For any matrix $A \in \mathbb{R}^{m \times n}$, $$\| A \|_{\infty} = \max_{i}\sum_{j=1}^{n} |a_{ij}|$$
The proof I saw found a upper bound for $\| A\|_{\infty}$:
$$\dfrac{\|Ax\|_{\infty}}{\| x\|_{\infty}} \leq \max_{i} \sum_{j=1}^{n} |a_{ij}|$$
I could not understand one step, where a vector $\tilde{x}$ is exhibited, such that
$$\tilde{x}=\left\{ \begin{array}{ll} 1 & \mbox{if } a_{kj} \geq 0, \ j=1,...,n \\ -1 & \mbox{if } a_{kj} < 0, \ j=1,...,n \end{array} \right.$$
$a_{kj}$ is the k-th line with maximum $\|.\|_1$, it follows that
$$\dfrac{\| A\tilde{x}\|_{\infty}}{\|\tilde{x}\|_{\infty}}=\max{i} |(A\tilde{x})_i|=\max_i |\sum_{j}^n a_{ij}\tilde{x}_j|\stackrel{?}=|\sum_{j}^n a_{kj}\tilde{x}_j|$$
Someone could explain me the last equality sign?
With the upper bound given above it is sufficient to find one vector $\bar x$ for which the inequality above is an equality. For that we take the line with maximal $\lVert \cdot \rVert_1$-norm which just means that we pick the line $k$ for which $$\sum_{j=1}^n \lvert a_{kj}\rvert$$ obtains a maximum out of all the $n$ rows. Now $$\sum_{j=1}^n \lvert a_{kj} \rvert = \sum_{j=1}^n \mathrm{sgn}(a_{kj}) a_{kj} = \sum_{j=1}^n \bar x_j a_{kj}$$ by the definition of the given $\bar x$. On top of that we have $\lVert \bar x \rVert_\infty = \max_{i=1, \dots, n} \lvert \bar x_i \rvert = 1$ and from that we can conclude $$\frac{\lVert A\bar x \rVert_\infty}{\lVert \bar x \rVert_\infty} = \lVert A\bar x \rVert_\infty = \max_{i=1, \dots, n} \lvert (A\bar x)_i\rvert =\max_{i=1, \dots, n} \left \lvert \sum_{j=1}^n \bar x_j a_{ij} \right \rvert.$$
By the triangle inequality the last line is less than or equal to $\max_{i=1, \dots, n} \sum_{j=1}^n \lvert a_{ij} \rvert = \sum_{j=1}^n \lvert a_{kj} \rvert$ by choice of $k$ as the row with maximal $\lVert \cdot \rVert_1$-norm. But we also have $$\sum_{j=1}^n \lvert a_{kj} \rvert=\left \lvert \sum_{j=1}^n \bar x_j a_{kj} \right \rvert \le \max_{i=1, \dots, n} \left \lvert \sum_{j=1}^n \bar x_j a_{ij} \right \rvert \le \max_{i=1, \dots, n} \sum_{j=1}^n \lvert a_{ij} \rvert = \sum_{j=1}^n \lvert a_{kj} \rvert$$ so in fact we have the equality. This means $$\lVert A \rVert_\infty =\sup_{x \ne 0 }\frac{\lVert A x \rVert_\infty}{\lVert x \rVert_\infty} \ge \max_{i=1, \dots, n} \lvert a_{ij}\rvert$$ and by the upper bound that you already know we have equality.