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Let $A$ be an $n×m$ matrix and $T : \mathbb{R}^m\to\mathbb{R}^n$ be the linear transformation $T: x\to Ax$. Let $N$ (resp. $M$) be a norm on $\mathbb{R}^n$ (resp. $\mathbb{R}^m$) equivalent to the $\infty$-norm. Define $\|A\|$ to be $$ \|A\|= \sup\limits_{x\neq 0}\left\{\frac{N(Ax)}{M(x)}\right\}. $$ Prove: If $B$ is an $m\times k$ matrix and $K$ is a norm on $\mathbb{R}^k$, then $||AB||≤||A||||B||$ (where the left-hand norm is defined using $(N,K)$ and the right-hand norms are defined using $(N,M)$ and $(M,K)$ respectively).

(Does this question mean $\sup\limits_{x\neq 0}\left\{\frac{N(ABx)}{M(x)}\right\} \le \sup\limits_{x\neq 0}\left\{\frac{N(Ax)}{M(x)}\right\}\sup\limits_{x\neq 0}\left\{\frac{N(Bx)}{M(x)}\right\}$? where in the left hanside $x$ is in $\mathbb{R}^k$ and right handside in $\mathbb{R}^m$ and $\mathbb{R}^k$ respectively? If so, is there any suggestion how to prove it?)

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No, the questions means to prove that $$ \sup_{x\not=0}\frac{N(ABx)}{K(x)} \le \sup_{y\not=0}\frac{N(Ay)}{M(y)} \ \sup_{x\not=0}\frac{M(Bx)}{K(x)}. $$ The fact that $$ \frac{N(ABx)}{K(x)} = \frac{N(ABx)}{M(Bx)} \ \frac{M(Bx)}{K(x)} $$ for $x\not=0$ and $Bx\not=0$ may help.

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