The matrix $\infty$-norm for $A \in {\Bbb{R}}^{m \times n}$ is defined as
$$\|A\|_{\infty} =\max_{1 \leq i \leq n}\|a^{i}\|_{1}$$
where $a^{i}$ is the i$^{th}$) row in matrix $A$. Show that
$$\|A\|_{\infty} =\max \left\{\|Ax\|_{\infty} : x_{\infty} \le 1\right\} =\max \left\{\|Ax\|_{\infty} : x_{\infty} = 1\right\}$$
I know that this is a property of subordinate matrix norms but I'm not sure how to go about with proving it.
Proof of "$\max_{1\le i \le n} \Vert a^i \Vert_1 = \max_{\Vert x \Vert_\infty=1}\Vert Ax\Vert_\infty$":
$\max_{\Vert x \Vert_\infty=1}\Vert Ax\Vert_\infty = \max_{\Vert x \Vert_\infty=1} \{ \max_{1 \le i \le n} \vert \sum_{j=1}^m a^i_j x_j \vert \} \le \max_{\Vert x \Vert_\infty=1} \{ \max_{1 \le i \le n} \sum_{j=1}^m \vert a^i_j \vert \vert x_j \vert \} \le \{ \max_{1 \le i \le n} \sum_{j=1}^m \vert a^i_j \vert\}$,
which proves "$\ge$".
W.L.O.G., $\max_{1\le i \le n} \Vert a^i \Vert_1 = \Vert a^1 \Vert_1$. Then we can choose $x^* =(x^*_1,\cdots,x^*_n)^T$ as $x^*_j=\frac{\vert a^1_j \vert}{a^1_j}$ if $a^1_j \not = 0$ else $x^*_j = 0$. Then $\Vert x^* \Vert =1$ and $\Vert Ax^*\Vert_\infty =\max_{1\le i \le n} \vert \sum_{j=1}^m a^i_jx^*_j\vert \ge \vert \sum_{j=1}^m a^1_jx^*_j\vert = \sum_{j=1}^m \vert a^i_j \vert = \max_{1\le i \le n} \Vert a^i \Vert_1$, which proves "$\le$".
Proof of "$\max_{\Vert x \Vert_\infty=1}\Vert Ax\Vert_\infty = \max_{\Vert x \Vert_\infty \le 1}\Vert Ax\Vert_\infty$":
"$\le$" is obvious.
If $\max_{\Vert x \Vert_\infty \le 1}\Vert Ax\Vert_\infty$ can be acheived at $x^*$ that $\Vert x^* \Vert_\infty <1$. Then $\Vert \frac{x^*}{\Vert x^* \Vert_\infty} \Vert_\infty =1 \le 1$ and $\Vert A\frac{x^*}{\Vert x^* \Vert_\infty} \Vert_\infty = \frac{\Vert Ax^*\Vert_\infty}{\Vert x^* \Vert_\infty} > \Vert Ax^*\Vert_\infty$, contradiction. Hence $\max_{\Vert x \Vert_\infty \le 1}\Vert Ax\Vert_\infty$ can be acheived only at $\{\Vert x \Vert_\infty = 1\}$, which proves "$\ge$".