Why is $L^p(S) \subseteq L^q(S)$?

Question

Let $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$.

Exercise: Let $p,q\in(1,\infty)$ and let $1\leq p<q<\infty$. Show that $L^p(S) \subseteq L^q(S)$.

What I've tried:

I know that $L^p(S) = \{f:S\to\mathbb{K}:\text{ $f$ measurable and }\int_S \left|f\right|^p d\mu < \infty\}$ and $L^q(S) = \{f:S\to\mathbb{K}:\text{ $f$ measurable and }\int_S \left|f\right|^q d\mu < \infty\}$. $L^p(S)\subseteq L^q(S)$ would mean that for every $f$ such that $\int_S \left|f\right|^p d\mu < \infty$ we would have $\int_S \left|f\right|^q d\mu < \infty$. I don't know how to show that this is true though.

Question: How do I solve this exercise?

If $\mu$ is a probability measure you can use Jansen's inequality. — Yanko, Dec 29 '17 at 13:41
Possible duplicate of https://math.stackexchange.com/questions/1397097/how-to-show-that-lp-norm-is-monotone-increasing — Yanko, Dec 29 '17 at 13:41
@yanko No, for a probability measure the inclusion is in the other direction. — David C. Ullrich, Dec 29 '17 at 15:27
This is not true, unless there's an assumption about $\mu$ that you haven't told us about... — David C. Ullrich, Dec 29 '17 at 15:29
The inclusion can go either way, depending on the measure space. Meaning of course also that there may be no inclusion at all. — ziggurism, Dec 29 '17 at 15:31
Suppose that $0 < p < q ≤ ∞$. Then $L^q(S, μ) ⊂ L^p(S, μ)$ iff $S$ does not contain sets of finite but arbitrarily large measure, and $L^p(S, μ) ⊂ L^q(S, μ)$ iff $S$ does not contain sets of non-zero but arbitrarily small measure.
https://en.wikipedia.org/wiki/Lp_space — ziggurism, Dec 29 '17 at 15:52

ziggurism · Answer 1 · 2018-01-10T21:21:02.680

In general there need not be any inclusion between $L^p(S,\mu)$ and $L^p(S,\mu)$, for an arbitrary measure space $(S,\mu).$

However if your measures are bounded, there are inclusions. See for example Wikipedia. Thus criteria for inclusion of $L^p$ spaces is, for $0<p<q\leq\infty$:

$L^q(S,\mu)\subseteq L^p(S,\mu)$ iff $S$ does not contain sets of finite but arbitrarily large measure,
$L^p(S,\mu)\subseteq L^q(S,\mu)$ iff $S$ does not contain sets of non-zero but arbitrarily small measure.

Let's see this with some examples.

For example, let $S$ be the positive natural numbers line $S=\{n\in\mathbb{N}|n>0\}$ with counting measure. This space has sets of arbitrarily large measure, and the sequence $1/n^{1/p}$ is in $L^q$ but not $L^p$, so we have $L^q\not\subseteq L^p.$

Or let $S$ be the real interval $(0,1)$ with the Lebesgue measure. This space has sets of arbitrarily small measure, and the function $1/x^{1/q}$ is in $L^p$ but not $L^q$, so we have $L^p\not\subseteq L^q.$

The interval $(0,\infty)$ has sets of arbitrarily large and small measure, so neither inclusion holds, as the functions $1/x^{1/q}$ and $1/x^{1/p}$ show.

If the measure is both bounded above and bounded below, then there are inclusions going both ways, and $L^p$ and $L^q$ are isomorphic as topological vector spaces. I think this only happens if $S$ is finite and so $L^p$ is finite-dimensional, where all norms are equivalent.

So the upshot is, if we would like to prove one or the other inclusion, we must adopt one of the assumptions about our measure space.

So let's adopt axiom 1, then $S$ must have finite measure. Now $p<q$, so $q/p>1$, so $x\mapsto x^{q/p}$ is a convex function, so by Jensen's inequality, we have

$$\left(\frac{1}{\mu(S)}\int\lvert f\rvert^p\right)^{q/p}\leq \frac{1}{\mu(S)}\int \lvert f\rvert^q,$$ so $\lVert f\rVert_p\leq \mu(S)^{\frac{1}{p}-\frac{1}{q}}\lVert f\rVert_q$ and so $L^q(S)\subseteq L^p(S).$

Alternatively, apply Hölder's inequality to $f^p\in L^{\frac{q}{p}}$ and $\chi_S\in L^{\frac{q}{q-p}},$ giving

$$\lVert\chi_S f^p\rVert_1\leq\lVert\chi_S\rVert_{\frac{q}{q-p}}\lVert f^p\rVert_{\frac{q}{p}},$$ and take the $p$th root.

For option 2, to answer your title question and show that $L^p(S)\subseteq L^q(S)$, let's assume we have a lower bound, so $\mu(E)\geq m$ for all $E\subseteq S.$

Given a simple function

$$f=\sum a_i\chi_{E_i}$$

where $E_i$ are disjoint measurable sets and $\chi_{E_i}$ are the indicator functions, we have

$$\lVert f\rVert_p^p=\sum \lvert a_i\rvert^p\mu(E_i)$$

and so

$$\frac{\sum \lvert a_i\rvert^p\mu(E_i)}{\lVert f\rVert_p^p}=1$$

but if a sum of nonnegative terms is equal to one, then each term is at most one, so

$$\frac{\lvert a_i\rvert^p\mu(E_i)}{\lVert f\rVert_p^p}\leq 1$$ as well as its $p$th root

$$ \frac{\lvert a_i\rvert\mu(E_i)^{1/p}}{\lVert f\rVert_p}\leq 1. $$

A fundamental fact of exponentiation, which is at the root of the identity we seek, with exponents $p<q$ is that if the base $x>1$ then it is increasing, $x^p<x^q$, while if the base $x<1$ it is decreasing, $x^p>x^q.$

Thus we have

$$ \frac{\lvert a_i\rvert^q\mu(E_i)^{q/p}}{\lVert f\rVert_p^q}\leq\frac{\lvert a_i\rvert^p\mu(E_i)}{\lVert f\rVert_p^p}\leq 1, $$ and therefore also the sum $$ \frac{\sum\lvert a_i\rvert^q\mu(E_i)^{q/p}}{\lVert f\rVert_p^q}\leq\frac{\sum\lvert a_i\rvert^p\mu(E_i)}{\lVert f\rVert_p^p}=1. $$

In order to transform the left-hand side of this inequality into the $\lVert\cdot\rVert_q$ norm, we use the boundedness $\mu(E_i)\geq m$ to write

$$\mu(E_i)^{q/p}=\mu(E_i)\mu(E_i)^{\frac{q-p}{p}}\geq\mu(E_i)m^{\frac{q-p}{p}},$$

so that our inequality becomes

$$ \frac{m^{\frac{q-p}{p}}\sum\lvert a_i\rvert^q\mu(E_i)}{\lVert f\rVert_p^q}\leq\frac{\sum\lvert a_i\rvert^q\mu(E_i)^{q/p}}{\lVert f\rVert_p^q}\leq\frac{\sum\lvert a_i\rvert^p\mu(E_i)}{\lVert f\rVert_p^p}=1. $$

Finally, clearing denominators and taking the $q$th root gives us

$$\lVert f\rVert_q\leq \frac{1}{m^{\frac{1}{p}-\frac{1}{q}}}\lVert f\rVert_p,$$ for simple functions $f$. And since the simple functions are dense, it gives us also the inclusion we desire $L^p(S)\subseteq L^q(S).$

Why is $L^p(S) \subseteq L^q(S)$?

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