# finding the Limit of a integral .

Can anyone help me to find the limit of the following problem ?

$$\lim_{r\to0^+} \frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy$$

What i think here is to use lesbegue differentiation theorem. I am not being successful. Can anyone give me hints to solve it explicitly.

$h$ is continuous .

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What are the hypothesis on $h$? –  student Jul 18 '12 at 15:54
Is $h$ continuous in $y$ at $(x,t)$? –  copper.hat Jul 18 '12 at 15:55
Did you mean to have $x$ as the limit variable or $r$? –  copper.hat Jul 18 '12 at 15:56
@copper.hat : i have edited . sorry for typo. –  Theorem Jul 18 '12 at 15:58
What topology do you have on your function space? The limit will be a function of $x$, so it's important to specify what it means for a family of functions to converge. Are we looking at the $L^p$ norm? Pointwise convergence? Uniform convergence? –  Alex Becker Jul 18 '12 at 16:02

If $h$ is continuous, you don't need any Lebesgue tools.

Since $h$ is continuous then $\forall \epsilon>0$, there exists $\delta>0$ such that if $|y-t|<\delta$, then $|h(x,y)-h(x,t)| < \epsilon$.

Then we have $\frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy = h(x,t)+\frac{1}{2r} \int^{r+t}_{-r+t} (h(x,y)-h(x,t)) dy$. If $r<\delta$, we can bound the second term using: $|\int^{r+t}_{-r+t} (h(x,y)-h(x,t)) dy| \leq \int^{r+t}_{-r+t} |(h(x,y)-h(x,t))| dy \leq \epsilon \int^{r+t}_{-r+t} dy = 2 r \epsilon$.

So, if $r < \delta$, we have $|\frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy - h(x,t)| \leq \epsilon$. Since $\epsilon>0$ was arbitrary, it follows that the limit is: $$\lim_{r\to0^+} \frac{1}{2r} \int^{r+t}_{-r+t} h(x,y) dy = h(x,t).$$

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what should be done if $h(x,y)$ belongs to some function space , say sobolev space . Is there a way –  Theorem Jul 18 '12 at 16:18
You need to specify a topology and some more details. I'm out of depth here, but possibly you could use Meyers/Serrin? –  copper.hat Jul 18 '12 at 16:33

If $\,h\,$ is continuous then it has a primitive function $\,H\,$ , say as a function of $\,y\,$ , so:

$$\lim_{r\to 0^+}\frac{1}{2r}\int_{-r+t}^{r+t}h(x,y)\,dy=\lim_{r\to 0^+}\frac{1}{2r}\left(H(x,r+t)-H(x,-r+t)\right)\stackrel{\text{L'Hospital!}}=$$ $$=\lim_{r\to 0^+}\frac{H'_r(x,r+t)+H'_r(x,-r+t)}{2}=\lim_{r\to 0^+}\frac{h(x,r+t)+h(x,-r+t)}{2}=h(x,t)$$

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