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*}

$\square$

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