A standard way to define the "complexification" $E_\mathbb{C}$ of a real Banach space $E$ is to define a complex linear structure on $E\times E$ by (1) $(x,y)+(u,v)=(x+u, y+v)$, (2) $(a+ib)(x,y)=(ax-by, bx+ay)$ and a norm by (3) $\|(x,y)\|^\mathbb{C}=\sup_{\theta\in [0,2 \pi]}\|\cos(\theta)x+\sin(\theta)y\|$. (That $\|\cdot\|^\mathbb{C}$ is a norm requires a little proving.)

This definition gives us what we want when going from real $C(K)$ or $l_p$, etc., to their complex versions.

I believe I can show that the (complex) Banach dual of $E_\mathbb{C}$ is the complexification of the real dual $E^*$, i.e., $(E_\mathbb{C})^* = (E^*)_\mathbb{c}$. But my proof is somewhat messy. This must surely be in the literature somewhere! Can anyone suggest a reference?

How about the converse? What if we know that $E_\mathbb{C} = V^*$ for some complex Banach space $V$. Is $E$ the (real) Banach dual of some Banach space? (Again, a reference would be appreciated).

[Edit Feb. 13, 2013] Based on Bill Johnson's comment below, perhaps I should motivate the question a bit, and revise question 2. For a compact space $K$, we know that the Banach space $C(K)$ over the real scalars is isometrically the dual of a real Banach space if and only if $C(K)$ over the complex scalars is isometrically the dual of some complex Banach space. A standard proof is particular to the situation, going through hyperstonian spaces $K$ and normal measures. I was wondering if this is just a special case of a more general fact. So here is a revision of 2.

Q3. Suppose $E$ is a real Banach lattice and the complexification $E_\mathbb{C}$ is a Banach dual space. Must $E$ then be the dual of some real Banach space?

Bases in Banach spaces, Icontains a discussion of the complexification in quite some detail on the first few pages. – Martin Feb 11 '13 at 2:58