An inequality related to real matrices

Question

Let $A=\{a_{i,j}\}_{1\leq i,j\leq n}$ be a square real-valued matrix and $M>0$ such that $$\left| \sum_{1\leq i,j\leq n}a_{i,j}t_{i}s_{j}\right|\leq M $$ holds for every real-valued sequences $\{t_{i}\}_{i=1}^{n}$ and $\{s_{j}\}_{j=1}^{n}$ satisfying $\left| t_{i}\right|\leq 1$ and $\left| s_{j}\right|\leq 1$ for all $1\leq i,j\leq n$.

My question is as follows:

Let $B=\{b_{i,j}\}_{1\leq i,j\leq n}$ be a real-valued matrix satisfying $\left| b_{i,j}\right|\leq 1$ for all $1\leq i,j\leq n$. Prove that $$\left| \sum_{1\leq i,j\leq n}a_{i,j}b_{i,j}t_{i}s_{j}\right|\leq M $$ holds for every real-valued sequences $\{t_{i}\}_{i=1}^{n}$ and $\{s_{j}\}_{j=1}^{n}$ satisfying $\left| t_{i}\right|\leq 1$ and $\left| s_{j}\right|\leq 1$ for all $1\leq i,j\leq n$.

I'm trying to express $b_{i,j}$ as $b_{i,j}=c_{i}d_{j}$ for all $1\leq i,j\leq n$. But, in general, this expression is incorrect unless $rank(B)=1$. Is there another way to prove above inequality?

Renew my question: Thank JimmyK4542 for giving the counterexample! Under what condition does above conclusion hold?

Answer 1

Counterexample: For $n = 2$, define the matrix $A$ by $a_{1,1} = a_{2,1} = a_{2,2} = 1$, and $a_{1,2} = -1$, define the matrix $B$ by $b_{1,1} = b_{2,1} = b_{2,2} = 1$, and $b_{1,2} = -1$, and pick $M = 2\sqrt{2}$.

For any $-1 \le t_1,t_2,s_1,s_2 \le 1$, we can use the Cauchy-Schwarz Inequality to get

$\displaystyle\left|\sum_{1 \le i,j \le 2}a_{i,j}t_is_j\right| = |t_1s_1-t_1s_2+t_2s_1+t_2s_2| = |t_1(s_1-s_2)+t_2(s_1+s_2)|$

$\le \sqrt{t_1^2+t_2^2} \cdot \sqrt{(s_1-s_2)^2+(s_1+s_2)^2} = \sqrt{t_1^2+t_2^2} \cdot \sqrt{2(s_1^2+s_2^2)} \le 2\sqrt{2} = M$.

However, for $t_1 = t_2 = s_1 = s_2 = 1$, we have

$\displaystyle\left|\sum_{1 \le i,j \le 2}a_{i,j}b_{i,j}t_is_j\right| = |t_1s_1+t_1s_2+t_2s_1+t_2s_2| = 4 > 2\sqrt{2} = M$.