A converse to the Lax Milgram theorem Assume that a banach space $X$ satisfies the Lax Milgram theorem. Must $X$ be isomorphic to a Hilbert space?

If you have a bounded and coercive bilinear form $a:X\to X$ (which is the pre-requisite in Lax-Milgram), e.g., $a$ satisfies $$ a(x,x) \ge c_1\|x\|_X^2, \ |a(x,y)| \le c_2 \|x\|_X\|y\|_Y \quad \forall x,y\in X, $$ then the Hermitian part of $a$ can be used as scalar product on $X$, $$ \langle x,y\rangle_a:=\frac12 \left( a(x,y) + \overline{a(y,x)}\right), $$ and the norm induced by $a$, $$ \|x\|_a:=\sqrt{ \langle x,x\rangle_a} $$ is equivalent to the norm $\|\cdot\|_X$. With this scalar product $X$ itself becomes a Hilbert space