Can anyone give me an example of $f(x) $ such that $ f \in L^2 ( \mathbb R)$ but $ x^{\frac{1}{2}} f \notin L^1 ( \mathbb R ) $. Thanks!
It seems that $f(x) = x^\alpha$ doesn't work...
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I thought $f \notin {L_1}$. I gave my answer based on that. – Mhenni Benghorbal May 02 '14 at 02:52
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Take $f(x)=x^{-1}\chi_{[1,\infty]}$
voldemort
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I think you need to add a $\log$ there, i.e. $f(x) = \log(x)x^{-1/2}\chi_{[1,\infty)}$. Unless I'm doing something stupid, your function is not in $L^2$ since $\int_1^\infty x^{-1} dx = \infty$. – snar May 02 '14 at 02:59
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@snarski: you are right. I actually meant $x^{-1}\chi_{[1,\infty]}$. Then the result of multiplying $x^{1/2}$ gives my function. Let me edit the answer. – voldemort May 02 '14 at 03:00