$L^p$ spaces and counting measure currently I am working on the following two exercises as a revision for my exam.
Let $\mu$ be the counting measure on $\mathbb N$.


*

*Show that if $1 \le p < s < \infty$ then $f \in L^p$ implies $f \in L^s$ and $\|f\|_s \le \|f\|_p$.

*Show that if $f \in L^p$ for some $p \ge 1$ then $\lim_{s \to \infty} \|f\|_s = \|f\|_\infty$.


My attempts so far:


*

*It is $f \in L^s$ iff $\sum_{k = 1}^\infty |f(k)|^s$ converges. Since $f \in L^p$ we know that $\sum_{k = 1}^\infty |f(k)|^p$ converges. Thus for any $\varepsilon > 0$ there is $N \in \mathbb N$ s.t.
\begin{align*}
k \ge N \Longrightarrow |f(k)|^p < \varepsilon \Longrightarrow |f(k)| < \varepsilon^{1/p}
\end{align*}
We have $\forall k \ge N$ that $|f(k)|^s < \varepsilon^{s/p}$. Originally I wanted to use the comparison test now, showing that
\begin{align*}
\sum_{k = N}^\infty \varepsilon^{s/p} < \infty,
\end{align*}
but that does not work since
\begin{align*}
\sum_{k = N}^\infty \varepsilon^{s/p} = \varepsilon^{s/p} \underbrace{\sum_{k \ge N} 1}_{= \infty}, 
\end{align*}
so I don't know how to go on here.

*Let $K := \|f\|_\infty$. We need to show that $\forall \varepsilon > 0$ $\exists M \in \mathbb N$ s.t.
\begin{align*}
s \ge M \Longrightarrow \left|\left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} - K\right| < \varepsilon.
\end{align*}
Choose $M$ s.t. $M > p$. Then $s > p$ and from part (i) we know that
\begin{align*}
\left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} \le \left(\sum_{k = 1}^\infty |f(k)|^p\right)^{1/p}.
\end{align*}
But how can I use this?

 A: Note that your problem in your first try was that you were using only the fact that $\lim\limits_{n\to\infty}|f(n)|=0$ and not merely the fact that the series converges. Now, following a similar approach, we know that given $\varepsilon>0$ there is some $N$ such that $n\geq N$ implies $|f(n)|\leq 1$. Since $s/p>1$ this implies that
$$
|f(n)|^{s/p}\leq |f(n)|\implies |f(n)|^s\leq |f(n)|^p
$$
Since this is valid for every $n\geq N$ we can add up this inequalities and obtain
$$
\sum_{n=N}^\infty |f(n)|^s\leq \sum_{n=N}^\infty |f(n)|^p<\infty
$$
Then, $f\in L^s$ and your statement follows.
For the second part, note that we already know that the limit
$$
\lim\limits_{s\to\infty}\Vert f\Vert_s
$$
exists and is finite (Why?). Let L be this limit. Since $f$ is bounded, then it belongs to $L^\infty$. Let $M=\Vert f\Vert_\infty$.  To show that $L=M$ we proceed as follows:
Given $\varepsilon>0$ there is an $n$ such that $M-\varepsilon<|f(n)|$. Then, if $s\geq p$, $(M-\varepsilon)^s<|f(n)|^s\leq \Vert f\Vert_s^s$ and this implies that $M-\varepsilon\leq \Vert f\Vert_s$. Since this is valid for every $\varepsilon$ and for every $s\geq p$, we should have $M\leq L$. 
To show the reverse inequality, you can use the fact that $L=\inf\limits_{s\geq p}(\Vert f\Vert_s)$ and a similar argument to show that $L\leq M$ and obtain your result.
A: This is not an answer. 
In the case of the counting measure on $\mathbb N$, the Borel sets are ALL the subsets of $\mathbb N$ and the measurable functions are the sequence $\{ a_n\}_{n\in\mathbb N}\subset \mathbb R$ or $\mathbb C$. Also, the corresponding $L^p$ space are the 
$$
\ell^p(\mathbb N)=\big\{\{a_n\}_{n\in\mathbb N}:\sum_{n\in\mathbb N}|a_n|^p<\infty\big\},
$$
with norm
$$
\big\|\{a_n\}_{n\in\mathbb N}\big\|_p=\left(\sum_{n\in\mathbb N}|a_n|^p\right)^{1/p},
$$
for $1\le p<\infty$, and $\|\{a_n\}_{n\in\mathbb N}\|_\infty=\sup_{n\in\mathbb N}|a_n|.$
A: Let $f\in L_p$.  The statement is trivial if $f=0$.
Suppose $f\neq 0$.  Let $g = f/\|f\|_p$.  Then $\|g\|_p =1$, and $|g(n)|\leq 1$ for all $n$.  So
$$\sum_{n=1}^\infty |g(n)|^s \leq \sum_{n=1}^\infty |g(n)|^p = \|g\|_p^p = 1.$$
Hence, $g\in L_s$ and
$$\|g\|_s = \left( \sum_{n=1}^\infty |g(n)|^s\right)^{1/s} \leq 1.$$
Then $f = \|f\|g\in L_s$ and $\|f\|_s = \|f\|_p \|g\|_s \leq \|f\|_p$.
