Upper-bound for nuclear norm of $A \circ (v \otimes v)$ in terms of operator norm (or nuclear norm) of matrix $A$ and $L_\infty$-norm of vector $v$.

Question

Let $A \in \mathbb R^{n \times }$ be a psd matrix such that $\|A\|_{op} \le r_1$ and $\|A\|_{*} \le r_2$. Let $v \in \mathbb R^n$ such that $\|v\|_\infty \le r_3$. Let $B:=A \circ V$ be the Hadamard product of $A$ and the outer-product $V := v \otimes v$ of $v$ with itself.

Question. Is there a good generic upper-bound for the nuclear norm $\|B\|_*$ of $B$ in terms of $r_3$ and $r_1$ (or $r_2$) ?

Answer 1

As initially observed by user Ben Grossmann in the comments, one has $\|B\|_\star \le r_2 r_3^2$.

Indeed, if $D = diag(v)$, then $B = DAD$ and so $$ \|B\|_\star = tr(B) = tr(DAD) = tr(AD^2) \le tr(A)\|D\|_{op}^2 = \|A\|_\star\|D\|_{op}^2 \le r_2 r_3^2, $$ where we have used the fact that $A$ and $D$ are symmetric psd matrices (thus so is $B$).

Moreover, this inequality is tight as can be seen by taking $v=1_n:=(1,1,\ldots,1) \in \mathbb R^n$, so that $B=A$.