$L^p$ spaces and counting measure

Question

currently I am working on the following two exercises as a revision for my exam.

Let $\mu$ be the counting measure on $\mathbb N$.

1. Show that if $1 \le p < s < \infty$ then $f \in L^p$ implies $f \in L^s$ and $\|f\|_s \le \|f\|_p$.
2. 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:

1. 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.
2. 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?

Answer 1

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.

Answer 2

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 $\|\{a_n\}_{n\in\mathbb N}\|_\infty=\sup_{n\in\mathbb N}|a_n|.$

Answer 3

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 $\|f\|_s = \|f\|_p \|g\|_s \leq \|f\|_p$.