Let $f$ be a continuous function on the closure of $U$ where $U=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2<1 \}$ and harmonic on $U$, If $$f(x,y)=x^2y^2 $$ on $\partial U$ I want to find $f(0,0)$.

Now I could of course solve the Laplace equation on the disk and then calculate the value at the origin, but I was wondering if I could apply the mean value property.

Now the mean value property holds if the function is harmonic on the closure of $U$, but here it is harmonic on only the interior, but I actually tried to compute the mean value and I got the same value that I got when I solved the Laplace equation on the disk and calculated the value at the origin, so I am guessing that the mean value property must hold.

Perhaps the fact that $f$ is continuous on the closure is helpful.

My question is, can I use the mean value property in this case?


1 Answer 1


The mean value property holds for any disk of radius $1-\varepsilon$ with $\varepsilon$ as close to zero as you want. Observe that all those means over the boundaries are equal, independent of $\varepsilon$, since all of them are $f(0,0)$. By (uniform) continuity, the limit of the integrals over circumferences of radius $1-\varepsilon$ is equal to the integral over the circumference of radius $1$, so the property holds.


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