Bound for Double Sum.

Question

For $i = 1,\cdots,n$ and $k=1,\cdots,m$, $a_{ij} > 0$ and $b_{ij} \geq 0$. Do we have the following bound, $$ \sum_{i=1}^n \sum_{j=1}^m a_{ij} b_{ij} \leq \Big( \max_i \sum_j a_{ij} \Big) \Big( \max_j \sum_i b_{ij} \Big) ?$$

Answer 1

No. For $n=m=2$ and $$ A = B = \begin{pmatrix} 1 & x \\ x & 1 \end{pmatrix} \quad $$ the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.

Answer 2

No, since $a_{ij}=b_{ij}=\delta_{ij}+\epsilon$ for small $\epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $\epsilon \to 0$.) To make your statement true, it should be $$ \sum_{i,j} a_{ij}b_{ij} \le \color{red}{\sqrt{nm}} \cdot\max_i\sum_j a_{ij} \max_j \sum_i b_{ij}. $$