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Can somebody tell me what's wrong with the following argument?

If $f$ is $L^1$ Lebesgue-integrable, say $f$ positive, then it is bounded almost everywhere by some bound $M$. Then $f^2 < M\cdot f$ which is in $L^1$, then $f$ is in $L^2$ and $L^1$ lies in $L^2$. It seems to me that the map $x^{-1/2}$ is $L^1$ but not $L^2$ on $(0,1)$, hence a counterexample...

So I'm a bit confused.

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Integrable functions may not be almost everywhere bounded. – Did Jul 26 '12 at 9:48
@superM: no, it's not. – Martin Argerami Jul 28 '12 at 22:25

Your argument shows that if $f\in L^1$ and it is bounded a.e., then it is in $L^2$. But, as your own example shows, not every function in $L^1$ is bounded.

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do you perhaps have any proof that any bounded $L^1$ function is in $L^2$ ? – Dominique May 19 '14 at 0:01
Yes. The easiest one is the one typed in the question above. – Martin Argerami May 19 '14 at 0:07
Yes, and it makes indeed sense to me. But I was wondering if you perhaps knew any sources which might proof this claim ? – Dominique May 19 '14 at 0:08
I don't get what you are saying. There is a very concrete proof by tina above. That's it. A more sophisticated way of stating tina's proof is using Hölder $1,\infty$, as in $$\int|f|^2\leq\|f\|_\infty\,\|f\|_1.$$ – Martin Argerami May 19 '14 at 4:35

$f \in L^2(a,b) \Rightarrow f \in L^1(a,b)$, but the converse is false! you just gave a counterexample.

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