A question about inclusion of $L^r(\mu)$ spaces for different $r$ and different measures $\mu$

Question

For some measures, the relation $r<s$ implies $L^r(\mu)\subseteq L^s(\mu)$ ; for others, the inclusion is reversed; and there are some for which $L^r(\mu)$ does not contain $L^s(\mu)$ if $r\ne s$. Give examples of these situations, and find conditions on $\mu$

I am thanked if you give some hint. I do not know how to start to think about this.

Answer 1

Hint 1: If you do not have sets of arbitrarily small measure, one of the inclusions $L^r \subset L^s$ holds. For instance, consider $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$. How does this relate to $\left(\int_X |f(x)|^p d\mu(x)\right)^{1/p}$?

Hint 2: If you do not have sets of arbitrarily large measure, the other inclusion holds. For instance, what can you say about $\frac{1}{x}$ and $\frac{1}{\sqrt{x}}$ integrated on $[0,1]$?

Hint 3: If you want something to belong to a single $L^p$, you can't have either of the inclusions. That is, you need a space with infinite total measure and sets of arbitrarily small measure, like $[0,\infty)$. You have to do some arts and crafts (gluing) on functions which behave like $\frac{1}{\sqrt{x}}$ (which is in $L^1([0,1])$ but not $L^1([1,\infty])$) and $\frac{1}{x^2}$ (which is in $L^1([1,\infty))$ but not $L^1([0,1])$).

Finally, you will need a "tipping" point, which can be achieved by using a function like $f(x) = \frac{1}{\sqrt{x}\log x}$. Notice that this $f \in L^2([0,1])$, but $\frac{1}{\sqrt{x}} \notin L^2([0,1])$.

Answer 2

If $\mu(X)<\infty$, then $$ p<q \quad\Longrightarrow\quad L^q(X)\subset L^p(X). $$ This is due to the fact that (Holder inequality) $$ \int_X \lvert\, f\rvert^p\,d\mu=\int_X \lvert\, f\rvert^p\cdot 1\,d\mu =\left(\int_X \lvert\, f\rvert^q\,d\mu\right)^{p/q} \left(\int_X 1^{q/(q-p)}\,d\mu\right)^{(q-p)/q}. $$ Hence $\|\,f\|_p\le \mu(X)^{\frac{q-p}{pq}}\|\,f\|_q$.

If $X=\mathbb N$, and $\mu(A)$ is the cardinality of $A$, then $$ p<q \quad\Longrightarrow\quad L^q(X)\supset L^p(X). $$ To understand this, note first that $$L^p(\mathbb N)=\big\{(a_n)_{n\in\mathbb N}: \textstyle\sum_{n\in\mathbb N}\lvert a_n\rvert^p<\infty\big\}.$$

Combine the two previous results to get an example where if $p\ne q$, then $L^p\not\subset L^q$.