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)
$\complex(E)$ is a TVS over $\complex$.
- (2)
$\iota: E \to \complex(E)$ is a continuous $\real$-linear map.
- (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)
$\complex(E) = \iota(E) \oplus i\iota(E)$ as a TVS over $\real$.
The pair $(\complex(E), \iota)$ is the complexification of $E$ as a topological vector space, and
- (5)
If $E$ is locally convex, then so is $\complex(E)$.
- (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.
- (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
then for any $\phi \in [0, 2\pi]$ and $x, y \in E$,
so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$,
Therefore $\iota: E \to \complex(E)$ is isometric.
(F): By (U) applied to $\iota \circ T$.$\square$
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