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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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    @superM: no, it's not.2012-07-28

2 Answers 2

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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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    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.$2014-05-19
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$f \in L^2(a,b) \Rightarrow f \in L^1(a,b)$, but the converse is false! you just gave a counterexample.