For which values $p, q$ does one have $L^{p} \subset L^{q}$?

Question

This question seems to suggest that one has $L^{p}([-\pi,\pi]) \subset L^{1}([-\pi,\pi])$ for every $p \in (1,2)$. There are some obvious cases where this does not hold: for instance, not every $L^{1}$-integrable function is also $L^4$-integrable, and common counterexamples are functions involving square roots and characteristic functions.

However, if $p,q$ are any real numbers (so that if, for instance, $p \in \mathbb{N}$, then the norm $\| \cdot \|_p$ on $L^p$ defines a quasi-norm rather than a norm), can we find pairs $(p,q)$, with $p \neq q$, such that one has $L^p([a,b]) \subset L^q([a,b])$ for any (or a particular choice of) $a$ and $b$, with $a < b$? If $p, q \in \mathbb{N}$, do any pairs exist, and if so, how many?

Is there an analogous result for $l^p$ spaces?

Answer 1

For any finite measure space $X$, one has $L^q(X)\subset L^p(X)$ for $0<p\leq q\leq\infty$.

This is a theorem in real analysis. (See for instance Folland's Real Analysis Proposition 6.12.) In this theorem, there is no assumption about whether $p,q$ are integers or not.

When $X=[-\pi,\pi]$, $L^p(X)\subset L^1(X)$ for any $p>1$, and in particular for $p\in (1,2)$.

"not every $L^1$ function is $L^4$" does not serve as a counterexample.

On the other hand, if $A$ is any set, and $0<p<q\leq\infty$, then $\ell^p(A)\subset\ell^q(A)$. See Proposition 6.11 in Follands real analysis.

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Answer 2

Actually, if $p<q$, you have $L^q(a,b)\subset L^p(a,b),$ when your interval is bounded. More generally, if $X$ is a space of finite measure, than also $L^q(X)\subset L^p(X).$

For the spaces $l^p$, it's the other way around: if $p<q$, then $l^p\subset l^q.$