Optimum is achieved when both variables are equal Consider the problem $\max_{y,z:\|y\|_\infty,\|z\|_\infty \leq 1}y^TBz$, where $B$ is symmetric, positive semidefinite, $y,z\in \mathbb{R}^n$, $\|z\|_\infty=\max_{i\in\{1,\ldots,n\}}|z_i|$.
It turns out the max is always achieved when $y=z$ (so we can simplify to the problem $\max_{z:\|z\|_\infty \leq 1}z^TBz$), but I'd like to convince myself this is true.
I am reading through some optimization notes and this claim was made. 
One route I am considering right now is to fix $z$ s.t. $\|z\|_\infty \leq 1$.  But I'm having justifying why $\max_{y:\|y\|_\infty \leq 1}y^TBz=z^TBz$
 A: Lemma: If $B$ is positive semidefinite then 
$$
\max_{\|y\|_\infty\le 1,\|z\|_\infty\le 1}y^TBz=\max_{\|z\|_\infty\le 1}z^TBz.
$$
Proof: 
$\fbox{$\le$}$ Since $(y-z)^TB(y-z)\ge 0$ we have for $\|y\|_\infty\le 1$ and $\|z\|_\infty\le 1$
$$
2y^TBz\le y^TBy+z^TBz\le 2\max_{\|z\|_\infty\le 1}z^TBz\quad
\Rightarrow\quad y^TBz\le \max_{\|z\|_\infty\le 1}z^TBz\quad
\Rightarrow\quad 
$$
$$
\max_{\|y\|_\infty\le 1,\|z\|_\infty\le 1}y^TBz\le\max_{\|z\|_\infty\le 1}z^TBz.
$$
$\fbox{$\ge$}$ For $\|z\|_\infty\le 1$ we have
$$
z^TBz\le \max_{\|y\|_\infty\le 1}y^TBz\le 
\max_{\|y\|_\infty\le 1,\|z\|_\infty\le 1}y^TBz\quad\Rightarrow\quad
\max_{\|z\|_\infty\le 1}z^TBz\le\max_{\|y\|_\infty\le 1,\|z\|_\infty\le 1}y^TBz.
$$

P.S. Actually, exactly the same proof works if we replace the conditions $\|y\|_\infty\le 1$ and $\|z\|_\infty\le 1$ with $y,z\in K\subset\mathbb{R}^n$ for any set $K$. So the maximum of a bilinear form being the same as of the corresponding quadratic form is a structural property of positive semidefinite matrices.
