Complex version of Lax-Milgram Theorem

I'm trying to prove Lax-Milgram Theorem in the complex case, i.e.

Let $X$ be complex Hilbert space and let $f\in X'$, its topological dual. If $a(\cdot,\cdot):X\times X\to \mathbb{C}$ is sesquilinear bounded coercive form, then there exists a unique $u\in X$ such that \begin{equation} a(u,v)=\langle f , v \rangle\qquad\forall\ v\in X. \end{equation}

I'm trying to proceed as in the real case. The only thing that differs is the statement of Riesz Representation Theorem. First of all, I rewrite the variational problem. By Riesz Representation Theorem there exists a unique $w\in X$ such that \begin{equation} \langle f, v \rangle =(v,w)_X\qquad\forall\ v\in X. \end{equation} We need to define a bounded linear bijection $A:X\to X$ such that \begin{equation} (v,w)_X=(v,Au)_X\qquad\forall\ v\in X, \end{equation} for a proper $u\in X$. Any help?

• May the statement of the theorem has to be rewritten as $$a(u,v)=\overline{\langle f,v \rangle}\qquad\forall\ v\in X ?$$ – avati91 Jul 7 '16 at 12:13