Problem involving Lebesgue Integral

Question

Suppose $f\in L^1(\mathbb{R}^2)$ is s.t. $||f||_{L^1}=1$. I want to prove that $\exists x\in\mathbb{R}^2$ with $|x|\leq 1$ AND $\int_{\mathbb{R}^2}f(y)|x-y|^{-1}dy<100$. I tried taking a double integral, switching them, expressing the integral over $\mathbb{R}^2$ as a sum and I can make the double integral $\int_{|y|\geq 2}\int_{|x|\leq 1}|f(y)||x-y|^{-1}dxdy$ smaller than $2\pi$. But I'm not sure how to make the integral over the ball of radius 2 small.

Answer 1

Let $y \in \mathbb R^2$. You can estimate $\displaystyle \int_{B(0,1)} \frac{1}{|x-y|} \, dx$ from above in a couple of ways.

1) If $y \notin B(0,2)$ then $|x-y| \ge 2$ for all $x \in B(0,1)$, and $\displaystyle \int_{B(0,1)} \frac{1}{|x-y|} \, dx \le \frac 12 \int_{B(0,1)} \, dx = \pi$.

2) If $y \in B(0,2)$ then $B(0,1) \subset B(y,3)$ and thus $$\int_{B(0,1)} \frac{1}{|x-y|} \, dx \le \int_{B(y,3)} \frac{1}{|x-y|} \, dx = \int_{B(0,3)} \frac{1}{|x|} \, dx = 6\pi.$$

(You may need to check that last integral; it looks like $6\pi$ to me...)

Assume without loss of generality that $f \ge 0$. Suppose to the contrary that $\displaystyle \int_{\mathbb R^2} \frac{f(y)}{|x-y|} \, dy \ge 100$ for all $x \in B(0,1)$. Then $$ \int_{B(0,1)} \int_{\mathbb R^2} \frac{f(y)}{|x-y|} \, dy dx\ge 200 \pi. $$ On the other hand, $$ \int_{B(0,1)} \int_{\mathbb R^2} \frac{f(y)}{|x-y|} \, dy dx = \int_{\mathbb R^2} f(y) \int_{B(0,1)} \frac{1}{|x-y|} \, dx dy \le 6\pi \|f\|_{L^1}. $$ Thus $\|f\|_{L^1} > 1$, a contradiction.