# use plancherel theorem to prove an integral inequality

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).

I'm sorry I don't know how to use Latex to post the problem. The origional problem is here: http://www.math.purdue.edu/~bell/MA598R/advanced1.pdf. 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.

• You don't need Plancherel. Use Cauchy-Schwarz to show that $f$ is integrable on $\{ x;\; \vert x\vert\geq 1\}$. Commented Feb 27, 2014 at 17:34
• Actually, f is integrable on {x;|x|≥1} is obvious, since it is dominated by xf. But f is integrable on R\{x;|x|≥1} is not obvious. Besides, the key point is how to give an estimate of the L^1 norm of f, just like the expression given in the original problem post. Commented Feb 27, 2014 at 17:50
• I don't understand your first sentence. I presume you wanted to say "$f$ is square integrable on $\{ \vert x\vert\geq 1\}$ is obvious". On the other hand, $f$ integrable on $[-1,1]$ is indeed obvious since $L^2(0,1)\subset L^1(0,1)$. Commented Feb 27, 2014 at 18:14

Set $a:=\Vert f\Vert_{L^2}$ and $b:=\Vert xf\Vert_{L^2}$.
For any $\varepsilon >0$, you may write, using Cauchy-Schwarz: \begin{eqnarray} \int_{\mathbb R} \vert f\vert&=&\int_{\{\vert x\vert\leq\varepsilon\}}\vert f\vert +\int_{\{\vert x\vert<\varepsilon\}}\vert f\vert\\ &\leq& \sqrt{2\varepsilon}\, a+b\times \left(\int_{\{\vert x\vert>\varepsilon\}} \frac{dx}{x^2}\right)^{1/2}\\&=&\sqrt{2\varepsilon}\, a+\sqrt{\frac2\varepsilon}\, b\, . \end{eqnarray}
Choosing $\varepsilon:=\frac{b}a$, this gives $$\int_{\mathbb R} \vert f\vert\leq 2\sqrt {2ab}\, .$$
So $f$ is in $L^1$, with the required estimate.