Definition 12.6.7 (Hilbert Space).label Let $(H, \norm{\cdot}_{H})$ be a normed vector space over $K \in \RC$, then the following are equivalent:
- (1)
There exists a unique inner product $\inp_{H}: H \times H \to K$ such that for every $x \in H$, $\norm{x}_{H} = \dpn{x, x}{H}^{1/2}$, and
- (i)
If $K = \real$, then
\[\dpn{x, y}{H}= \frac{1}{4}\braks{\norm{x + y}_H^2 - \norm{x - y}_H^2}\] - (ii)
If $K = \complex$, then
\[\dpn{x, y}{H}= \frac{1}{4}\sum_{k = 0}^{3} i^{k} \normn{x + i^ky}_{H}^{2}\]
- (2)
(Parallelogram Law) For each $x, y \in H$,
\[\norm{x + y}_{H}^{2} + \norm{x - y}_{H}^{2} = 2\norm{x}_{H}^{2} + 2\norm{y}_{H}^{2}\]
If the above holds, then $(H, \inp_{H})$ is an inner product space. If in addition $H$ is complete, then $(H, \inp_{H})$ is a Hilbert space.
Proof, [Theorem I.5.1, Yos12]. (2) $\Rightarrow$ (1, Real): For each $x, y \in H$, let
then for every $x, y, z \in H$,
By (2) applied to the pairs $((x + y)/2 + z, (x - y)/2)$ and $((x + y)/2 - z, (x - y)/2)$,
Thus
and
- (H2)
By repeated application of the above equation, $\dpn{\mu x, z}{H}= \mu \dpn{x, z}{H}$ for all dyadic rational numbers $\mu \in \mathbb{D}$. Since the norm is continuous, $\dpn{\mu x, z}{H}= \mu \dpn{x, z}{H}$ for all $\mu \in \real$.
- (H1)
By (H2), $\dpn{x + y, z}{H}= \dpn{x, z}{H}+ \dpn{y, z}{H}$.
- (H3)
By definition, $\inp_{H}$ is symmetric.
- (I)
By non-negativity of the norm,
\[\dpn{x, x}{H}= \frac{1}{4}\braks{\norm{2x}_H^2 + 0}= \norm{x}_{H}^{2} \ge 0\]with equality if and only if $x = 0$.
Therefore $\inp_{H}$ is a real-valued inner product with $\norm{x}_{H} = \dpn{x, x}{H}^{1/2}$ for all $x \in H$. By the polarisation identity, this inner product is unique.
(2) $\Rightarrow$ (1, Complex): For each $x, y \in H$, let
and
then for any $x, y, z \in H$,
- (H1)
By (H1) of the real case, $\dpn{x + y, z}{H}= \dpn{x, z}{H}+ \dpn{y, z}{H}$.
- (H3)
By absolute homogeneity of the norm, $\dpn{ix, iy}{H_\real}= \dpn{x, y}{H_\real}$, so
\begin{align*}\dpn{x, y}{H}&= \dpn{x, y}{H_\real}+ i \dpn{x, iy}{H_\real}\\&= \dpn{x, y}{H_\real}- i \dpn{ix, y}{H_\real}\\&= \dpn{y, x}{H_\real}- i \dpn{y, ix}{H_\real}= \ol{\dpn{y, x}{H}}\end{align*} - (H2)
By absolute homogeneity of the norm,
\begin{align*}\dpn{ix, y}{H}&= \dpn{ix, y}{H_\real}+ i \dpn{ix, iy}{H_\real}\\&= -\dpn{x, iy}{H_\real}+ i\dpn{x, y}{H_\real}= i\dpn{x, y}{H}\end{align*}so $\inp_{H}$ satisfies (H2) by (H2) of the real case.
- (I)
By absolute homogeneity of the norm,
\begin{align*}\dpn{x, ix}{H_\real}&= \frac{1}{4}\braks{\norm{x + ix}_H^2 - \norm{x - ix}_H^2}\\&= \frac{|1 + i|^{2} - |1 - i|^{2}}{4}\norm{x}_{H}^{2} = 0\end{align*}so
\[\dpn{x, x}{H}= \dpn{x, x}{H_\real}+ i \dpn{x, ix}{H_\real}= \dpn{x, x}{H_\real}= \norm{x}_{H}^{2}\]
Therefore $\inp_{H}$ is a complex-valued inner product with $\norm{x}_{H} = \dpn{x, x}{H}^{1/2}$ for all $x \in H$. By the polarisation identity, this inner product is unique.
(1) $\Rightarrow$ (2): For each $x, y \in H$,
so
By the Cauchy-Schwarz inequality,
so $\norm{\cdot}_{H}$ satisfies the triangle inequality, and is a norm.$\square$
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