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 $\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2$ 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.