Riesz potential-inequality

Question

Does the following hold?

There is a $C>0$ such that

$$\left|\left|\int_{\mathbb{R}^3} \frac{|f(y)|^2}{|x-y|} dy\right|\right|_{L^3(\mathbb{R}^3)} \leq C||f||^2_{L^2(\mathbb{R}^3)}$$ for all $f \in L^2(\mathbb{R}^3).$

Any advice would be appreciated.

Answer 1

You can apply this famous result and the fact that $$\||f|^2\|_{L^{1}(\Bbb{R}^3)}=\|f\|_{L^{2}(\Bbb{R}^3)}^2.$$

Answer 2

Let's introduce Riesz potential $$ I_\alpha f(x) = \int_{\mathbb{R}^n} \frac{f(y)}{|x-y|^{n - \alpha}} \; dy. $$ According to the Hardy-Littlewood-Sobolev frcational integration theorem if $$ \frac{1}{q} = \frac{1}{p} - \frac{\alpha}{n}, $$ for $1 < p < n/\alpha$, then Riesz potential acts as a bounded operator from $L^p$ to $L^q$ $$ ||I_\alpha f||_{L^q(\mathbb{R}^n)} \le c ||f||_{L^p(\mathbb{R}^n)}. $$

Now just take $f^2$ instead of $f$ and set $q = 3$, $\alpha = 2$, $n = 3$, $p = 1$.

And everything seems correct, but wait... You cannot apply fractional integration theorem for $p = 1$! And the counterexample for this is famous as well: we need to take a characteristic function $$ f = \chi_{(0,1)^3} $$ Then for $|x| > 1$ $$ I_2 f(x) = \int_{(0,1)^3}\frac{dy}{|x-y|} \ge \frac{1}{2}\frac{1}{|x|} $$ So we have $$ ||I_2 f||_{L^3 (\mathbb{R}^3)}^3 \ge \frac{1}{2} \int_{|x| > 1} \frac{dx}{|x|^3} = \infty. $$ On the other side $$ ||f||_{L^2(\mathbb{R}^3)} = 1 $$ as the volume of unite cube. So this inequality does not hold even for so simple function.