We're asked to show that Holder's inequality (for the case when $1/p + 1/q = 1$) holds for the case when $p=\infty$ and $q=1$. The inequality is given to us in the following form.
$\sum\limits_{i=0}^\infty \vert a_ix_i \vert \leq \vert \vert a \vert \vert_q \vert \vert x \vert \vert_p$
Here's my proof and I'd like to make sure that the logic is sound.
$$
\begin{align*}
\sum \vert a_i x_i \vert &\leq \vert \vert a \vert \vert_\infty \vert \vert x \vert \vert_1 &&\text{plug in variables} \\
&= \max(\vert a \vert) \vert \vert a \vert \vert_1 &&\text{value of infinite norm}
\end{align*}
$$
Now divide by $\max(\vert a \vert)$
$$
\sum \cfrac{\vert a_i x_i \vert}{\max(\vert a \vert)} \leq \vert \vert x \vert \vert_1
$$
We know that $\sum\limits_{i=o}^\infty \cfrac{\vert a_i x_i \vert}{\max(\vert a \vert)} \leq \sum \vert x_i \vert $ because $\cfrac{|a_i|}{\max(|a|)} \leq 1\, \forall a_i \in a \in \ell_\infty $.
Now we have that $\sum \vert x_i \vert \leq \vert \vert x \vert \vert_1$. These are, in fact, equal to each other. So we see that our case when $p=\infty$ and $q=1$ holds.
If there something I ought to do to make the proof easier to understand, please let me know.
Best Answer
The proof is not completely correct because the $\|\cdot\|_{\infty}$ norm is not the maximum of the absolute value of the sequence, but the supremum. For example, if $a_n=1-\frac{1}{n}$, then $\|a\|_{\infty}=1$, but there is no maximum.
To prove the inequality note that for every $n\geq 1$, $$|a_nx_n|=|a_n|\cdot |x_n|\leq\sup_n|a_n|\cdot |x_n|=\|a\|_{\infty}|x_n| $$
Therefore, for every $N$,
$$\sum_{n=1}^N|a_nx_n|\leq\|a\|_{\infty}\sum_{n=1}^N|x_n|\leq\|a\|_{\infty}\|x\|_1$$ Letting $N\to\infty$ the desired result follows.