$
\newcommand{\norm}[1]{\| #1 \|}
\newcommand{\inorm}[1]{\norm{#1}_\infty}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\R}{\mathbb{R}}
$Context: This ultimately ties back to a homework assignment. (Specifically it's from Fundamentals of Matrix Calculations by Watkins, Exercise $2.1.30$ from $\S2.1$ but that's not really a huge deal, seems like a fairly common exercise.)
The goal is to show that, knowing matrix $p$-norms are induced by the corresponding vector norms, $\forall A \in \R^{n\times n}$,
$$\|A\|_\infty = \max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$$
I have an approach and was just curious as to how valid it is; I just don't feel very sure of myself. I feel like I made some sort of small-yet-critical mistake somewhere, but I can't figure out where…
My Attempt:
Ultimately, we will show that
$$\max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \le \|A\|_\infty \le \max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$$
which will let me conclude with the desired equality ($a \le b \le a \implies a=b$).
From the definition of the $\infty$-norm, with $A := (a_{i,j})_{1 \le i,j \le n} \in \R^{n \times n}$ and $x := (x_i)_{1 \le i \le n} \in \R^n$,
$$
\inorm{Ax} = \max_{1 \le i \le n} \abs{ \sum_{k=1}^n a_{i,k} x_k }
$$
Applying the triangle inequality yields
$$
\inorm{Ax} \le \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k} x_k| = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \cdot | x_k|
$$
Note that
$$
|x_k| \le \max_{1 \le j \le n} |x_j| =: \norm{x}_\infty
$$
and thus
$$
\inorm{Ax} \le \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \cdot \norm{x}_\infty
$$
$\norm{x}_\infty$ is independent of $i$, so we factor it out and conclude
$$
\inorm{Ax} \le \norm{x}_\infty \para{ \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| }
$$
Thus,
$$
\inorm{A} = \max_{x \ne \vec 0} \frac{\inorm{Ax}}{\inorm{x}} \le \frac{\displaystyle \norm{x}_\infty \para{ \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| }}{\norm{x}_\infty} = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$
Thus, with $\inorm{A}$ less than or equal to the desired expression, we need to find an $\hat x$ such that equality is achieved. Suppose in row $r$ the maximum is achieved; for every $k$, let
$$
\hat x_k = \begin{cases}
+1 & \text{if } a_{r,k} \ge 0 \\
-1 & \text{if } a_{r,k} < 0
\end{cases} = \mathrm{sign}(a_{r,k})
$$
Define $\hat x := (\hat x_i)_{1 \le i \le n}$ as defined above; clearly $\inorm{\hat x}=1$. This ensures $a_{r,k}\hat x_k = |a_{r,k}|$. Then
\begin{align*}
\norm{A}_\infty
&= \max_{x \ne \vec 0} \frac{\inorm{Ax}}{\inorm{x}} \tag{def. of matrix norm} \\
&\ge \frac{\inorm{A \hat x}}{\inorm{\hat x}} \tag{def. of maximum}\\
&= \inorm{A \hat x} \tag{$\inorm{\hat x} = 1$} \\
&= \max_{1 \le i \le n} \abs{ \sum_{k=1}^n a_{i,k} \hat x_k } \tag{definition}\\
&= \abs{ \sum_{k=1}^n a_{r,k} \hat x_k } \tag{definition of $r$}\\
&= \abs{ \sum_{k=1}^n |a_{r,k}| } \tag{choice of $\hat x_k$, $r$}\\
&= \sum_{k=1}^n |a_{r,k}| \tag{$|z| \ge 0 \implies \sum_i |z_i| \ge 0$}\\
&= \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \tag{def. of $r$}
\end{align*}
Thus what we have seen is that
$$
\max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \le \norm{A}_\infty \le \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$
thus letting us conclude equality:
$$
\norm{A}_\infty = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$
Best Answer
Here is an issue. In the part of showing $\big\lVert A \big\rVert_\infty \geq \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$ you've write the equality$$ \max_{1 \le i \le n} \Bigg\lvert \sum_{k=1}^n a_{i,k} \hat x_k \Bigg\rvert = \Bigg\lvert \sum_{k=1}^n a_{r,k} \hat x_k \Bigg\rvert $$
is true because of the definition of $r$. Recall that $r$ is taken as an integer between $1$ and $n$ such that $$\max_{1\leq i\leq n} \sum_{k=1}^n |a_{i,k}|= \sum_{k=1}^n |a_{r,k}|$$holds, so the equality does not follow immediately. I think it would be better to write as follows:
\begin{align*} & \max_{1 \le i \le n} \Bigg\lvert \sum_{k=1}^n a_{i,k} \hat x_k \Bigg\rvert \\ & \geq \Bigg\lvert \sum_{k=1}^n a_{r,k} \hat x_k \Bigg\rvert \tag{definition of maximum}\\ &= \Bigg\lvert \sum_{k=1}^n |a_{r,k}| \Bigg\rvert \tag{choice of $\hat x_k$, $r$}\\ &= \sum_{k=1}^n |a_{r,k}| \\ &= \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \tag{definition of $r$} \end{align*} Other parts seem good to me.