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 $x^2+y^2\to \infty$