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Let f be a function on the real line R such that both f and xf are in L^2(R). Prove that f ∈ L^1(R) and the L^1 norm of f(x) is less than or equal to 8 times (the L^2 norm of f(x)) times the L^2 norm of xf(x).

I'm sorry I don't know how to use Latex to post the problem. The origional problem is here: It's Problem 18. It's one of the problems created by a teacher to help students prepare qualifying exams of real analysis.

I think the purpose of the question is to ask students to apply Plancherel Theorem, but I don't know where to start. I know all the properties of fourier transforms, but I couln't relate them here. I think the trick of the problem is to break f into two parts, but I couldn't figure out how to do it. Also, the form of the inequality is somewhat similar to the uncertainty principle, but I couldn't get any hints from the proof of the uncertainty principle beacause the hypothesis is different here.

I believe the question is not hard unless one knows where to start. I've tried my best and I'm still blind. Could anyone tell me how to do it? Any hints would be appreciated.

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up vote 1 down vote accepted

It is just an application of the Cauchy-Swartz inequality. For any $a>0$ $$\begin{align} \int_\mathbb{R}|f|&=\int_{|x|\le a}|f|+\int_{|x|>a}|x\,f|\,\frac{1}{|x|}\\ &\le\Bigl(\int_{|x|\le a}|f|^2\Bigr)^{1/2}\Bigl(\int_{|x|\le a}1\Bigr)^{1/2}+ \Bigl(\int_{|x|>a}|x\,f|^2\Bigr)^{1/2}\Bigl(\int_{|x|>a}\frac{1}{|x|^2}\Bigr)^{1/2}\\ &\le\sqrt{2\,a}\,\|f\|_2+\frac{\sqrt2}{\sqrt{a}}\|x\,f\|_2. \end{align}$$ Choose $a=b^2\|x\,f\|_2/(2\,\|f\|_2)$ to get $$ \|f\|_1\le\Bigl(b+\frac2b\Bigr)\sqrt{\strut\|f\|_2\,\|x\,f\|_2}\ . $$ Choose $b=\sqrt2$ and square to get $$ \|f\|_1^2\le8\,\|f\|_2\,\|x\,f\|_2\ . $$

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Thanks a lot. I didn't realize using this way to tell f is integrable. But could you please show me how to get the estimate like in the original problem posted? – student Feb 27 '14 at 18:41
See the edited answer. – Julián Aguirre Feb 27 '14 at 22:12
Strangely enough, the same question was posed one hour before, with more or less the same answer:… – Etienne Feb 27 '14 at 22:59
@Etienne: I'm sorry the question I posted twice, because I posted the earlier one without logging on my account. – student Feb 28 '14 at 0:07
@Julian Aguirre: Thank you very much for your elegant proof. I was kind of nerd in solving these problems. I wanted to try to apply Plancherel Theorem because somebody confidently told me that I should do the problem like that, but I failed. It's just like there are three methods to prove Hardy's Inequality: Fourier analysis, Jensen's Inequality and nondecreasing rearrangement of a function. Hardy's original proof was using Fourier analysis. But if someone told Hardy to prove it using nondecreasing rearrangement of a function, I suspect Hardy cannot give a proof. – student Feb 28 '14 at 0:13

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