# When does the limit of the mean values of a function around a point approach the value of the function at that point ?

When does the limit of the mean values of a function around a point approach the value of the function at that point ? We can prove it if the function is continuous. But are there general classes of functions for which this holds ?

In precise language , what is the most general class of functions for which the following hold ?

$\frac{1}{n\alpha(n)\epsilon^{n-1}}\int_{\partial B(x,\epsilon)}f(y)dS(y) \rightarrow f(x)$ as $\epsilon \rightarrow 0$

• If the function is harmonic, this comes straight from the definition, perhaps more context is needed – Ellya Jun 11 '14 at 8:14
• yes but f(y) is not harmonic it comes from poisson equation delta u = -f.how it comes from definition? – Raja Sekar Jun 11 '14 at 8:27
• if it were harmonic, the mean integral over a ball centred at $x$ is the function evaluated at $x$. – Ellya Jun 11 '14 at 8:31
• ok sorry mean value property.again sorry sometimes it happens.but can you please check this out and give me some explanation.(math.stackexchange.com/questions/828836/…) – Raja Sekar Jun 11 '14 at 8:37
• Pointwise it can be absolutely horrible, note that the limit holds if you replace $f(y)$ by $f(y) + g(y)$ where around every ball $\partial B(x,\epsilon)$ the integral of $g$ vanishes. In particular, $g(y)$ doesn't even need to be Lebesgue integrable on $B(x,1)$. If you want this to hold everywhere, then the question may get more interesting. – Willie Wong Jun 23 '14 at 11:16

• It doesn't look to me like the original question is asking about functions for which this holds almost everywhere. I guess as long as the derivative of the integral exists everywhere for a Lebesgue integrable function $f$, then what you can say is that it is almost everywhere equal to a unique representative function which has the property OP asks for. – Dustan Levenstein Jun 23 '14 at 6:06
• This has been used even before introducing MVT in Page 25 of Evans when there are no restrictions on $f$. – Srinivas K Jun 28 '14 at 12:13