Let $A\in \mathbb{R}^{n\times n}$ be a symmetric matrix, and consider the $l_p$ norm ($p\geq 2$). Can we prove that the following problems are equivalent: $$\max_{\|x\|_p=\|y\|_p=1} \left| \langle x, Ay\rangle \right|$$ and $$\max_{\|x\|_p=|} \left| \langle x, Ax\rangle \right| $$
Can the result be generalized to symmetric tensor and symmetric multilinear form? In particular, if $F: R^n\times\cdots\times R^n$ is a $m$-order symmetric multilinear form, can we prove the problem $$\max_{\|x_1\|_p=\cdots=\|x_m\|_p=1} \left| F(x_1,\ldots,x_m) \right|$$ is equivalent to $$\max_{\|x\|_p=1} \left| F(x,\ldots,x) \right| $$ If not, can we prove if for any special case? (the case $p\neq 2$ would be more interesting.)
Thanks for any help!
I think this is a proof when $p=2$; though there may be some mistakes. I agree that the case $p \neq 2$ is more interesting, but I'm having trouble generalizing this. It is straight-forward to show that we get the same result if we take the suprema over vectors with norm $\le 1$. Define $$S_x := \sup_{\| x \|_2 \le 1} \lvert \langle x, Ax \rangle\rvert, \,\,\,\,\, S_{x,y} := \sup_{\| x \|_2, \| y \|_2 \le 1} \lvert \langle x, Ay \rangle\rvert.$$ It is clear that $S_x \le S_{x,y}$. Since $A$ is symmetric, its eigenvectors $\{x_1, \ldots, x_n \}$ can be taken to form an orthonormal basis for $\mathbb R^n$. Say the corresponding eigenvalues (which are real since $A$ is symmetric) are $\{\lambda_1, \ldots, \lambda_n \}$ respectively. For $x, y \in \mathbb R^n$, $\| x \|_2, \| y \|_2 \le 1$, there are $\alpha_j$ and $\beta_j$ such that $$x = \sum_j \alpha_j x_j, \,\,\,\,\,\,\,\,\,\, y = \sum_j \beta_j x_j.$$ The condition that $\| x \|_2, \| y \|_2 \le 1$ is equivalent to $$\sum_j \alpha_j^2 \le 1, \,\,\,\,\,\,\,\, \sum_j \beta_j^2 \le 1$$ Then $$\lvert \langle x, Ay \rangle \rvert = \left \lvert \sum_j \lambda_j \alpha_j \beta_j \right \rvert \le \sum_j \left \lvert \lambda_j \alpha_j \beta_j \right \rvert = \langle \overline x, A \overline x \rangle$$ where $$\overline x = \sum_j \text{sign}(\lambda_j) \sqrt{ \lvert \alpha_j \beta_j \rvert } x_j$$ and $$\| \overline x \|_2^2 = \sum_j \rvert\alpha_j \beta_j\lvert \le \sqrt{ \sum_j \rvert\alpha_j \lvert^2 } \sqrt{ \sum_j \rvert\beta_j\lvert^2 } \le 1.$$ This shows that for any $x,y \in \mathbb R^n$, $\| x \|_2, \| y \|_2 \le 1$, we can find $\overline x \in \mathbb R^n$ with $\| \overline x \|_2 \le 1$ such that $$\lvert \langle x, Ay \rangle \rvert \le \lvert \langle \overline x , A \overline x \rangle \rvert.$$ Taking suprema, this gives $$S_{x,y} \le S_x$$ so we get $S_x = S_{x,y}.$