Extending real functional to complex functional? Do I need Hahn-Banach? I'm trying to extend a real functional $f:V \rightarrow \mathbb{R}$ ($V$ is a complex vector space, boundedness of $f$ not known a priori) to the complex functional $f_{\mathbb{C}}: V \rightarrow \mathbb{C}$.
If I need H-B, then how can I discover an upper bound needed for Hahn-Banach theorem? Particularly, I need to display that $f$ is dominated by some function $p$, i.e. that $f \leq p$ on a subspace of $V$.
How to do this? Does it follow from subspace properties?
 A: If $V$ is a normed space and $f:V\to\mathbb{R}$ is a bounded linear functional, then 
$$
u(x) = f(x)-if(ix)
$$
is a bounded linear functional from $V$ to $\mathbb{C}$ and
$$
\lVert f\rVert = \lVert u\rVert.
$$
The above result is from proposition 5.5. in Folland's Real Analysis book and does not rely on Hahn-Banach.
Note: Conversely, if $u : V\to\mathbb{R}$ is a linear functional such that $\Re(u) = f$, then $u(x)= f(x)-if(ix)$. 
A: Maybe I don't understand your question, but what do you mean by "extending real functional to complex functional". Can you be more precise on what you are doing in particular? 
One of the versions of Hahn-Banach theorem says that given a real (or complex) continuous functional on vector subspace $W$ of normed vector space $V$, you can extend it in a continuous functional defined on the whole of $V$.
Here continuity refers to boundedness, as more generally continuity of linear operator in linear topological space is equivalent to boundedness. In our case this means that you should be able to find a constant $C>0$ such that for each $x\in W$ one has $\|f(x)\| \leq C \|x\|$.
