I'm thinking about how we can bound every element of a $n\times n$ symmetric positive semidefinite matrix $A$ with $\operatorname{Tr}(A)$? More formally:
Given a nonnegative integer $n$. Is there any non-negative number $k$ such that for every positive semidefinite matrix $A$ and any entry $a_{i,j}$ of $A$ we have:
\begin{align}
\left| a_{i,j}\right| \leq k\cdot\operatorname{Tr}(A)\quad ?
\end{align}
Best Answer
Every 2 by 2 submatrix of form $\begin{pmatrix}a_{ii}&a_{ij}\\a_{ji}&a_{jj}\end{pmatrix}$ (for $i\ne j$) is p.s.d. and hence $a_{ij}^2\le a_{ii} a_{jj}$ and hence $$a_{ij}\le\sqrt{a_{ii} a_{jj}}\le\frac{a_{ii}+a_{jj}}2\le\frac{\operatorname{Tr}(A)}2.$$
A different argument covers the case where $i=j$, left as an exercise.