Limit of a Nonnegative Continuous Function

  • POTW
  • Thread starter Euge
  • Start date
In summary, the limit of a nonnegative continuous function is the value that the function approaches but never quite reaches at a point. It is calculated by evaluating the function at values approaching the given point from both sides, and if the values approach the same value, then that value is the limit. The limit of a nonnegative continuous function is important because it helps us understand the behavior of the function at a particular point and is crucial in many mathematical and scientific applications. It cannot be negative and is represented on the graph as the horizontal asymptote, which helps us understand the overall behavior and shape of the function.
  • #1
Euge
Gold Member
MHB
POTW Director
2,057
215
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##.
 
  • Like
Likes jbergman and topsquark
Physics news on Phys.org
  • #2
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##
 

Similar threads

  • Math POTW for University Students
Replies
15
Views
1K
  • Math POTW for University Students
Replies
3
Views
724
  • Math POTW for University Students
Replies
1
Views
3K
  • Math POTW for Graduate Students
Replies
1
Views
806
  • Math POTW for Graduate Students
Replies
2
Views
767
  • Math POTW for University Students
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
169
Replies
2
Views
997
  • Calculus
Replies
11
Views
2K
  • Math POTW for University Students
Replies
1
Views
2K
Back
Top