Realification of Topological Vector Space is Topological Vector Space

Theorem
Let $\struct {X, \tau}$ be a topological vector space over $\C$.

Let $X_\R$ be the realification of $X$.

Then $\struct {X_\R, \tau}$ is a topological vector space over $\R$.

Proof
Since $\struct {X, \tau}$ is a topological vector space, the vector addition map $+_X : X \times X \to X$ is continuous.

Note that the underlying sets and topologies of $X_\R$ and $X$ are identical, we are only restricting the scalar field.

So the vector addition map $+_{X_\R} : X_\R \times X_\R \to X_\R$ is also continuous.

Again since $\struct {X, \tau}$ is a topological vector space, the scalar multiplication map $\circ_X : \C \times X \to X$.

By the definition of realification, the scalar multiplication map $\circ_{X_\R} : \R \times X \to X$ on the realification is precisely the restriction of $\circ_X$ to $\R \times X$.

From Restriction of Continuous Mapping is Continuous, $\circ_{X_\R}$ is therefore continuous.

The topology on $X$ and $X_{\R}$ are the same, so we conclude that $\struct {X_\R, \tau}$ is Hausdorff.

So $\struct {X_\R, \tau}$ is a topological vector space over $\R$.