# Prove that: $\displaystyle\int_\Omega (Av)\cdot (Av)\,dx\leq \int_\Omega (Av)\cdot (\|A\|_{L^\infty(\Omega)}v)\,dx$

$A$ is a symmetric tensor in $L^\infty(\Omega)^{n\times n}$, with $\Omega$ a bounded domain in $\mathbb{R}^n$.

Prove that

$\displaystyle\int_\Omega (Av)\cdot (Av)\,dx\leq \int_\Omega (Av)\cdot (\|A\|_{L^\infty(\Omega)}v)\,dx$

where $v\in L^2(\Omega)^n$.

• I edited my question with a specific space where live the vector. – yemino Aug 16 '16 at 21:52

This boils down to the question, whether for symmetric $A$ $$\|Ax\|_2^2\le \|A\|_\infty \cdot x^TAx$$ holds for all $x$.
This is not true without positive definiteness of $A$ (take $A=-I$).
It is true for positive definite $A$: $$\|Ax\|_2^2 =\|A^{1/2}A^{1/2}x\|_2^2 \le \|A^{1/2}\|_2^2 \cdot \|A^{1/2}x\|_2^2 \le \|A^{1/2}\|_2^2\cdot x^TAx.$$ Since $A^{1/2}$ is symmetric positive definite $$\|A^{1/2}\|_2^2 = \lambda_\max(A^{1/2})^2 = \lambda_\max(A) \le \|A\|_\infty,$$ which is the claim. Note that we can use any matrix norm on $\mathbb R^{n,n}$ instead of $\|\cdot\|_\infty$.