# $L^2\subset L^1$ in case if the measure is finite [closed]

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

• Try Hölder's inequality. May 16, 2022 at 5:32

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 $$a \le a^2$$.

In fact, if $$(X,\mu)$$ is a finite measure space, you have a stronger result, for all $$1\le q then $$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.

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