Limit of a Nonnegative Continuous Function

[Euge]

Let ##D_r\subset \mathbb{R}^2## be the disk of radius ##r## centered at the origin. If ##f : \mathbb{R}^2 \to [0,\infty)## is uniformly continuous such that ##\sup_{0 < r< \infty} \iint_{D_r} f(x,y)\, dx\, dy < \infty##, show that ##f(x,y) \to 0## as ##x^2 + y^2 \to \infty##.

 

[Office_Shredder]

Since ##f## is uniformly continuous for every ##\epsilon >0## there exists ##\delta_{\epsilon} > 0## s.t. ##||(x_1,y_1)-(x_2,y_2)||<\delta ## implies ##|f(x_1,y_1)-f(x_2,y_2)|<\epsilon##.

Suppose that ##f## does not go to zero as ##x^2+y^2## goes to infinity. Since ##f## is non negative this means there exists some ##\epsilon >0## and a sequence of points ##(x_i,y_i)## such that ##x_i^2+y_i^2\to \infty ## and ##f(x_i,y_i)> 2 \epsilon##. Restrict to a subsequence such that ##\sqrt{x_i^2+y_i^2} > 2\delta_\epsilon \sqrt{x_{i-1}^2+y_{i-1}^2}##. Then the ball of radius ##\delta_{\epsilon}## around each ##(x_i,y_i)## do not intersect. We will call these balls ##B_i##

furthermore, on each ##B_i##, we have ##f> \epsilon## at every point, since it can be no more than ##\epsilon## smaller than ##f(x_i,y_i)\geq 2\epsilon##. Hence ##\int \int_{B_i}f(x,y)dx dy \geq \pi \delta_\epsilon^2 \epsilon##.

pick a sequence ##r_k## such that ##r_k>\sqrt{x_k^2+y_k^2}+\delta_{\epsilon}##. Then ##D_{r_k}## contains ##B_1,..., B_k## entirely, and since ## f## is non negative,

$$ \int\int_{D_{r_k}} f(x,y)dx dy \geq \sum_{i=}^k \int \int_{B_i} f(x,y)dx dx\geq k\pi \delta_{\epsilon} ^2\epsilon $$

the last expression is just some constant multiplied by ##k##, which goes to infinity as ##k## does. This contradicts the assumption that the supremum of the integral over all radii choices is finite. Hence ##f## must go to zero as ##x^2+y^2\to \infty##