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Thanks to Riesz representation theorem, a continues bilinear (sesquilinear) form on Hilbert space $$a: \mathcal H\times \mathcal H\rightarrow\mathbb R \ \ (\text{or} \ \ \mathbb C)$$ can be represented by a linear and continuous operator $S: \mathcal H \rightarrow \mathcal H$, ie $$a(u,v)=(Su,v) \ \ \forall u,v\in \mathcal H$$ Often I read that bilinear form $a$ is symmetric if and only if the operator $S$ is self-adjoint but, evidently, it is a well known result because I never find its proof. Where can I find the proof of this statement?

Thanks in advance.

Mark
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    In the complex case, $a$ must be sesquilinear, not bilinear. A proof is simple, just look at $a(u,v) - \overline{a(v,u)}$ (ignore the conjugation for the real case). – Daniel Fischer Jun 30 '15 at 15:52

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In the complex case, you have $(Su,v)=\overline {(Sv,u)} $ for all $u,v$. Then $$ (Su,v)=\overline {(Sv,u)}=(u,Sv)=(S^*u,v). $$ So $(\, (S-S^*)u,v)=0$ for all $v $, which implies $(S-S^*)u=0$. As this occurs for all $u $, $S-S^*=0$.

For the real case, just remove the bars.

Martin Argerami
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