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/Part 3: Functional Analysis/Chapter 11: Topological Vector Spaces

11.2 Complexification

Definition 11.2.1 (Complexification).label Let $E$ be a vector space over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:

  1. (1)

    $\complex(E)$ is a vector space over $\complex$.

  2. (2)

    $\iota: E \to \complex(E)$ is a $\real$-linear map.

  3. (U)

    For any pair $(F, T)$ satisfying (1) and (2), there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:

    \[\xymatrix{ \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\ E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} & }\]

  4. (4)

    $\complex(E) = \iota(E) \oplus i\iota(E)$ as a vector space over $\real$. For each $z \in \complex(E)$ with $z = x + iy$, $x = \text{Re}(x)$ and $y = \text{Im}(y)$ are the real and imaginary parts of $z$.

The pair $(\complex(E), \iota)$ is the complexification of $E$, and

  1. (F)

    For any vector space $F$ over $\real$ and $\real$-linear map $T: E \to F$, there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:

    \[\xymatrix{ \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\ E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota} }\]

    which is given by

    \[\complex(T)(x + iy) = Tx + iTy\]

Proof. (1): Let $\complex(E) = E \times E$ with coordinate-wise addition. For each $a, b \in \real$ and $x, y \in E$, let

\[(a + bi)(x, y) = (ax - by, bx + ay)\]

then $\complex(E)$ is a vector space over $\complex$.

(2): Let $\iota: E \to \complex(E)$ be defined by $\iota(x) = (x, 0)$, then $\iota$ is $\real$-linear.

(U): Let

\[\complex(T): \complex(E) \to F \quad (x, y) \mapsto Tx + iTy\]

then $\complex(T)$ is the unique $\complex$-linear map such that the given diagram commutes.

(F): By (U) applied to $\iota \circ T$.$\square$

Definition 11.2.2 (Complex Conjugation).label Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a $\real$-linear map, then $*$ is a complex conjugation if:

  1. (C1)

    For each $\lambda \in \complex$, $(\lambda x)^{*} = \ol \lambda x^{*}$.

  2. (C2)

    For each $x \in E$, $x^{**}= x$.

In which case, $\text{Re}(E) = \bracs{x \in E| x^* = x}$ is the real part of $E$.

Proposition 11.2.3.label Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a complex conjugation, then:

  1. (1)

    $E = \complex(\text{Re}(E))$.

  2. (2)

    For each $x \in E$,

    \[\text{Re}(x) = \frac{x + x^{*}}{2}\quad \text{Im}(x) = \frac{x - x^{*}}{2i}\]

  3. (3)

    For each $x \in E$, $x^{*} = \text{Re}(x) - i\text{Im}(x)$.

Proof. (2): By properties of the complex conjugation, $\text{Re}(x), \text{Im}(x) \in \text{Re}(E)$.

(1): For any $x, y \in \text{Re}(x)$ with $x = iy$, $x = -iy$ as well by (2) of the complex conjugation, so $x = y = 0$. Thus if $z = x + iy = x' + iy'$, then $x = x'$ and $y = y'$, and the decomposition is unique.$\square$

Definition 11.2.4 (Hermitian).label Let $E$ be a vector space over $\real$, $*: \complex(E) \to \complex(E)$ be the canonical complex conjugation map, and $\phi \in \hom(\complex(E); \complex)$, then the following are equivalent:

  1. (1)

    $\phi|_{E} \in \hom(E; \real)$.

  2. (2)

    For each $x \in E$, $\dpn{x, \phi}{\complex(E)}= \ol{\dpn{x^*, \phi}{\complex(E)}}$.

If the above holds, then $\phi$ is Hermitian.

Definition 11.2.5 (Complexification of Topological Vector Space).label Let $E$ be a TVS over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:

  1. (1)

    $\complex(E)$ is a TVS over $\complex$.

  2. (2)

    $\iota: E \to \complex(E)$ is a continuous $\real$-linear map.

  3. (U)

    For any pair $(F, T)$ satisfying (1) and (2), there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:

    \[\xymatrix{ \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\ E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} & }\]

  4. (4)

    $\complex(E) = \iota(E) \oplus i\iota(E)$ as a TVS over $\real$.

The pair $(\complex(E), \iota)$ is the complexification of $E$ as a topological vector space, and

  1. (5)

    If $E$ is locally convex, then so is $\complex(E)$.

  2. (F)

    For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:

    \[\xymatrix{ \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\ E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota} }\]

    which is given by

    \[\complex(T)(x + iy) = Tx + iTy\]

    Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))}= \norm{T}_{L(E; F)}$.

Proof. (1), (2): Let $(\complex(E), \iota)$ be the complexification of $E$ as a vector space, and equip it with the direct sum topology.

(U): By (U) of the complexification, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the direct sum.

(4): By Proposition 12.8.3, the direct sum and product of finitely many locally convex spaces coincide. By Proposition 12.7.1, this topology is locally convex.

(F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$.

For the isometry,

\begin{align*}\norm{\complex(T)}_{L(\complex(E); \complex(F))}&= \norm{\complex(T)}\end{align*}

$\square$

Definition 11.2.6 (Complexification of Normed Spaces).label Let $E$ be a normed vector space over $\real$, then

\[\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_{E}\]

is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric.

Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))}= \norm{T}_{L(E; F)}$.

Proof. For any $\phi \in [0, 2\pi]$ and $x, y \in E$,

\begin{align*}\normn{e^{i \phi}(x, y)}_{\complex(E)}&= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)}\\&= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_{E} \\&= \norm{(x, y)}_{\complex(E)}\end{align*}

so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$,

\[\norm{\iota x}_{\complex(E)}= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_{E} = \norm{x}_{E} \\\]

Therefore $\iota: E \to \complex(E)$ is isometric.

Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then

\begin{align*}\norm{\complex(T)}_{L(\complex(E); \complex(F))}&= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1}\\&\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1}= \norm{T}_{L(E; F)}\end{align*}

On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that

\begin{align*}\normn{\complex(T)(x, y)}_{\complex(F)}&= \normn{(Tx, Ty)}_{\complex(F)}= \norm{\cos(\theta)Tx + \sin(\theta)Ty}_{F} \\&\le \norm{T}_{L(E; F)}\cdot \norm{\cos(\theta)x + \sin(\theta)y}_{E} \\&\le \norm{T}_{L(E; F)}\cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_{E} \\&= \norm{T}_{L(E; F)}\cdot \norm{(x, y)}_{E}\end{align*}

$\square$

Proposition 11.2.7.label Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_{E} = \normn{x^*}_{E}$ for all $x \in E$, and $\phi \in E^{*}$ be a Hermitian functional, then

\[\norm{\phi}_{E^*}= \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}\]

Proof. Since $\bracsn{x \in E|x = x^*}\subset E$, $\norm{\phi}_{E^*}\ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$.

On the other hand, let $x \in E$ with $\norm{x}_{E} = 1$. Assume without loss of generality that $\dpn{x, \phi}{E}\in \real$, then

\begin{align*}\dpn{x, \phi}{E}&= \dpn{\text{Re}(x), \phi}{E}+ \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real}= \dpn{\text{Re}(x), \phi}{E}\\&\le \norm{\text{Re}(x)}_{E} \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}\end{align*}

where $\norm{\text{Re}(x)}_{E} = \norm{{(x + x^*)}/{2}}_{E} \le \norm{x}_{E}$. As the above holds for all $x \in E$,

\[\norm{\phi}_{E^*}\le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}\]

$\square$

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