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Let $(X, \mathcal B, \mu)$ be a finite measure space. Prove that $L^2(X, \mathcal B, \mu) \subset L^1(X, \mathcal B, \mu)$.

I will be glad for any idea, comment, hint or advice.

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    $\begingroup$ Try Hölder's inequality. $\endgroup$
    – copper.hat
    May 16, 2022 at 5:32

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Hint: Let $f: X \to \mathbb{R}$ be non-negative and in $L^2$. Separate $f = f|_{\{f\le 1\}}+f|_{\{f> 1\}}$.

For the first part, notice that the measure space is finite. For the second part, notice that for some (which ones?) $\mathbb{R}\ni a>0$ we have $a \le a^2$.

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In fact, if $(X,\mu)$ is a finite measure space, you have a stronger result, for all $1\le q<p \le\infty$ then $L^p(\mu)\subset L^q(\mu)$, in particular, $q=1$ and $p=2$.

I will provide the proof for your specific case, but think about the general case... Just use Hölder's inequality.

What you need to prove is that if given a function $f\in L^2$ then $f\in L^1$. We have from the holder inequality that

$$\lVert f g \rVert_1 \le \lVert f \rVert_p \lVert g \rVert_q$$ for $1/p +1/q=1$. Note that for $p=q=2$, we can use the inequality. Consider $g=1$.

$$\int_X|f\cdot1|~d\mu\le \left(\int_X|f|^2~d\mu\right)^{1/2}\cdot \left(\int_X|1|^2~d\mu\right)^{1/2}$$

$$\int_X|f|~d\mu\le \lVert f\rVert_2\cdot \mu(X)^{1/2}$$

As $\mu(X)<\infty$ and $f\in L^2$, then we have $$\lVert f \rVert_1=\int_X |f|~d\mu<\infty$$ so $f \in L^1$.


For the general case, the strategy is similar, instead of using $f$ you should use $|f|^q$ and $g=1$. Now you must also change the exponents, consider $s = \frac{p}{q}$ and $r = \frac{p}{p-q}$. I suggest you finish the argument for this case as well.

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  • $\begingroup$ How about if the measure is not finite, but rather $\sigma$- finite? $\endgroup$
    – user996159
    Oct 10, 2023 at 6:50

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