When are two $L^p$ spaces equal

Question

Q) Consider a probability space i.e. $\mu(X)=1$. Under what conditions is $L^p(\mu)=L^q(\mu)$ for $0<p<q\leq\infty$?

I can see when $||f||_p=||f||q$ by looking at the fact the Holder's inequality is an equality if and only if $\alpha |f|^q=\beta$ a.e. for some $\alpha,\beta$ both not zero which leads to the fact that $f$ is constant a.e.

But I'm not sure under what conditions the whole of $L^p$ space is equal to $L^q$?

Answer 1

Following my comment, here is a sketch. Let's assume that $$ \inf_{\mu(E)\neq 0}\mu(E) > 0. \qquad (*) $$ Given $f \in L^q(\mu)$, let $E_n = \{ \vert f \vert^p > n \}$. Then $$ \mu(E_n) \leq \frac{1}{n}\Vert f \Vert_p^p \to 0 $$ as $n \to \infty$. Therefore there is some $N$ for which $\vert f \vert \leq N^{1/p}$ a.e., hence $f \in L^q(\mu)$.

(This shows that $(*)$ implies $L^p(\mu) \subset L^q(\mu)$. One should also be able to show the opposite: $L^p(\mu) \subset L^q(\mu)$ implies $(*)$.)

Answer 2

If I am not wrong, they are not equal, except trivial case in which $\mu$ is atomic and has only a finite number of atoms.
If it is not the case, you can choose a sequence of disjoint measurable sets $E_1,E_2,..$ such that: $$ \mu(E_n)=a_n>0 \forall n$$ So it's only the work of creating a measurable function , which belongs to $L^p$, but not $L^q$, of form: $$ f= \sum_{n \ge 1} b_n \mathbb{1}_{E_n}$$ For simplicity, let $p=1$; $q=2$, I can choose : $$b_n= \frac{1}{ \sqrt{R_n} } $$ where $R_n$ is the residu of the series $\sum_{n \ge 1} a_n$, that is: $$R_n= \sum_{k \ge n} a_n$$ Then clearly , $$ \int f d\mu \le \sum_{n \ge 1} 2\left( \sqrt{R_{n}}-\sqrt{R_{n+1}}\right) < \infty$$ and $$ \int f^2 d\mu = \sum \dfrac{a_n}{R_n} = \infty$$

So the inclusion $ L^q( \mu ) \subset L^p( \mu)$ is strict.
The same idea can so apply for other $p,q$, we just need to adjust the power of $R_n$

Disclaimer: I'm just not so sure about my argument about finite atom measures in the beginning, but I guess having well presented the essence of solution.

Answer 3

Here is another proof (child Rudin Exercise 5(c), Chapter 3).

Claim: There do not exist countably infinite disjoint $E_1, E_2, \dots$ such that $\mu(E_i) > 0$ and $\sum_i \mu(E_i) = 1$.

Proof: Suppose otherwise and denote $a_i = \mu(E_i)$, we can find $n_1, n_2, \dots$ and $k_1, k_2, \dots$ such that

$$\frac12 < \sum_{i=1}^{n_1} a_i < 1,$$ $$\frac{1}{2^{k_1+1}} < \sum_{i=n_1 + 1}^{n_2} a_i < \frac{1}{2^{k_1}},$$ $$ \dots$$ Take $f = \sum_{i=0}^\infty b_i \mathbb{1}(E_{n_i + 1} \cup \dots E_{n_{i+1}})$ with $b_i = 1.5^{k_i+1}$, $n_0 = k_0 = 0$. Then $$\int fd \mu < 1.5\sum_{i=0}^\infty \frac{1.5^{k_i}}{2^{k_i}} < \infty,$$ and $$\int f^2 d \mu \geq \frac{1.5^2}{2}\sum_{i=0}^\infty \frac{2.25^{k_i}}{2^{k_i}} = \infty.$$

The claim tells us that $\mu$ must be atomic with finite atoms (Theorem 2.2 here). This condition is also sufficient.