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/Part 3: Functional Analysis/Chapter 12: Normed Vector Spaces

12.6 Hilbert Spaces

Definition 12.6.1 (Sesquilinear Form).label Let $E$ be a vector space over $K \in \RC$ and $\lambda: E \times E \to \complex$, then $\lambda$ is a sesquilinear form if:

  1. (1)

    For each $x, y, z \in E$ and $\mu \in K$, $\lambda(\mu x + y, z) = \mu\lambda(x, z) + \lambda(y, z)$.

  2. (2)

    For each $x, y, z \in E$ and $\mu \in K$, $\lambda(x, \mu y + z) = \ol \mu\lambda(x, y) + \lambda(x, z)$.

Proposition 12.6.2 (Polarisation Identity).label Let $E$ be a vector space over $\complex$, $\lambda: E \times E \to \complex$ be a sesquilinear form, and

\[q: E \to K \quad x \mapsto \lambda(z, z)\]

be its quadratic form, then for any $x, y \in E$,

\begin{align*}\lambda(x, y)&= \frac{1}{4}\braks{q(x + y) - q(x - y) + iq(x + iy) - iq(x - iy)}\\&= \frac{1}{4}\sum_{k = 0}^{3} i^{k} q(x + i^{k}y)\end{align*}

Proof. On the real side,

\begin{align*}q(x + y)&= \lambda(x, x) + \lambda(x, y) + \lambda(y, x) + \lambda(y, y) \\ q(x - y)&= \lambda(x, x) - \lambda(x, y) - \lambda(y, x) + \lambda(y, y) \\ q(x + y) - q(x - y)&= 2[\lambda(x, y) + \lambda(y, x)]\end{align*}

On the imaginary side,

\begin{align*}q(x + iy)&= \lambda(x, x) - i\lambda(x, y) + i\lambda(y, x) + \lambda(y, y) \\ q(x - iy)&= \lambda(x, x) + i\lambda(x, y) - i\lambda(y, x) + \lambda(y, y) \\ q(x + iy) - q(x - iy)&= -2i[\lambda(x, y) - \lambda(y, x)] \\ iq(x + iy) - iq(x - iy)&= 2[\lambda(x, y) + \lambda(y, x)]\end{align*}

Therefore

\[\sum_{k = 0}^{3} i^{k} q(x + i^{k}y) = 4\lambda(x, y)\]

$\square$

Definition 12.6.3 (Hermitian Form).label Let $E$ be a vector space over $K \in \RC$ and $\lambda: E \times E \to K$, then $\lambda$ is a Hermitian form if

  1. (H1)

    For each $x, y, z \in E$, $\lambda(x + y, z) = \lambda(x, z) + \lambda(y, z)$.

  2. (H2)

    For any $x, y \in E$ and $\mu \in K$, $\lambda(\mu x, y) = \mu \lambda(x, y)$.

  3. (H3)

    For every $x, y \in E$, $\lambda(x, y) = \ol{\lambda(y, x)}$.

Proposition 12.6.4 (Polarisation Identity (Real)).label Let $E$ be a vector space over $\real$, $\lambda: E \times E \to \real$ be a Hermitian form, and

\[q: E \to K \quad x \mapsto \lambda(z, z)\]

be its quadratic form, then for any $x, y \in E$,

\[\lambda(x, y) = \frac{1}{4}\braks{q(x + y) - q(x - y)}\]

Proof. For each $x, y \in E$, by (H3),

\begin{align*}q(x + y)&= \dpn{x, x}{E}+ 2 \dpn{x, y}{E}+ \dpn{y, y}{E}\\ q(x - y)&= \dpn{x, x}{E}- 2 \dpn{x, y}{E}+ \dpn{y, y}{E}\end{align*}

so

\[q(x + y) - q(x - y) = 4 \dpn{x, y}{E}\]

$\square$

Definition 12.6.5 (Inner Product).label Let $E$ be a vector space over $K$ and $\inp_{E}: E \times E \to K$, then $\inp_{E}$ is an inner product if:

  1. (H1)

    For each $x, y, z \in E$, $\angles{x + y, z}_{E} = \dpn{x, z}{E}+ \dpn{y, z}{E}$.

  2. (H2)

    For any $x, y \in E$ and $\mu \in K$, $\dpn{\mu x, y}{E}= \mu \dpn{x, y}{E}$.

  3. (H3)

    For every $x, y \in E$, $\dpn{x, y}{E}= \ol{\dpn{y, x}{E}}$.

  4. (I)

    For each $x \in E$, $\dpn{x, x}{E}\ge 0$, with equality if and only if $x = 0$.

Proposition 12.6.6 (Cauchy-Schwarz Inequality).label Let $H$ be a vector space over $K \in \RC$ and $\inp_{H}: E \times E \to K$ be an inner product, then for any $x, y \in H$, $\dpn{x, y}{H}\le \norm{x}_{H}\norm{y}_{H}$.

Proof, [Theorem 5.19, Fol99]. Assume without loss of generality that $\dpn{x, y}{H}> 0$, then for each $t \in \real$,

\[0 \le \dpn{x - ty,x - ty}{H}= \norm{x}_{H}^{2} - 2t\dpn{x, y}{H}+ t^{2}\norm{y}_{H}^{2}\]

which is a quadratic function of $t$ with absolute minumum at $t = \dpn{x, y}{H}/\norm{y}_{H}^{2}$, so

\[0 \le \norm{x}_{H}^{2} - 2\dpn{x, y}{H}^{2}/\norm{y}_{H}^{2} + \dpn{x, y}{H}^{2}/\norm{y}_{H}^{2} = \norm{x}_{H}^{2} - \dpn{x, y}{H}^{2}/\norm{y}_{H}^{2}\]

$\square$

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. (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

    1. (i)

      If $K = \real$, then

      \[\dpn{x, y}{H}= \frac{1}{4}\braks{\norm{x + y}_H^2 - \norm{x - y}_H^2}\]

    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. (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

\[\dpn{x, y}{H}= \frac{1}{4}\braks{\norm{x + y}_H^2 - \norm{x - y}_H^2}\]

then for every $x, y, z \in H$,

\begin{align*}\dpn{x, z}{H}+ \dpn{y, z}{H}&= \frac{1}{4}\braks{\norm{x + z}_H^2 - \norm{x - z}_H^2}\\&+ \frac{1}{4}\braks{\norm{y + z}_H^2 - \norm{y - z}_H^2}\end{align*}

By (2) applied to the pairs $((x + y)/2 + z, (x - y)/2)$ and $((x + y)/2 - z, (x - y)/2)$,

\begin{align*}\norm{x + z}_{H}^{2} + \norm{y + z}_{H}^{2}&= 2\braks{\norm{\frac{x + y}{2} + z}_H^2 + \norm{\frac{x - y}{2}}_H^2}\\ \norm{x - z}_{H}^{2} + \norm{y - z}_{H}^{2}&= 2\braks{\norm{\frac{x + y}{2} - z}_H^2 + \norm{\frac{x - y}{2}}_H^2}\end{align*}

Thus

\begin{align*}\dpn{x, z}{H}+ \dpn{y, z}{H}&= \frac{1}{2}\braks{\norm{\frac{x + y}{2} + z}_H^2 - \norm{\frac{x + y}{2} - z}_H^2}\\&= 2\dpb{\frac{x + y}{2}, z}{H}\end{align*}

and

  1. (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 $\mu \in \real$.

  2. (H1)

    By (H2), $\dpn{x + y, z}{H}= \dpn{x, z}{H}+ \dpn{y, z}{H}$.

  3. (H3)

    By definition, $\inp_{H}$ is symmetric.

  4. (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

\[\dpn{x, y}{H_\real}= \frac{1}{4}\braks{\norm{x + y}_H^2 - \norm{x - y}_H^2}\]

and

\[\dpn{x, y}{H}= \frac{1}{4}\sum_{k = 0}^{3} i^{k} \normn{x + i^ky}_{H}^{2} = \dpn{x, y}{H_\real}+ i \dpn{x, iy}{H_\real}\]

then for any $x, y, z \in H$,

  1. (H1)

    By (H1) of the real case, $\dpn{x + y, z}{H}= \dpn{x, z}{H}+ \dpn{y, z}{H}$.

  2. (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*}

  3. (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.

  4. (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 $x, y \in H$,

\begin{align*}\norm{x + y}_{H}^{2}&= \dpn{x + y, x + y}{H}\\&= \norm{x}_{H}^{2} + \dpn{x, y}{H}+ \dpn{y, x}{H}+ \norm{y}_{H}^{2} \\ \norm{x - y}_{H}^{2}&= \norm{x}_{H}^{2} - \dpn{x, y}{H}- \dpn{y, x}{H}+ \norm{y}_{H}^{2}\end{align*}

so

\[\norm{x + y}_{H}^{2} + \norm{x - y}_{H}^{2} = 2\norm{x}_{H}^{2} + 2\norm{y}_{H}^{2}\]

By the Cauchy-Schwarz inequality,

\begin{align*}\norm{x + y}_{H}^{2}&= \norm{x}_{H}^{2} + 2\text{Re}\dpn{x, y}{H}+ \norm{y}_{H}^{2} \\&\le \norm{x}_{H}^{2} + 2\norm{x}_{H}\norm{y}_{H} + \norm{y}_{H}^{2} \\&= (\norm{x}_{H} + \norm{y}_{H})^{2}\end{align*}

so $\norm{\cdot}_{H}$ satisfies the triangle inequality, and is a norm.$\square$

Definition 12.6.8 (Orthogonal).label Let $H$ be an inner product space and $x, y \in H$, then $x$ and $y$ are orthogonal, denoted $x \perp y$, if $\dpn{x, y}{H}= 0$.

For any $A \subset H$, $A$ is pairwise orthogonal if for any $x, y \in A$, $x \perp y$.

Definition 12.6.9 (Orthogonal Projection).label Let $H$ be an inner product space and $P \in L(H; H)$, then $P$ is an orthogonal projection if:

  1. (1)

    $P$ is idempotent.

  2. (2)

    For any $x, y \in H$, $\dpn{Px, y}{H}= \dpn{x, Py}{H}$.

Definition 12.6.10 (Orthogonal Complement).label Let $H$ be an inner product space and $A \subset H$ be a subspace, then the closed subspace

\[A^{\perp} = \bracsn{x \in H| \dpn{x, y}{H} = 0 \forall y \in A}\]

is the orthogonal complement of $H$.

Theorem 12.6.11 (Pythagorean Theorem).label Let $H$ be an inner product space over $K \in \RC$ and $\seqf{x_j}\subset H$ be pairwise orthogonal, then

\[\norm{\sum_{j = 1}^n x_j}_{H}^{2} = \sum_{j = 1}^{n} \norm{x_j}_{H}^{2}\]

Theorem 12.6.12.label Let $H$ be a Hilbert space over $K \in \RC$ and $A \subset H$ be a closed subspace, then:

  1. (1)

    For each $x \in H$ and $y \in A$, then $\norm{x - y}_{H} = d(x, A)$ if and only if $x - y \perp A$.

  2. (2)

    For any $x \in H$ and $y, z \in A$,

    \[\norm{y - z}_{H}^{2} \le 2\norm{y - x}_{H}^{2} + 2\norm{z - x}_{H}^{2} - 4d(x, A)^{2}\]

  3. (3)

    For each $x \in H$, there exists a unique $x_{A} \in A$ such that $\norm{x - x_A}_{H} = d(x, A)$, and the mapping $P_{A}: H \to A$ with $x \mapsto x_{A}$ is an orthogonal projection.

  4. (4)

    $H = A \oplus A^{\perp}$ as a direct sum of closed subspaces, where for every $x \in H$, $x = P_{A}x + P_{A^\perp}x$.

Proof, [Theorem 5.24, Fol99]. (1): Suppose that $\norm{x - y}_{H} = d(x, A)$. Let $z \in A$ and assume without loss of generality that $\dpn{y, z}{H}> 0$, then

\[f(t)= \norm{(x - y) + tz}_{H}^{2} = \norm{(x-y)}_{H}^{2} + 2t \dpn{(x-y), z}{H}+ t^{2}\norm{z}_{H}^{2}\]

is a quadratic function. Since $y - tz \in A$ for all $t \in \real$, it has a minimum at $t = 0$. Therefore $\dpn{(x - y), z}{H}= 0$, and $x - y \perp A$.

Now suppose that $x - y \perp A$, then for any $z \in A$,

\[\norm{x - z}_{H}^{2} = \norm{x - y}_{H}^{2} + \norm{y - z}_{H}^{2} \ge \norm{x - y}_{H}^{2}\]

by the Pythagorean Theorem.

(2): By the parallelogram law,

\[2(\norm{x - y}_{H}^{2} + \norm{x - z}_{H}^{2}) = \norm{y - z}_{H}^{2} + \norm{y + z - 2x}_{H}^{2}\]

Since $(y + z)/2 \in A$,

\begin{align*}\norm{y - z}_{H}^{2}&= 2(\norm{x - y}_{H}^{2} + \norm{x - z}_{H}^{2}) - 4\norm{(y + z)/2 - x}_{H}^{2} \\&\le 2\norm{x - y}_{H}^{2} + \norm{x - z}_{H}^{2} - 4d(x, A)^{2}\end{align*}

(3): Let $\seq{y_n}\subset A$ with $\norm{y_n - x}_{H} \to d(x, A)$ as $n \to \infty$, $\seq{y_n}$ is a Cauchy sequence by (2). Since $H$ is complete and $A$ is a closed subspace, there exists $x_{A} \in A$ such that $y_{n} \to x_{A}$ as $n \to \infty$ by Proposition 6.7.3. In which case, by continuity of the norm, $\norm{x_A - x}_{H} = d(x, A)$. By (1), $x_{A}$ is the unique element of $A$ such that $\norm{x_A - x}_{H} = d(x, A)$.

Let $x, y \in H$ and $\lambda \in K$, then $(\lambda x + y) - (\lambda P_{A}x + P_{A}y) \perp A$, so by (1), $P_{A}(\lambda x + y) = \lambda P_{A}x + P_{A}y$, and $P_{A}$ is linear. In addition,

\begin{align*}\dpn{P_Ax, y}{H}&= \dpn{P_Ax, P_Ay}{H}+ \dpn{P_Ax, y - P_Ay}{H}= \dpn{P_Ax, P_Ay}{H}\\&= \dpn{P_Ax, Py}{H}+ \dpn{x - P_Ax, P_Ay}{H}= \dpn{x, P_Ay}{H}\end{align*}

so $P_{A}$ is self-adjoint.

(4): By (1), for each $x \in H$, $P_{A^\perp}x = x - P_{A}x$. Since $A \cap A^{\perp} = \bracs{0}$, $H = A \oplus A^{\perp}$.$\square$

A significant property of Hilbert spaces is that every closed subspace is complemented by another closed subspace. It turns out that this is a defining property:

Theorem 12.6.13 (Lindenstrauss-Tzafriri, 1971).label Let $E$ be a Banach space. If for every closed subspace $A \subset E$, there exists a closed subspace $A^{\perp} \subset E$ such that $E = A \oplus A^{\perp}$, then $E$ is isomorphic to a Hilbert space.

Proof. See [LT71].$\square$

Theorem 12.6.14 (Riesz Representation Theorem).label Let $H$ be a Hilbert space over $K \in \RC$. For each $x \in H$, let

\[\phi_{x}: H \to \complex \quad \phi_{x}(y) = \dpn{y, x}{E}\]

then the mapping $H \to H^{*}$ defined by $x \mapsto \phi_{x}$ is an isometric conjugate linear isomorphism.

Proof. By the Cauchy-Schwarz inequality and definition of the norm, $x \mapsto \phi_{x}$ is an isometric conjugate linear map.

Let $\phi \in H^{*}$, then $\ker \phi$ is a closed subspace of codimension one. By Theorem 12.6.12, $(\ker \phi)^{\perp} \ne \emptyset$. Let $y \in (\ker \phi)^{\perp}$ with $\norm{y}_{H} = 1$, then for any $x \in H$,

\begin{align*}\phi(x)&= \phi(P_{\ker \phi}x) + \phi(P_{(\ker \phi)^\perp}x) \\&= \dpn{P_{\ker \phi}x, \phi(y)y}{H}+ \dpn{P_{(\ker \phi)^\perp}x, \phi(y)y}{H}= \dpn{x, \phi(y)y}{H}\end{align*}

$\square$

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