Is the matrix norm $\mu(A)=n\max_{i,j}|a_{ij}|$ for $A\in\mathbb R^{n\times n}$ NOT compatible with any vector norm on $\mathbb R^n$

Question

Define $\mu(A)=n\max_{i,j}|a_{ij}|$ for $A\in\mathbb R^{n\times n}$. We can easily verify it is a matrix norm:
\begin{align}
\mu(AB) &=n\max_{i,j}|\sum_{k}a_{ik}b_{kj}| \\
& \leq n\max_{i,j}\sum_{k=1}^n|a_{ik}||b_{kj}|\\
&\leq n(n\max_{i,j,k,k'}|a_{ik}||b_{k'j}|)\\
&=(n\max_{i,k}|a_{ik}|)(n\max_{k',j}|b_{k'j}|)\\
&\leq\mu(A)\mu(B)
\end{align}
Is this matrix norm NOT compatible with any vector norm? I am trying to construct a counter example as follows: $A$ is a square matrix consisting of all ones, and $x$ is a column vector consisting of all ones. Then $Ax=(n,\cdots,n)^T$. However, if we use any $p$-norm, then we have
$$||Ax||_p=n^{1+1/p},\quad ||x||_p=n^{1/p},\quad \mu(A)=n$$
but this satisfies $||Ax||_p\leq\mu(A)||x||_p$ and is not a counterexample. Any help is appreciated.

Answer 1

There are no counterexamples, because the inequality holds for all $1\le p\le\infty$. For $p\ge1$, (letting $\|A\|_\infty=\max_{ij}|A_{ij}|$) \begin{align} \|Ax\|_p&=\left(\sum_i|\sum_jA_{ij}x_j|^p\right)^{1/p}\\ &\le\|A\|_\infty\left(\sum_i\sum_j|x_j|^p\right)^{1/p}\\ &=\|A\|_\infty\ n^{1/p}\|x\|_p\le \mu(A)\|x\|_p \end{align}

For $p=\infty$, \begin{align} \|Ax\|_\infty&=\max_i|\sum_jA_{ij}x_j|\\ &\le\max_i\sum_j|A_{ij}x_j|\\ &\le\|A\|_\infty\sum_j|x_j|\\ &\le\|A\|_\infty\ n\|x\|_\infty=\mu(A)\|x\|_\infty \end{align}

Any other norm is equivalent to the standard norms, $$c\|x\|_2\le\|x\|\le d\|x\|_2$$ so continuity of $A$ is assured: $$\|Ax\|\le d\|Ax\|_2\le dn^{1/2}\|A\|_\infty\|x\|_2\le\frac{d}{c n^{1/2}}\mu(A)\|x\|$$

If the question is specifically about the inequality $\|Ax\|\le\mu(A)\|x\|$, then one must choose, for a counterexample, a norm such that $d>>c$. For example, take $\|(x_1,x_2)\|:=\sqrt{9x_1^2+x_2^2}$, $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $x=(0,1)$, to get $\|Ax\|=3$, $\mu(A)=2$, $\|x\|=1$.