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Show that $L^4([0,1]) \subset L^2([0,1])$

I have problems understanding the (rather simple it seems) proof. It goes likes this ($\int$ denotes $\int\limits_{0}^{1}$):

Let $f\in L^4([0,1])$, then, by Hölder's inequality:

$\int |f|^2 = \int |1|^2|f|^2 \le 1\ (\int |f|^4)^{1/2} < \infty$

My problem is in the $\le$ step. I do not see how this follows from Hölder's. No matter how much I try, I can only prove that

$$L^4([0,1]) \subset L^1([0,1])$$

what am I missing? If I assume that $f\in L^4([0,1])$ implies that $f^2 \in L^2([0,1])$, the result follows trivially. Is that assumption true though?

JustANoob
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  • We even have $|f|{L^p([0, 1])} \nearrow |f|{L^{\infty}([0, 1])}$ as $p \nearrow \infty$. You can use Holder's inequality or Jensen's inequality to prove this. – Mason Aug 21 '22 at 17:05

1 Answers1

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Basically, you have that: $$\int_{0}^{1} |f|^2 \ d\lambda = \left|\left| |f|^2 \right|\right|_1^2 = \left|\left| 1\cdot |f|^2 \right|\right|^2_1 \leq \left|\left| 1 \right|\right|^2_2\cdot \left|\left| |f|^2 \right| \right|^2_2$$

I'm just applying Hölder's Inequality here. Then: $$\left|\left| 1 \right|\right|_2^2 = \int_{0}^{1} 1 \ d\lambda = 1$$ and: $$\left|\left| |f|^2 \right|\right|^2_2 = \int_{0}^{1} |f|^4 \ d\lambda$$ Since $f \in L^4[0,1]$, it follows that: $$\int_{0}^{1} |f|^2 \ d\lambda \leq \int_{0}^{1} |f|^4 \ d\lambda < \infty$$ and that proves what you wanted. Alternatively, you can also see that: $$|f|^2 = 1 \cdot |f|^2 \leq \frac{1+|f|^4}{2}$$

by the AM-GM inequality. Therefore: $$\int_{0}^{1} |f|^2 \ d\lambda \leq \int_{0}^{1} \frac{1}{2} + \frac{1}{2} |f|^4 \ d\lambda$$

and since the right-hand side is finite, it follows that $f \in L^2[0,1]$.

Abhijeet Vats
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