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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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    @AlexBecker : I am particularly not assuming that it belongs to particular function space or a particular convergence . I would definitely appreciate if you point out for which function space does it actually make sense .2012-07-18

2 Answers 2

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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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    You need to specify a topology and some more details. I'm out of depth here, but possibly you could use Meyers/Serrin?2012-07-18
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If $\,h\,$ is continuous then it has a primitive function $\,H\,$ , say as a function of $\,y\,$ , so:

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