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I'm trying to show that $ {L^{2}}([0,1]) $ is contained in $ {L^{1}}([0,1]) $. This is what I have so far:

Since $ f \in {L^{2}}([0,1]) $, then we have that $(\int_0^1 |f|^2)^{\frac{1}{2}}<\infty$. Thus we have $\int_0^1 |f|^2=M<\infty$, making $|f|^2$ integrable over $[0,1]$. This means that we can write $\int_{A_1} |f|^2+\int_{A_2} |f|^2$, where $A_1=\lbrace x\in[0,1]||f|^2>|f|\rbrace$ and $A_2=\lbrace x\in[0,1]||f|^2\leq |f|\rbrace$. Thus we have that $\infty > M=\int_{A_1} |f|^2+\int_{A_2} |f|^2\geq \int_{A_1} |f|$. Thus $f$ is integrable on $A_1$, and since $x\in A_2$ implies $f(x)<1$, then $\int_{A_2}f$ is integrable as well, hence $\int_{A_1} |f|+\int_{A_2} |f|=\int_0^1 f$ is integrable.

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This looks fine. The crux of the matter here is that $\int_{A_2}f$ is integrable here because $m(A_2)<\infty$, which fails to be true when considering $L_2[0,\infty)$ –  Alex R. Jan 9 '13 at 3:38
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Dear Frank, Aren't you showing that $L_2$ is contained in $L_1$, rather than the other way round? Regards, –  Matt E Jan 9 '13 at 3:50
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Consider the function $f(x)=\frac{1}{\sqrt{x}}$. This is in $L_1[0,1]$ but not in $ L_2[0,1] $. –  Mhenni Benghorbal Jan 9 '13 at 4:01
    
@Mhenni: Good counterexample! –  Haskell Curry Jan 9 '13 at 4:15
    
@HaskellCurry: Thank you. –  Mhenni Benghorbal Jan 9 '13 at 4:32

2 Answers 2

up vote 1 down vote accepted

You should be proving that $ {L^{2}}([0,1]) \subseteq {L^{1}}([0,1]) $ instead.

Due to an earlier typo error in the title, there was some minor confusion as to the correctness of the OP's proof, which is actually correct.

Let $ f \in {L^{2}}([0,1]) $. Define $$ A := \{ x \in [0,1] ~|~ |f(x)| \leq 1 \} \quad \text{and} \quad B := \{ x \in [0,1] ~|~ |f(x)| > 1 \}. $$ For each $ x \in B $, we have $ |f(x)| < |f(x)|^{2} $, so $$ \int_{B} |f|^{2} ~d{\mu} < \infty \Longrightarrow \int_{B} |f| ~d{\mu} < \infty. $$

Next, notice that $ \displaystyle \int_{A} |f| ~d{\mu} < \infty $ because $ |f| $ is bounded by the value $ 1 $ on $ A $, which has finite measure. As $ \{ A,B \} $ partitions the interval $ [0,1] $, we see that $$ \int_{[0,1]} |f| ~d{\mu} = \int_{A} |f| ~d{\mu} + \int_{B} |f| ~d{\mu} < \infty. $$ Therefore, $ f \in {L^{1}}([0,1]) $.

Conclusion: $ {L^{2}}([0,1]) \subseteq {L^{1}}([0,1]) $.

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Dear Haskell, While this is correct, it is also the solution written in the question. (I think one should put the reversed title down to a typo.) Regards, –  Matt E Jan 9 '13 at 4:24

Here is a lemma that will be of great use to you in the future.

Lemma. Let $1 \leq p \leq q \leq \infty$ and $\mu(X) < \infty$. If $f \in L^q$, then $f \in L^p$ and $\|f\|_p \leq \mu(X)^{1/p-1/q}\|f\|_q$.

To prove it note that the case $p=1$ is trivial, and for $p>1$ note that $q/p$ and $q/(q-p)$ are Holder conjugates so that $\|f\|_p^p \leq \|f\|_{q/p}\|1\|_{(q-p)/q} = \|f\|_q^p \mu(X)^{(q-p)/q}.$

This of course implies $L^q(X) \subseteq L^p(X)$ when $\mu(X) < \infty$ and $p\leq q$.

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