Here are Hahn-Banach theorem and corollary of this.
Hahn-Banach theorem
$X$: linear space on $\mathbb R$
$p:X\to \mathbb R$ satisfies $p(x+y)\leqq p(x)+p(y)$ and $p(ax)=ap(x) (a>0)$
$L\subset X$ : linear subspace
$f : L\to \mathbb R$ : linear mapping
$f(x)\leqq p(x)$ for $x\in L$
Then, there exists linear mapping $F : X\to \mathbb R$ s.t. $F|_{L}=f$ and $F(x)\leqq p(x)$ for $x\in X$.
corollary
$X$: linear space on $K=\mathbb R$ or $\mathbb C$
$p:X\to \mathbb R$ : seminorm i.e., $p$ satisfies $p(x)\geqq 0$ and $p(ax)=|a|p(x)$ and $p(x+y)\leqq p(x)+p(y)$ for $x,y\in X, a\in K$.
$L\subset X$ : linear subspace
$f : L\to K$ ; linear mapping
$|f(x)|\leqq p(x)$ for $x\in L$
Then, there exists a linear mapping $F : X\to K$ s.t. $F|_L=f$ and $|F(x)|\leqq p(x)$ for $x\in X$.
I'm trying to understand the proof of this corollary using Hahn-Banach theorem, but there is a point I don't understand.
Proof of corollary
If $K=\mathbb R$, the statement is almost obvious.(I understand this.)
Consider the case $K=\mathbb C$.
Then, define $g : L\to \mathbb R$ as $x\mapsto$ Re $f(x)$.
$g$ is linear about scalar $\mathbb R$, i.e., $g(x+y)=g(x)+g(y)$ for $x,y\in X$ and $g(ax)=ag(x)$ for $a\in \mathbb R, x\in X$.
Now, restrict the scalar of $X$ to $\mathbb R.$
Then,
$X$: linear space on $\mathbb R$
$p:X\to \mathbb R$ : seminorm
$L\subset X$ : linear (about $\mathbb R$) subspace
$g : L\to \mathbb R$ : linear (about $\mathbb R$) mapping
$g(x)=$ Re$f(x) \leqq |f(x)|\leqq p(x)$ for $x\in L$
Thus, from Hahn-Banach, there exists linear mapping $G : X\to \mathbb R$ s.t. $G|_{L}=g$ and $G(x)\leqq p(x)$ for $x\in X$.
Then, define $F : X\to \mathbb C$ as $F(x)=G(x)-i G(ix)$. $\cdots (\ast)$
This $F$ works.
I don't know why we can define $G(ix)$ in $(\ast)$.
$G$ is defined as linear mapping from $X$ to $\mathbb R$, and this $X$ should be seen as the space on $\mathbb R$ (since we restrict scalar of $X$ to $\mathbb R$).
In order to define $G(ix)$, $ix$ has to be in $X$ (since the domain of $G$ is $X$), but now, $i$ is not real number and $X$ is linear space on $\mathbb R$, I think $ix$ is not in $X$.
Could you explain about this confusion ?
Each vector space over $\Bbb C$ can be treated as a vector space over $\Bbb R$. We don't change the set $X$, we just restrict our field.
For example, the set $\Bbb C$ is a one-dimensional space over $\Bbb C$, but can be also treated as a two-dimensional space over $\Bbb R$. Indeed, if we have $u,v\in \Bbb C$ and $a,b\in\Bbb R$ then $au+bv\in\Bbb C$. This shows that $\Bbb C$ is indeed a space over $\Bbb R$.
Generalisation: the space $\Bbb C^n$ is an $n$-dimensional space over $\Bbb C$ but it as well can be considered as a $2n$-dimensional space over $\Bbb R$.
In order to use Hahn-Banach theorem we have to treat the space $X$ as the vector space over $\Bbb R$. Then we are in a position of Hahn-Banach theorem and we obtain $G$. Then we start treating $X$ as a space over $\Bbb C$ and define $F$.