Hadamard product: Optimal bound on operator norm Let $A,B$ be $n\times n $ matrices and denote by $A\star B$ the Hadamard product $(A\star B)(i,j)=A(i,j)B(i,j)$ (pointwise matrix multiplication). For $A$ positive definite it is known that 
$$\|A\star B\| \leq \sup_{i,j} |A(i,j)| \|B\|.$$
My question is what happens if we drop the positive definiteness assumption, i.e. what is the best constant $C>0$ such that
$$\|A\star B\| \leq C \sup_{i,j} |A(i,j)| \|B\|$$
holds for arbitrary $n\times n$ matrices $A,B$. Is the constant $C$ independent of the size of the matrix $n$? 
 A: The optimal such $C$ is $\sqrt{n}$, and so in particular there does not exist any such $C$ that is independent of $n$.
First, let me prove that $C=\sqrt{n}$ works.  Let $e_1,\dots,e_n$ be the standard basis vectors, let $v=\sum v_ie_i$ be any vector, and let $a=\sup_{i,j}|A(i,j)|$.  Note that for each $i$, $\|(A\star B)e_i\|\leq a\|B\|$, since $\|Be_i\|\leq \|B\|$ and $(A\star B)e_i$ is obtained from $Be_i$ by multiplying each entry by a scalar of size at most $a$. Thus $$\|(A\star B)v\|\leq\sum|v_i|\|(A\star B)e_i\|\leq a\|B\|\sum|v_i|.$$  By Cauchy-Schwarz, $\sum|v_i|\leq \sqrt{n}\|v\|$, so we conclude that $\|A\star B\|\leq a\|B\|\sqrt{n}$ and $C=\sqrt{n}$ works.
To prove $C=\sqrt{n}$ is optimal, let $\omega$ be a primitive $n$th root of unity and let $B(i,j)=\frac{\omega^{ij}}{\sqrt{n}}$.  The columns of $B$ are orthonormal, so $B$ is unitary and $\|B\|=1$.  Now let $A(i,j)=\omega^{-ij}$, so that $A\star B$ is the matrix whose entries are all $\frac{1}{\sqrt{n}}$.  We have $\|A\star B\|=\sqrt{n}$ (the vector of all $1$s is an eigenvector of $A\star B$ with eigenvalue $\sqrt{n}$).  Since $|A(i,j)|=1$ for all $i,j$, this means we must have $C\geq \sqrt{n}$.
