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

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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

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

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