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/Part 3: Functional Analysis/Chapter 10: Topological Vector Spaces

10.2 Complexification

Definition 10.2.1 (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 10.2.2 (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. (1)

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

  2. (2)

    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 10.2.3.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 $x \in E$,

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

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 10.2.4 (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. (6)

    If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric.

  3. (F)

    For any 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

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

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 11.7.3, the direct sum and product of finitely many locally convex spaces coincide. By Proposition 11.6.1, this topology is locally convex.

(5): Let $\norm{\cdot}_{E}: E \to [0, \infty)$ be the norm of $E$, and define

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

then 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 $\norm{(x, y)}_{\complex(E)}$ is a norm. 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.

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

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