There is this post in Math Overflow which asks about an inequality given in Wikipedia, which is
$$\lVert A \rVert_2 \leq \sqrt{\lVert A \rVert_1 \lVert A \rVert_\infty}$$
In the accepted answer, an elementary proof of a more general inequality is given using Hölder's inequality, which is
$$\lVert A \rVert_p \leq \lVert A \rVert_1^{1/p} \lVert A \rVert_\infty^{1/q}$$
where $1/p+1/q=1$. However I'm having trouble following the proof. I copied the proof and put question marks.
\begin{eqnarray*}
\|Ax\|_p^p & = & \sum_i\left|\sum_ja_{ij}{\color{red}{x_j (?)}}\right|^p\stackrel{\color{red}{??}}{\le}\sum_i\left(\sum_j|a_{ij}|\right)^{p/q}\sum_k|a_{ik}x_k|^p \\
& \le & \|A\|_\infty^{p-1}\sum_{i,k}|a_{ik}x_k|^p\stackrel{\color{red}{???}}{=}\|A\|_\infty^{p-1}\sum_k|x_k|^p\sum_i|a_{ik}|\le\|A\|_\infty^{p-1}\|A\|_1\|x\|_p^p.
\end{eqnarray*}
$\color{red}{?}$ I believe there is $x_j$ missing in here?
$\color{red}{??}$ How does Hölder's inequality give this one?
$\color{red}{???}$ Where did ^p of $|a_{ik}|$ go?
Edit: Typos in the MO have been corrected.
Best Answer
Thera are indeed a few typos in the posting from MO. Here I tried to fill in all the missing parts. Throughout $\frac1p+\frac{1}{p'}=1$. Notice that $\frac{p}{p'}=p-1$.
\begin{align} \|Ax\|_p^p &= \sum_i\Big|\sum_ja_{ij}x_j\Big|^p\leq \sum_i\Big(\sum_j|a_{ij}||x_j|\Big)^p=\sum_i\left(\sum_j|a_{ij}|^{\tfrac{1}{p'}}|a_{ij}|^{\tfrac1p}|x_j|\right)^p\\ &\leq\sum_i\Big(\sum_j|a_{ij}|\Big)^{p/p'}\Big(\sum_k|a_{ik}||x_k|^p\Big) \\ & \le \|A\|_\infty^{p-1}\sum_{i,k}|a_{ik}||x_k|^p=\|A\|_\infty^{p-1}\sum_k|x_k|^p\sum_i|a_{ik}|\le\|A\|_\infty^{p-1}\|A\|_1\|x\|_p^p. \end{align}
Hölder’s inequality is used in the second line as follows: for each $i$ fixed
$$\sum_j|a_{ij}|^{\tfrac{1}{p'}}|a_{ij}|^{\tfrac1p}|x_j|\leq \Big( \sum_j|a_{ij}|\Big)^{1/p’}\Big(|a_{ik}|\,|x_j|^p\Big)^{1/p}$$