Compact inclusion in $L^p$

Question

Is it true that there is a compact inclusion from $L^p$ to $L^q$ whith $q<p$?

What is the counterexample if what I said is wrong?

Thank you.

Answer 1

It is well known that $L^p \subseteq L^q$ with $1 \leq q < p < \infty$ on a measure space $(X,\Sigma,m)$ if and only if $m(X) < \infty$.

Compactness, however, doesn't have to hold.

Example: Let $X = [0,1]$ with the usual Lebesgue measure. Let $f_n(x) = \sin(2\pi n x)\in L^\infty(X) \subset L^p(X)$ for all $1\leq p$. So in particular it is a bounded sequence in $L^p(X)$.

Now, observe that if $n \neq m$ $$ \int_0^1 |f_n(x) - f_m(x)|^2 \mathrm{d}x = \int_0^1 f_n^2(x) + f_m^2(x) + f_n(x) f_m(x) \mathrm{d}x = 1 $$ and hence the sequence cannot have any Cauchy subsequence in $L^2$. (Similarly you can show that there exists no Cauchy subsequence in any $L^p$ using Holder's inequality and various interpolation inequalities.) Hence you have no compactness.

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In fact, we have the following:

Theorem If a measure space $(X,\Sigma,m)$ is such for some $1 \leq q < p < \infty$, the inclusion $L^p(X)\subset L^q(X)$ holds and is compact, then every non-negligible subset of $X$ contains an atom.

The proof basically goes the same as the above example. By a theorem of Sierpinski's a non-atomic measure space can be evenly divided into two halves. So if $X$ contains a non-atomic subset $Y$ with positive measure, we can associate to every finite binary string $a_1a_2\cdots a_n$ a subset $Y_{a_1a_2\cdots a_n} \subset Y$ such that

• For any two distinct strings of the same length, $Y_{a_1a_2\cdots a_n} \cap Y_{b_1b_2\cdots b_n} = \emptyset$
• For any string of length $n$, $m(Y_{a_1 a_2\cdots a_n}) = \frac1{2^n} m(Y)$
• For any string of length $n$, $Y_{a_1a_2\cdots a_n} \subsetneq Y_{a_1a_2\cdots a_{n-1}}$.

Defining $$f_n = \sum_{\text{strings of length } n} (2a_n - 1) \mathbf{1}_{Y_{a_1a_2\cdots a_n}}$$ we see that $f_n$ cannot admit any Cauchy subsequence, much in the same way as the example above.

On the other hand, if $X$ is made up only of atoms, then the inclusion can indeed be compact. For example, if $X$ is a finite set and $m$ is the counting measure on $X$. Then since the unit ball in $L^p(X,m)$ now is compact, automatically the inclusion $L^p(X) \subset L^q(X)$ is compact.

Answer 2

Note that, $L^{2}([0, 1])\subset L^{1}([0,1])$(using holders inequality); but take $f:(0, 1]\rightarrow \mathbb R$ such that $f(x)=x^{-\frac{1}{2}}$, $f\in L^{1}$ but $f\notin L^{2}$.