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I am having a hard time to understand how "elementary considerations about bilinear forms" can imply the following result:

Let $E$ be a function space, 1 the constant function $x\mapsto 1$ and let $Q:E\times E\to\Bbb R$ be a bilinear form such that $Q(1,1)=0$ and $Q(\phi,\phi)\geq0$ for all $\phi\in E$. Then $Q(1,\phi)=0$ for all $\phi\in E$.

The reference is the paper Rigidity of Area-Minimizing Free Boundary Surfaces in Mean Convex Three-Manifolds by Lucas C. Ambrozio, page 6:

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  • $\begingroup$ Is $Q$ supposed symmetrical ? $\endgroup$ – NAC Nov 17 '17 at 2:53
  • $\begingroup$ A quick guess is that if $Q$ is supposed symmetrical then by CS $|Q(1,\phi)|^2 \leq Q(1,1)Q(\phi,\phi) = 0$ which implies $Q(1,\phi)=0$. $\endgroup$ – NAC Nov 17 '17 at 2:57
  • $\begingroup$ @NAC I am not sure but I will verify. If this is the case I will let you know, so your comment can turn into a short answer. :-) $\endgroup$ – Filburt Nov 17 '17 at 5:12
  • $\begingroup$ @NAC Yeap, the Second Variation of Area is symmetric. So in fact it follows from CS. I would accept it as an answer if you post it. $\endgroup$ – Filburt Nov 17 '17 at 15:32
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Since $Q$ seems symmetrical, from CS we have : $$ |Q(1,\phi)|^2 \leq Q(1,1)Q(\phi,\phi) = 0 $$ which implies $Q(1,\phi)=0$.

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