# Advanced Problem #18

Another fun problem, as requested by Qinfeng Li in MA598R: Measure Theory

Let \(f \in L_2(\mathbb{R})\) so that \(g(x)=xf(x) \in L_2(\mathbb{R})\) too. Show that \(f \in L_1(\mathbb{R})\) and \(\lVert f \rVert_1^2 \leq 8 \lVert f \rVert_2 \lVert g \rVert_2.\)

I got this problem by picking some strictly positive value \(\lambda\) and breaking the integral \(\int_{\mathbb{R}} \lvert f \rvert\, d\mu\) as follows:

Let us examine now the factors and \( \big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 \) above:

We have thus proven that \(f \in L_1(\mathbb{R})\) with \(\lVert f \rVert_1 \leq \sqrt{2\lambda} \lVert f \rVert_2 + \sqrt{\tfrac{2}{\lambda}} \lVert g \rVert_2.\) At this point, all you have to do is pick \(\lambda = \lVert g \rVert_2 / \lVert f \rVert_2\) (provided the denominator is not zero) and you are done.