}} \sqrt{\int {| 𝑔 |}^{2}}$ ), we have

\begin{aligned} | ℎ (𝑧) | & \leq \frac{1}{π 𝑟^{2}} \sqrt{\int_{𝐷 (𝑧, 𝑟)} {| ℎ (𝜁) |}^{2} d 𝜉 \land d 𝜂} \sqrt{\int_{𝐷 (𝑧, 𝑟)} d 𝜉 \land d 𝜂} \\ = \frac{1}{𝑟 \sqrt{π}} ‖ ℎ ‖ . □ \end{aligned}

With Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ in mind, we now proceed to prove the completeness of the metric space.

Let $𝐾 ⋐ Ω$ be arbitrary and let ${𝑓_{𝑛}}_{𝑛 \in ℕ} \subseteq 𝐴^{2} (Ω)$ be any Cauchy sequence in the inner product space. Then there is a $𝑟 > 0$ such that $\forall 𝑧 \in 𝐾$ , $\overline{𝐷 (𝑧, 𝑟)} ⋐ Ω$ . Then $\forall 𝜀 > 0$ , $\exists 𝑁 > 0$ such that $\forall 𝑚, 𝑛 > 𝑁$ , $‖ 𝑓_{𝑛} - 𝑓_{𝑚} ‖ < 𝜀 𝑟 \sqrt{π}$ . By Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ applied to $ℎ \equiv 𝑓_{𝑛} - 𝑓_{𝑚} \in 𝐴^{2} (Ω)$ ,

sup_{𝑧 \in 𝐾} | 𝑓_{𝑛} (𝑧) - 𝑓_{𝑚} (𝑧) | \leq \frac{1}{𝑟 \sqrt{π}} ‖ 𝑓_{𝑛} - 𝑓_{𝑚} ‖ < 𝜀,

hence giving that ${𝑓_{𝑛}}$ uniformly converges in $𝐾$ . Since $𝐾$ was arbitrary, ${𝑓_{𝑛}}$ locally uniformly converges in $Ω$ , and converges by the Weierstrass Convergence Theorem (Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ) to a holomorphic function $𝑓$ on $Ω$ . It remains to show that $𝑓 \in 𝐴^{2} (Ω)$ and $𝑓_{𝑛} \to 𝑓$ with respect to $‖ \cdot ‖$ .

By Lemma 7.5.3 $Lemma 7.5.3 Lemma 7.5.3$ , choose a nested exhaustion ${𝑈_{𝑗}}_{𝑗 \in ℕ}$ of $Ω$ by bounded grid regions. Fix $𝜀 > 0$ and choose $𝑁 \in ℕ$ such that

‖ 𝑓_{𝑛} - 𝑓_{𝑚} ‖ < 𝜀 . (\forall 𝑚, 𝑛 > 𝑁)

Fix $𝑛 > 𝑁$ . For every $𝑗 \in ℕ$ , since $𝑓_{𝑚} ⇉ 𝑓$ on $\overline{𝑈_{𝑗}}$ ,

\iint_{𝑈_{𝑗}} {| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) |}^{2} d 𝐴 = lim_{𝑚 \to \infty} \iint_{𝑈_{𝑗}} {| 𝑓_{𝑛} (𝑧) - 𝑓_{𝑚} (𝑧) |}^{2} d 𝐴 < 𝜀^{2},

because $\iint_{𝑈_{𝑗}} {| 𝑓_{𝑛} - 𝑓_{𝑚} |}^{2} d 𝐴 \leq {‖ 𝑓_{𝑛} - 𝑓_{𝑚} ‖}^{2} < 𝜀^{2}$ for every $𝑚 > 𝑁$ . By the definition of the improper integral,

\iint_{Ω} {| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) |}^{2} d 𝐴 = lim_{𝑗 \to \infty} \iint_{𝑈_{𝑗}} {| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) |}^{2} d 𝐴 < 𝜀^{2} .

Thus $‖ 𝑓_{𝑛} - 𝑓 ‖ < 𝜀$ for $𝑛 > 𝑁$ and $𝑓_{𝑛} - 𝑓 \in 𝐴^{2} (Ω)$ . Since $𝑓_{𝑛} \in 𝐴^{2} (Ω)$ , it follows that $𝑓 \in 𝐴^{2} (Ω)$ since ${| 𝑓 |}^{2} \leq 2 {| 𝑓_{𝑛} |}^{2} + 2 {| 𝑓 - 𝑓_{𝑛} |}^{2}$ by the Parallelogram Law (Theorem 10.1.18 $Theorem 10.1.18 Theorem 10.1.18$ ). Therefore, $𝑓_{𝑛} \to 𝑓$ in $𝐴^{2} (Ω)$ with respect to $‖ \cdot ‖$ , proving completeness.

⁠ $□$

The study of the Bergman space $𝐴^{2} (Ω)$ will become important later on as we aim to prove the following:

If $𝜕 Ω_{1}$ and $Ω_{2}$ are $𝐶^{\infty}$ Jordan curves, then $𝜑$ extends continuously and injectively to $𝜕 Ω_{1}$ and is $𝐶^{\infty}$ on $\overline{Ω_{1}}$ .

(Recall that this was briefly mentioned in Section 9.3 $Section 9.3 Section 9.3$ ; we have already proven the two other main cases, namely when the boundaries are merely Jordan and when the boundaries are real-analytic.)

For the remainder of this section, we will develop general properties of Hilbert spaces. Note that in a Hilbert space closedness and boundedness do not necessarily imply compactness (recall that compactness is most generally defined by the notion of finite subcovers), although compactness will always imply closedness and boundedness.

Example 10.1.17.

Show that there exists a Hilbert space

𝐻

such that there is a closed, bounded¹ set

𝑆 \subseteq 𝐻

that is not compact.

Solution to Example 10.1.17 $Example 10.1.17 Example 10.1.17$ .

Let $𝐻 = 𝔻$ with the inner product

⟨ 𝑧_{1}, 𝑧_{2} ⟩ = \sqrt{{[ℜ𝔢 (𝑧_{1}) ℜ𝔢 (𝑧_{2})]}^{2} + {[ℑ𝔪 (𝑧_{1}) ℑ𝔪 (𝑧_{2})]}^{2}} .

Then the associated norm

‖ \cdot ‖

is simply the Euclidean distance. Then under the induced metric topology,

𝐻 \subseteq 𝐻

is closed and bounded, but not compact, as it can be covered by the open cover

{𝐷 (0, 1 - \frac{1}{𝑛})}_{𝑛 = 2}^{\infty}

, which does not admit a finite subcover. ⁠

□

The following properties we define will be useful in serving as a replacement for compactness:

Figure 10.1: A geometric visualization of the Parallelogram Law.

Theorem 10.1.18 (Parallelogram Law).

Let $𝐻$ be an inner product space (or a Hilbert space). Then for any $𝑥, 𝑦 \in 𝐻$ ,

2 {‖ 𝑥 ‖}^{2} + 2 {‖ 𝑦 ‖}^{2} = {‖ 𝑥 + 𝑦 ‖}^{2} + {‖ 𝑥 - 𝑦 ‖}^{2} .

(More generally, the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals; see Figure 10.1

Figure 10.1 Figure 10.1

)

Proof.

By Part 4 $Part 4 Part 4$ of Proposition 10.1.2 $Proposition 10.1.2 Proposition 10.1.2$ ,

{‖ 𝑥 + 𝑦 ‖}^{2} = {‖ 𝑥 ‖}^{2} + {‖ 𝑦 ‖}^{2} + 2 ℜ𝔢 ⟨ 𝑥, 𝑦 ⟩

and

{‖ 𝑥 - 𝑦 ‖}^{2} = {‖ 𝑥 ‖}^{2} + {‖ 𝑦 ‖}^{2} - 2 ℜ𝔢 ⟨ 𝑥, 𝑦 ⟩ .

Adding the two equations gives the desired result. ⁠

□

Figure 10.2: A set which is convex (left) and one which is not (right).

Definition 10.1.19 (Convexity).

A set

𝐹

in a Hilbert space

𝐻

is said to be convex iff

\forall 𝑥, 𝑦 \in 𝐹

and

\forall 𝑡 \in (0, 1)

𝑥 + 𝑡 (𝑦 - 𝑥) \in 𝐹

. In other words, convexity means that the straight line segment between any two points lies in the set (see Figure 10.2

Figure 10.2 Figure 10.2

Note that if $𝐹 \subseteq 𝐻$ is a vector space, then $𝐹$ is trivially convex by the definition of a vector space.

Definition 10.1.20 (Orthogonality).

Two vectors

𝑥, 𝑦

in a Hilbert space

𝐻

are said to be orthogonal (

𝑥 ⟂ 𝑦

) iff their inner product is

0

Definition 10.1.21.

Let

𝐻

be a Hilbert space. A subspace

𝑆 \subseteq 𝐻

is said to be orthogonal to a vector

𝑥 \in 𝐻

(denoted by

𝑥 ⟂ 𝑆

) iff

\forall 𝑦 \in 𝑆

𝑥 ⟂ 𝑦

Then rather intuitively, we have

Theorem 10.1.22.

𝑥 \in 𝐻

is a vector in a Hilbert space, and

𝑥 = 𝑥_{1} + 𝑥_{2}

such that

𝑥_{1} ⟂ 𝑥_{2}

, then

‖ 𝑥 ‖ = \sqrt{{‖ 𝑥_{1} ‖}^{2} + {‖ 𝑥_{2} ‖}^{2}}

Proof.

This follows immediately from

⟨ 𝑥, 𝑥 ⟩ = ⟨ 𝑥_{1} + 𝑥_{2}, 𝑥_{1} + 𝑥_{2} ⟩ = {‖ 𝑥_{1} ‖}^{2} + {‖ 𝑥_{2} ‖}^{2} + 2 ℜ𝔢 ⟨ 𝑥_{1}, 𝑥_{2} ⟩ = {‖ 𝑥_{1} ‖}^{2} + {‖ 𝑥_{2} ‖}^{2} . □

Definition 10.1.23 (Orthogonal Complement).

Let $𝑥 \in 𝐻$ , where $𝐻$ is a Hilbert space, then for $𝑥 \in 𝐻$ , define the orthogonal component of $𝑥$ to be $𝑥^{⟂} = {𝑦 \in 𝐻 : 𝑥 ⟂ 𝑦}$ . Let $𝑉 \subseteq 𝐻$ be a subset (not necessarily a vector space). Then the orthogonal complement of $𝑉$ , given by

𝑉^{⟂} = {𝑥 \in 𝐻 : \forall 𝑦 \in 𝐻, 𝑥 ⟂ 𝑦},

is the space of vectors orthogonal to every vector in

𝑉

It is simple to verify that the orthogonal complement of a vector or set is always a vector space.

Definition 10.1.24 (Vector Projection).

Let $𝑦 \in 𝐻 ∖ {0}$ , where $𝐻$ is a Hilbert space. For any $𝑥 \in 𝐻$ , the projection of $𝑥$ onto $𝑦$ is the unique vector $𝜆 𝑦$ (where $𝜆$ is a scalar) such that $⟨ 𝑥 - 𝜆 𝑦, 𝑦 ⟩ = 0$ , which is given by

𝜆 𝑦 = \frac{⟨ 𝑥, 𝑦 ⟩}{{‖ 𝑦 ‖}^{2}} 𝑦

(immediate from additivity).

We can make geometric sense of this definition in the case where $⟨ \cdot, \cdot ⟩$ is the Euclidean dot product.

Proposition 10.1.25.

Let $𝐻$ be a Hilbert space and let $𝐹 \subseteq 𝐻$ be a convex set. Let $𝑚 = {inf}_{𝑥 \in 𝐹} ‖ 𝑥 ‖$ . Then

{‖ 𝑥 - 𝑦 ‖}^{2} \leq {‖ 𝑥 ‖}^{2} + {‖ 𝑦 ‖}^{2} - 4 𝑚^{2} . (where 𝑥, 𝑦 \in 𝐹)

Proof.

By the Parallelogram Law (Theorem 10.1.18 $Theorem 10.1.18 Theorem 10.1.18$ ),

{‖ 𝑥 - 𝑦 ‖}^{2} = {‖ 𝑥 ‖}^{2} + {‖ 𝑦 ‖}^{2} - {‖ 𝑥 + 𝑦 ‖}^{2}

for any $𝑥, 𝑦 \in 𝐹$ . By convexity, the midpoint $\frac{𝑥 + 𝑦}{2} \in 𝐹$ , giving

{‖ \frac{𝑥 + 𝑦}{2} ‖}^{2} \geq 𝑚^{2} ⟹ {‖ 𝑥 + 𝑦 ‖}^{2} \geq 4 𝑚^{2} .

Then

{‖ 𝑥 - 𝑦 ‖}^{2} \leq {‖ 𝑥 ‖}^{2} + {‖ 𝑦 ‖}^{2} - 4 𝑚^{2} . □

The following result shows how the convexity can be used in place of traditional compactness arguments:

Proposition 10.1.26.

Let

𝐹 \subseteq 𝐻

be a closed, convex set in a Hilbert space

𝐻

and let

𝑚 = {inf}_{𝑥 \in 𝐹} ‖ 𝑥 ‖

. Then

\exists! 𝑥^{'} \in 𝐹

such that

‖ 𝑥^{'} ‖ = 𝑚

Proof.

By the definition of infimum, either the infimum is attained (in which case we are done), or the infimum is an accumulation point in $ℝ_{\geq 0}$ of norms of vectors in $𝐹$ . Assume the latter case. Then there is a sequence ${𝑥_{𝑛}}_{𝑛 \in ℕ}$ such that $‖ 𝑥_{𝑛} ‖ \to 𝑚$ . Then for all $𝜀 > 0$ , $\exists 𝑁 \in ℕ$ such that ${sup}_{𝑛 > 𝑁} | {‖ 𝑥_{𝑛} ‖}^{2} - 𝑚^{2} | < \frac{𝜀}{4}$ . Then by Proposition 10.1.25 $Proposition 10.1.25 Proposition 10.1.25$ ,

sup_{𝑛, 𝑚 > 𝑁} {‖ 𝑥_{𝑛} - 𝑥_{𝑚} ‖}^{2} \leq sup_{𝑛, 𝑀 > 𝑁} 2 {‖ 𝑥_{𝑛} ‖}^{2} + 2 {‖ 𝑥_{𝑚} ‖}^{2} - 4 𝑚^{2} < 𝜀 .

Therefore, $𝑥_{𝑛}$ is a Cauchy sequence, and converges to some $𝑥^{'} \in 𝐻$ . Since $𝐹$ is closed in $𝐻$ , $𝑥^{'} \in 𝐹$ . Moreover, by the reverse triangle inequality,

| ‖ 𝑥^{'} ‖ - ‖ 𝑥_{𝑛} ‖ | \leq ‖ 𝑥^{'} - 𝑥_{𝑛} ‖ \to 0

by convergence. Then $‖ 𝑥_{𝑛} ‖ \to ‖ 𝑥^{'} ‖$ , giving $‖ 𝑥^{'} ‖ = 𝑚$ .

To show that $𝑥^{'}$ is the unique element of least norm, assume $𝑥^{″} \in 𝐻$ is another vector with norm $𝑚$ . Then once again applying Proposition 10.1.25 $Proposition 10.1.25 Proposition 10.1.25$ ,

‖ 𝑥^{'} - 𝑥^{″} ‖ \leq {‖ 𝑥^{'} ‖}^{2} + {‖ 𝑥^{″} ‖}^{2} - 4 𝑚^{2} = 0 ⟹ 𝑥^{'} = 𝑥^{″} . □

Remark.

To make sense of uniqueness from a geometric viewpoint, we consider a closed convex set $𝐹 \subseteq 𝐻 = ℂ$ with $𝑚 = min {| 𝑧 | : 𝑧 \in 𝐹} > 0$ . Then $𝐷 (0, 𝑚) \cap 𝐹 = ⌀$ .

Then $𝑚 e^{i 𝜃_{1}} \in 𝐹$ for some $𝜃_{1}$ . If there is another point $𝑚 e^{i 𝜃_{2}} \in 𝐹$ , then the straight segment between them lies in $𝐹$ by convexity. This straight line must intersect $𝐷 (0, 𝑚)$ , which is a contradiction.

Proposition 10.1.27.

The intersection between any vector subspace

𝑉

of a Hilbert space

𝐻

and its orthogonal complement

𝑉^{⟂}

{0}

Proof.

For

𝑥 \in 𝑉

and

𝑥 \in 𝑉^{⟂}

to be simultaneously true, we must have

⟨ 𝑥, 𝑥 ⟩ = {‖ 𝑥 ‖}^{2} = 0

, implying

𝑥 = 0

. ⁠

□

Theorem 10.1.28 (Orthogonal Decomposition).

Let

𝐾 \subseteq 𝐻

be a closed vector subspace of a Hilbert space

𝐻

. Then

\forall 𝑥_{0} \in 𝐻

\exists! 𝑥_{1} \in 𝐾 : \exists! 𝑥_{2} \in 𝐾^{⟂} : 𝑥_{0} = 𝑥_{1} + 𝑥_{2}

. Moreover,

𝑥_{2} \in 𝑥_{0} + 𝐾 = {𝑥_{0} + 𝑘 : 𝑘 \in 𝐾}

and is the unique element satisfying

‖ 𝑥_{2} ‖ = {inf}_{𝑥 \in 𝑥_{0} + 𝐾} ‖ 𝑥 ‖

Proof.

Observe that $𝑥_{0} + 𝐾$ is convex as $𝐾$ is convex; then $\exists! 𝑥_{2} \in 𝑥_{0} + 𝐾$ attaining the least norm $𝑚 \geq 0$ (by Proposition 10.1.26 $Proposition 10.1.26 Proposition 10.1.26$ ). Let $𝑘_{2} = 𝑥_{2} - 𝑥_{0} \in 𝐾$ and let $𝑥_{1} = 𝑥_{0} - 𝑥_{2} = - 𝑘_{2}$ . We claim that these are the desired values for $𝑥_{1}, 𝑥_{2}$ .

We obtain immediately that $𝑥_{1} \in 𝐾$ since $𝑘_{2} \in 𝐾$ . For any $𝑦 \in 𝐾$ , projecting $𝑥_{2}$ onto $𝑦$ gives

{proj}_{𝑦} 𝑥_{2} = \frac{⟨ 𝑥_{2}, 𝑦 ⟩}{{‖ 𝑦 ‖}^{2}} 𝑦 \in 𝐾 .

Let $𝑝 = 𝑥_{2} - {proj}_{𝑦} 𝑥_{2} = 𝑥_{0} + 𝑘_{2} - {proj}_{𝑦} 𝑥_{2} \in 𝑥_{0} + 𝐾$ . By the minimality of $𝑚$ ,

\begin{aligned} {‖ 𝑥_{2} ‖}^{2} & \leq {‖ 𝑝 ‖}^{2} = {‖ 𝑥_{2} - {proj}_{𝑦} 𝑥_{2} ‖}^{2} \\ = {‖ 𝑥_{2} ‖}^{2} + {‖ \frac{⟨ 𝑥_{2}, 𝑦 ⟩}{{‖ 𝑦 ‖}^{2}} 𝑦 ‖}^{2} - 2 ℜ𝔢 ⟨ 𝑥_{2}, \frac{⟨ 𝑥_{2}, 𝑦 ⟩}{{‖ 𝑦 ‖}^{2}} 𝑦 ⟩ \\ = {‖ 𝑥_{2} ‖}^{2} + {| \frac{⟨ 𝑥_{2}, 𝑦 ⟩}{{‖ 𝑦 ‖}^{2}} |}^{2} {‖ 𝑦 ‖}^{2} - 2 \frac{{| ⟨ 𝑥_{2}, 𝑦 ⟩ |}^{2}}{{‖ 𝑦 ‖}^{2}} \\ = {‖ 𝑥_{2} ‖}^{2} - \frac{{| ⟨ 𝑥_{2}, 𝑦 ⟩ |}^{2}}{{‖ 𝑦 ‖}^{2}} \\ \leq {‖ 𝑥_{2} ‖}^{2}, \end{aligned}

which means all inequalities are equalities, giving that $⟨ 𝑥_{2}, 𝑦 ⟩ = 0$ . Since $𝑦 \in 𝐾$ was arbitrary, $𝑥_{2}$ is orthogonal to the entire vector space $𝐾$ , hence $𝑥_{2} \in 𝐾^{⟂}$ .

If $𝑥 = 𝑥_{1}^{'} + 𝑥_{2}^{'}$ is another decomposition, then

𝐾 ∋ 𝑥_{1}^{'} - 𝑥_{1} = 𝑥_{2} - 𝑥_{2}^{'} \in 𝐾^{⟂} .

Since $𝐾 \cap 𝐾^{⟂} = {0}$ , the uniqueness follows. ⁠ $□$

Corollary 10.1.28.1.

The orthogonal complement of any closed proper subspace of a Hilbert space contains a nonzero element.

Proof.

Let the Hilbert space be

𝐻

, let the subspace be

𝐾

. Choose

𝑥 \in 𝐻 ∖ 𝐾

. Then

𝑥 = 𝑥_{1} + 𝑥_{2}

, where

𝑥_{1} \in 𝐾

𝑥_{2} \in 𝐾^{⟂}

(by Theorem 10.1.28

Theorem 10.1.28 Theorem 10.1.28

). Then

𝑥_{2} \neq 0

as otherwise

𝑥 = 𝑥_{1} \in 𝐾

. ⁠

□

Definition 10.1.29 (Linear Functional).

Let $𝑉$ be a vector space over a field $𝔽$ . Then a linear functional is a mapping

𝜙 : 𝑉 \to 𝔽

that is linear: for

𝑥, 𝑦 \in 𝐻

𝜙 (𝛼 𝑥) = 𝛼 𝜙 (𝑥)

(homogeneous) for

𝛼 \in 𝔽

and

𝜙 (𝑥 + 𝑦) = 𝜑 (𝑥) + 𝜑 (𝑦)

(additive).

(We will only consider the case if $𝑉 = 𝐻$ is a Hilbert space. Then linear functionals map to $ℂ$ .)

Definition 10.1.30.

Let $𝐻$ be a Hilbert space. Then a linear functional $𝜙$ on $𝐻$ is bounded iff $\exists 𝐶 > 0$ such that

| 𝜙 (𝑥) | \leq 𝐶 ‖ 𝑥 ‖ . (\forall 𝑥 \in 𝐻)

Definition 10.1.31.

Let

𝐻

be a Hilbert space. Then a linear functional

𝜙

𝐻

is continuous iff for any convergent sequence

{𝑥_{𝑛}} \subseteq 𝐻

(

𝑥_{𝑛} \to 𝑥_{\infty} \in 𝐻

𝜙 (𝑥_{𝑛}) \to 𝜙 (𝑥_{\infty})

In mathematical literature, “continuous” and “bounded” are often interchangeably used:

Theorem 10.1.32.

A linear functional on a Hilbert space is bounded iff it is continuous.

Proof.

(Denote the functional by $𝜙$ , the Hilbert space by $𝐻$ .)

Continuity implies boundedness (assume continuity).
Assume the boundedness does not hold. Then for any $𝑛 > 0$ there exists $𝑥 \in 𝐻$ such that
$| 𝜑 (𝑥) | > 𝑛 ‖ 𝑥 ‖ ⟹ | 𝜑 (\frac{𝑥}{𝑛 ‖ 𝑥 ‖}) | > 1 . (by homogeneity)$
Therefore, $\forall 𝑛 \in ℕ$ , $\exists 𝑥_{𝑛} \in 𝐻$ with $‖ 𝑥_{𝑛} ‖ = \frac{1}{𝑛}$ such that $| 𝜑 (𝑥_{𝑛}) | > 1$ ; then $𝑥_{𝑛} \to 𝑥_{\infty} = 0$ . By continuity, $‖ 𝜙 (𝑥_{𝑛}) - 𝜙 (𝑥_{\infty}) ‖ = ‖ 𝜙 (𝑥_{𝑛}) ‖ \to 0$ , is an impossibility, since $‖ 𝜙 (𝑥_{𝑛}) - 𝜙 (𝑥_{\infty}) ‖ > 1$ for each $𝑛 \in ℕ$ .
Then by contradiction, boundedness must hold.
Boundedness implies continuity.
For any sequence ${𝑥_{𝑛}}_{𝑛} \subseteq 𝐻$ converging to $𝑥_{\infty} \in 𝐻$ ,
$| 𝜑 (𝑥_{𝑛}) - 𝜑 (𝑥_{\infty}) | = | 𝜑 (𝑥_{𝑛} - 𝑥_{\infty}) | \leq 𝐶 ‖ 𝑥_{𝑛} - 𝑥_{\infty} ‖ \to \infty . □$

Definition 10.1.33 (Operator Norm).

For a bounded linear functional

𝜙 : 𝐻 \to ℂ

, define its operator norm

‖ 𝜙 ‖

to be the infimum of all such

𝐶

’s in Definition 10.1.29

Definition 10.1.29 Definition 10.1.29

Proposition 10.1.34.

Let $𝐻$ be a Hilbert space and fix $𝑥_{0} \in 𝐻$ . Then

𝜙 : 𝑥 \mapsto ⟨ 𝑥, 𝑥_{0} ⟩ (for 𝑥 \in 𝐻)

is a bounded linear functional on

𝐻

Proof.

The additivity and homogeneity are evident, and moreover, by Cauchy–Schwarz (Theorem 10.1.9 $Theorem 10.1.9 Theorem 10.1.9$ ),

| 𝜙 (𝑥) | \leq ‖ 𝑥_{0} ‖ ‖ 𝑥 ‖,

giving boundedness. ⁠

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Definition 10.1.35 (Kernel).

For any linear functional $𝜙$ on a Hilbert space $𝐻$ , define its null space or kernel to be the set on which $𝜙$ vanishes:

ker 𝜙 = {𝑥 \in 𝐻 : 𝜙 (𝑥) = 0} .

Moreover,

ker 𝜙

is a vector subspace of

𝐻

Theorem 10.1.36 (Riesz Representation Theorem).

Any bounded linear functional $𝜙$ on a Hilbert space $𝐻$ can be expressed in the form of

𝑥 \mapsto ⟨ 𝑥, 𝑥_{0} ⟩ for all 𝑥 \in 𝐻, (10.1 . 2)

where

𝑥_{0} \in 𝐻

is unique and lies in the orthogonal complement to the null space of

𝜙

Proof.

If $𝜙 \equiv 0$ , then letting $𝑥_{0} = 0$ gives the desired result.

Otherwise, observe that $ker 𝜙$ is closed in $𝐻$ ; indeed, for any sequence ${𝑘_{𝑛}}_{𝑛} \subset ker 𝜙$ converging to $𝑘_{\infty} \in 𝐻$ ,

| 𝜙 (𝑘_{\infty}) | = | 𝜙 (𝑘_{\infty}) - 𝜙 (𝑘_{𝑛}) | = | 𝜙 (𝑘_{\infty} - 𝑘_{𝑛}) | \leq ‖ 𝜙 ‖ ‖ 𝑘_{\infty} - 𝑘_{𝑛} ‖ \to \infty,

implying that $𝜙 (𝑘_{\infty}) = 0$ ; consequently, $𝑘_{\infty} \in 𝐻$ .

There exists a $𝑥_{0} \in ker 𝜙^{⟂}$ such that

𝜑 (𝑥_{0}) = ⟨ 𝑥_{0}, 𝑥_{0} ⟩ = ‖ 𝑥_{0} ‖ .

Indeed, $\exists 𝑞 \in ker 𝜙^{⟂} ∖ {0}$ by Corollary 10.1.28.1 $Corollary 10.1.28.1 Corollary 10.1.28.1$ ; then letting

𝑥_{0} = \frac{\overline{𝜙 (𝑞)}}{{‖ 𝑞 ‖}^{2}} 𝑞 \in ker 𝜙^{⟂},

it follows that

𝜙 (𝑥_{0}) = \frac{{| 𝜙 (𝑞) |}^{2}}{{‖ 𝑞 ‖}^{2}} = ⟨ \frac{\overline{𝜙 (𝑞)}}{{‖ 𝑞 ‖}^{2}} 𝑞, \frac{\overline{𝜙 (𝑞)}}{{‖ 𝑞 ‖}^{2}} 𝑞 ⟩ = {‖ 𝑥_{0} ‖}^{2}, (10.1 . 3)

and the assertion follows.

Then for any $𝑥 \in 𝐻$ , letting $𝑦 = \frac{𝜙 (𝑥)}{{‖ 𝑥_{0} ‖}^{2}} 𝑥_{0} \in ker 𝜙^{⟂}$ gives

\begin{aligned} 𝜙 (𝑥 - 𝑦) & = 𝜙 (𝑥) - 𝜙 (𝑦) = 𝜙 (𝑥) - \frac{𝜙 (𝑥)}{{‖ 𝑥_{0} ‖}^{2}} 𝜙 (𝑥_{0}) \\ = 𝜙 (𝑥) - \frac{𝜙 (𝑥)}{{‖ 𝑥_{0} ‖}^{2}} {‖ 𝑥_{0} ‖}^{2} = 0 . (by (10.1.3) (10.1.3) (10.1.3)) \end{aligned}

Then $𝜙 (𝑥 - 𝑦) = 0$ , implying $𝑥 - 𝑦 \in ker 𝜙^{⟂}$ , so $⟨ 𝑥 - 𝑦, 𝑥_{0} ⟩ = 0$ , and

⟨ 𝑥, 𝑥_{0} ⟩ = ⟨ 𝑥 - 𝑦, 𝑥_{0} ⟩ + ⟨ 𝑦, 𝑥_{0} ⟩ = 𝜑 (𝑥) .

If there exists another representative $𝑥_{1}$ such that $𝜙 \equiv ⟨ \cdot, 𝑥_{1} ⟩ \equiv ⟨ \cdot, 𝑥_{0} ⟩$ , then

⟨ 𝑥, 𝑥_{1} - 𝑥_{0} ⟩ = 0 . for all 𝑥 \in 𝐻

Letting $𝑥 = 𝑥_{1} - 𝑥_{0}$ implies $𝑥_{1} - 𝑥_{0} = 0$ . ⁠ $□$

Corollary 10.1.36.1.

𝜙

is a continuous linear functional on a Hilbert space

𝐻

, then the orthogonal complement to the null space of

𝜙

has dimension

1

Proof.

By the proof of Theorem 10.1.36

Theorem 10.1.36 Theorem 10.1.36

, for any

𝑞 \in ker 𝜙^{⟂} ∖ {0}

, the Riesz representer

𝑥_{0}

may be constructed by a scaling

𝑞

by a scalar

𝛼_{𝑞} \in ℂ

. Then any

𝑞 \in ker 𝜙^{⟂}

can be expressed as

\frac{𝑥_{0}}{𝛼_{𝑞}}

; then

𝜙^{⟂}

is one-dimensional. ⁠

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