I have the following question: Let $f=f(x,y)$ be continuous at $(0,0)$ and integrable there. Calculate:$$\lim\limits_{r\to 0^+}\dfrac 1 {\pi r^2}\iint\limits_{|(x,y)|\le r} f(x,y) \, dS$$ Since $f(0,0)$ is not given, I can only conclude from what's given, that $f$ is bounded in the circular vicinity of $(0,0)$, and that doesn't seem to help. Can anyone give me a direction?


Take $\lambda_r = \sup\limits_{|(x,y)|\leq r} f(x,y)$ and $\mu_r = \inf\limits_{|(x,y)|\leq r} f(x,y)$. Observe we have that for all $r$, $$ \mu_r \leq \frac{1}{\pi r^2}\int_{B(0,r)} f(x,y) \, dx \, dy \leq \lambda_r.$$ Because of continuity we have that $\lambda_r \to f(0,0)$ and $\mu_r \to f(0,0)$ as $r \to 0$. Hence the limit follows.

  • $\begingroup$ but $f(0,0)$ is not given as i've said. Otherwise it would've been trivial that the integral is $\le f(0,0)*$size of the ball. $\endgroup$ – CodeHoarder Jan 18 '17 at 6:25
  • $\begingroup$ I think you might be confused. The integral is not necessarily bounded above by $f(0,0)\pi r^2$, think about when $f(0,0)$ is a minimum. In regards to not knowing what $f(0,0)$is, it is some value and because of continuity we know that in small neighborhoods around it the function starts approaching that value. $\endgroup$ – Leon Sot Jan 18 '17 at 6:28
  • $\begingroup$ By definition if f is continuous , it's bounded by $f(0,0)+\epsilon$. Therefore, the integral is lower equal to the integral of $f(0,0)+\epsilon$, since that's a constant it equals to $(f(0,0)+\epsilon)\pi r^2$, That's what i've meant. $\endgroup$ – CodeHoarder Jan 18 '17 at 6:31
  • $\begingroup$ @CodeHoarder Yes you can do it that way as well, but remember that is only valid for $r \leq \delta$. Also, in your question you said integrable at $(0,0)$, I assume you meant integrable on some neighbourhood containing $(0,0)$? $\endgroup$ – Leon Sot Jan 18 '17 at 6:34
  • $\begingroup$ Yes, that's what i've meant, thanks. $\endgroup$ – CodeHoarder Jan 18 '17 at 10:30

HINT If $f(x,y)$ is as nice as you say, it will be approximately constant and equal to $f(0,0).$ in a small enough disk. You can pull $f(0,0)$ out of the integral.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.