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## 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:

$\begin{array}{rl} \displaystyle{\int_{\mathbb{R}} \lvert f \rvert\, d\mu} &= \displaystyle{\int_{\lvert x \rvert \leq \lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \lvert f \rvert\, d\mu} \\ \\ &= \displaystyle{\int_{-\lambda}^{\lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \tfrac{1}{\lvert x \rvert} \lvert g(x) \rvert\, d\mu} \\ \\ &= \displaystyle{\int_\mathbb{R} \lvert f \rvert \cdot \boldsymbol{1}_{\lvert x \rvert \leq \lambda}\, d\mu + \int_{\mathbb{R}} \lvert g \rvert \cdot \big( \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda}\big) \, d\mu} \\ \\ &\leq \lVert f \rVert_2 \cdot \big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 + \lVert g \rVert_2 \cdot \big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 \end{array}$

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:

$\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 = \sqrt{2\lambda}$

$\big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 = \displaystyle{ \sqrt{ 2\int_\lambda^\infty \dfrac{dx}{x^2}} = \sqrt{ \dfrac{2}{\lambda}} }$

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.