# 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
• $a$ need not be Hermitian, but a scalar product must. You should take the Hermitian part of $a$. – martini Feb 13 '15 at 9:47
• @daw what about symmetric property? May be $a(x,y)+a(y,x)$? – Ali Taghavi Feb 13 '15 at 9:48
• thanks for the hint, I have been using mostly symmetric $a$'s. – daw Feb 13 '15 at 9:52