When do $L^p$ and $L^q$ contain the same functions?

Question

From Rudin:

Given that X is a finite measure space, when do $L^p$ and $L^q$ contain the same functions?

What I know is that $L^p$ contains $L^q$ if p<q.

I was thinking

How to make a function which is in $L^p$ but not in $L^q$, assuming it is possible to do so? And what fails when it is not possible.

I appreciate any help.

Answer 1

The usual "measure level" characterization is that with $1 \leq p<q<\infty$:

• $L^q \subset L^p$ if and only if the space has no sets of arbitrarily large finite measure, i.e. there exists $M>0$ such that for all measurable $A$, if $\mu(A)<\infty$ then $\mu(A) \leq M$. For most practical purposes this condition is the same as "the space has finite measure", but it is slightly more general. Extending this to $q=\infty$ requires you to actually assume the space has finite measure.
• $L^p \subset L^q$ if and only if the space has no sets of arbitrarily small positive measure, i.e. there exists $\epsilon>0$ such that for every measurable $A$, if $\mu(A)>0$ then $\mu(A)>\epsilon$. For most practical purposes this is the same as "the space is discrete". This condition extends to $q=\infty$ with no issue.

Thus to have $L^p=L^q$ you must have both of these conditions holding.

Side remarks:

• The intuition for the first condition is that if there are large measure sets, then there are heavy-tailed functions whose integrability can be improved by raising them to larger powers. You can explore the family $x^r \chi_{[1,\infty)}(x)$ (which are integrable for $r<-1$) to get a feel for this.
• The intuition for the second condition is if there are small measure sets, then there are functions with singularities whose integrability can be improved by raising them to smaller powers. You can explore the family $x^r \chi_{(0,1)}(x)$ (which are integrable for $r>-1$) to get a feel for this.
• A fun exercise on this topic: with $X$ being the Lebesgue measure space on $\mathbb{R}$, and given $p_0 \in [1,\infty)$, construct a function $f$ which is in $L^p$ for only $p=p_0$.

Answer 2

I am not sure if this completely determines when $L^p=L^q$, but it is a case. Throughout all of this we can assume without loss of generality that $\mu(X)=1$ and that $p < q$. Using Holders we can show that $||f ||_p \leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}||f||_q=||f||_q$. This shows that $L^q\subset L^p$.

Assume that $$L^p\subset L^\infty $$. Then for any $f\in L^p$ we have that $$||f||_{p+1}^{p+1}=\int_\Omega |f|^{p+1}d\mu\leq \int_{\Omega}|f|^p ||f||_\infty d\mu = ||f||_p^p ||f||_\infty$$ Since $f\in L^p\subset L^\infty$, then $f\in L^{p+1}$. We can actually show by induction using this argument that $f\in L^{p+n}$ for any $n\in\mathbb{N}$. Now we can find an $n\in\mathbb{N}$ so that $q < p+n$. Then since $||f||_q\leq ||f||_{p+n}$ and $f\in L^{p+n}$, it follows that $||f||_q<\infty$ and so $f\in L^q$. Hence $L^q=L^p$.

So the condition I found is that if we are on a finite measure space, $p<q$ and $L^p\subset L^\infty$, then $L^p=L^q$.