# Understanding a conclusion through “elementary considerations about bilinear forms”

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: • Is $Q$ supposed symmetrical ? – NAC Nov 17 '17 at 2:53
• 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 $Q(1,\phi)=0$. – NAC Nov 17 '17 at 2:57
• @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. :-) – Filburt Nov 17 '17 at 5:12
• @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. – Filburt Nov 17 '17 at 15:32

Since $Q$ seems symmetrical, from CS we have : $$|Q(1,\phi)|^2 \leq Q(1,1)Q(\phi,\phi) = 0$$ which implies $Q(1,\phi)=0$.