$L^p \subset L^q$

Question

Let $(X,M,\mu)$ be a measure space.

Let $\Omega \subset X$ be a measurable set.

We have $L^2(\Omega) \subset L^1(\Omega)$ .

Can we have that $\mu(\Omega)< \infty $ ?

Answer 1

Note that every set mentioned below is assumed to be measurable.

Presumably you know that if $\mu(\Omega)<\infty$ then $L^2\subset L^1$, and meant to ask about the converse.

As @zhw comments, letting $X$ be a singleton of infinite measure gives a counterexample. In fact any counterexample must be wacky in an analogous way:

Theorem Suppose $\mu$ is a measure on $\Omega$. Then $L^2(\mu)\subset L^1(\mu)$ if and only if $\Omega=A\cup B$ where $A\cap B=\emptyset$, $\mu(A)<\infty$, and $\mu(E)$ is $0$ or $\infty$ for every $E\subset B$.

Proof: One direction is trivial: If $\Omega=A\cup B$ as above and $f\in L^2$ then $f=0$ almost everywhere on $B$ and Cauchy-Schwarz shows that $$\int_A|f|\le\mu(A)^{1/2}||f||_2.$$

Suppose then that $L^2\subset L^1$. Let $$\alpha=\sup\{\mu(E):\mu(E)<\infty\}.$$The key fact is that $$\alpha<\infty.$$It's not that hard to show this directly, but possibly tedious. Here's a cute proof: The Closed Graph Theorem shows that there exists $c$ so that $||f||_1\le c||f||_2$ for every $f\in L^2$. Now if $\mu(E)<\infty$ let $f=\chi_E$; then $||f||_1\le c||f||_2$ says $\mu(E)\le c^2$.

Now let $(A_n)$ be a sequence of sets with $$\alpha-\frac1n<\mu(A_n)<\infty.$$Since $\mu\left(\bigcup_{n=1}^N\mu(A_n)\right)<\infty$, the definition of $\alpha$ shows that $$\mu\left(\bigcup_{n=1}^N\mu(A_n)\right)\le\alpha.$$So if we let $A=\bigcup_{n=1}^\infty A_n$ then $\mu(A)=\alpha$. Let $B=\Omega\setminus A$, and it's easy to verify that $A$ and $B$ have the properties claimed.

Answer 2

If $\mu(\Omega)<\infty$ then $L^2(\Omega) \subset L^1(\Omega)$. One proof of this is to use Cauchy-Schwarz. This still works when we replace $2$ with $p>1$, but now the inequality is attributed to Holder instead.

If $\mu(\Omega)=\infty$ then either there is an atom of infinite measure or there are sets of arbitrarily large finite measure. (This "or" is inclusive.) An atom of infinite measure creates no distinctions between $L^p$ spaces; any $L^p$ function must be zero on the atom regardless of what $p$ is. So if that's the only reason that $\mu(\Omega)=\infty$, then $L^2(\Omega)$ is still a subset of $L^1(\Omega)$.

Now we have the actual claim: if there are sets of arbitrarily large finite measure, then $L^2 \setminus L^1$ is nonempty. To see why that "should" be the case, suppose $A_n$ are sets with $m(A_n) \to \infty$ and define $f_n$ to be $\frac{1}{m(A_n)}$ on $A_n$ and zero elsewhere. Then $\int |f_n|=1$ but $\int f_n^2=\frac{1}{m(A_n)}$. So $\frac{\int |f_n|}{\int f_n^2}=m(A_n)$. Thus $f \mapsto \frac{\int |f|}{\int f^2}$ is not a bounded function on $(L^1 \cap L^2) \setminus \{ 0 \}$. Now we can't just take a pointwise limit of such things to get something in $L^2 \setminus L^1$, because the pointwise limit will just be zero. So we have to be a bit more clever about it.

But we can work around that, by "remembering" what we have already done rather than "forgetting" it, which is what would happen if we just took a pointwise limit in the way above. Obtain a sequence of disjoint sets of finite measure but whose measures are growing so rapidly that the sum of their reciprocals converges. For instance we can take $m(A_n) \geq 2^n$. Then make $f=\frac{1}{m(A_n)}$ on $A_n$ and zero elsewhere. You have $\int |f|=\sum 1 = \infty$ but $\int f^2 \leq \sum 2^{-n}<\infty$.