Characterisation of operator norm on $l^2$ Let $A$ be a bounded operator on $l^2(\mathbb C)$ and let $\{e_i\}$ be some orthonormal basis. Is there a sharp characterisation of the operator norm $\|A\|_{\mathcal L(l^2(\mathbb C))}$ in terms of the matrix elements $A_{ij}:=\langle e_i,A e_j\rangle$?

It is straightforward to show that 
$$\sup_k\sum_i|A_{ik}|^2\leq \|A\|_{\mathcal L(l^2(\mathbb C))}^2 \leq \sum_{i,k}|A_{ik}|^2, $$

but simple counterexamples show that neither of the two inequalities can be sharp.
There isn't, even in finite dimension. For instance, let us consider $4\times 4$ matrices where four entries are $1$ and the rest are zero. Let 
$$
A_1=\begin{bmatrix} 
1&amp;0&amp;0&amp;0\\ 
0&amp;1&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;1\\
\end{bmatrix},
\ \ 
A_2=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
0&amp;0&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;1\\
\end{bmatrix},
$$
$$
A_3=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
0&amp;1&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;0\\
\end{bmatrix},
\ \  
A_4=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
1&amp;1&amp;0&amp;0\\
0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0\\
\end{bmatrix}.
$$
Then
$$
\|A_1\|=1,\ \ \ \|A_2\|=\sqrt2,\ \ \ \|A_3\|=\sqrt{\frac{3+\sqrt5}{2}},\ \ \|A_4\|=2.
$$
There isn't, even in finite dimension. For instance, let us consider $4\times 4$ matrices where four entries are $1$ and the rest are zero. Let 
$$
A_1=\begin{bmatrix} 
1&amp;0&amp;0&amp;0\\ 
0&amp;1&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;1\\
\end{bmatrix},
\ \ 
A_2=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
0&amp;0&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;1\\
\end{bmatrix},
$$
$$
A_3=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
0&amp;1&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;0&amp;0\\
\end{bmatrix},
\ \  
A_4=\begin{bmatrix} 
1&amp;1&amp;0&amp;0\\ 
1&amp;1&amp;0&amp;0\\
0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0\\
\end{bmatrix}.
$$
Then
$$
\|A_1\|=1,\ \ \ \|A_2\|=\sqrt2,\ \ \ \|A_3\|=\sqrt{\frac{3+\sqrt5}{2}},\ \ \|A_4\|=2.
$$