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I'm reading a computation that shows that the mean value property for a function implies that function is harmonic. It includes the following computation:

\begin{align}\int_{B(x,r)}\Delta h(y) dy &= \int_{\partial B(x,r)} \nabla h(\vec{y})\cdot \vec{\nu}\text{ }dS(\vec{y}) \\ &= \int_{\partial B(x,r)} \nabla h(\vec{y})\cdot \frac{\vec{y}-\vec{x}}{r}\text{ }dS(\vec{y}) \\ &= \int_{\partial B(0,1)} \nabla h(\vec{x}+r\vec{z})\cdot \vec{z} \underbrace{\quad \quad}_{\text{why no factor of r?: } d\vec{y} = rd\vec{z}} \text{ }dS(\vec{z}) \\ &= \underbrace{\frac{d}{dr}}_{\text{why can this be puled through?}} \int_{\partial B(0,1)} h(\vec{x}+r\vec{z}) \text{ }dS(\vec{z}) \end{align}

Can anyone answer the questions posed in lines 3 and 4?

yoshi
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  • What dimension are you working in? In general, $dy = r^{n-1}dz$ in $\mathbb{R}^n$. Also, what regularity is assumed for $h$? I suppose $C^2$, which allows the exchange of $d/dr$ with the integral. – Jeff Jul 02 '18 at 01:46
  • I'm working in n-dims. Here I set $\frac{\vec{y}-\vec{x}}{r} = \vec{z}$, how do you get $r^{n-1}$ then? – yoshi Jul 03 '18 at 22:09
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    If you set $y=rz$ in $\mathbb{R}^n$, you are really setting $y_1=rz_1,y_2=rz_2,\dots,y_n=rz_n$, and so $dy = dy_1\cdots dy_n = r^n dz_1\cdots dz_n=r^ndz$. Since the surface of the sphere is $n-1$ dimensional you get $r^{n-1}$ instead, since the change of variables is really taking place in $\mathbb{R}^{n-1}$. – Jeff Jul 04 '18 at 01:46
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    In general, if $y=Az$ for a matrix $A$, then the change of variables formula is $dy = |det(A)|dz$, where $|det(A)|$ is usually called the Jacobian of the map. – Jeff Jul 04 '18 at 01:48
  • Ah, thanks! I wrote everything out explicitly and see it now. Also $u \in C^2$. Why should if follow that $d/dr$ pulls through? – yoshi Jul 04 '18 at 14:28
  • this is by Dominated Convergence Theorem: https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proofs – yoshi Jul 04 '18 at 15:49
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    Yes, it follows by dominated convergence, but also from the uniform convergence of the difference quotients. You only need $h\in C^1$ to exchange the $d/dr$ with the integral. – Jeff Jul 06 '18 at 00:14
  • For example, see https://math.stackexchange.com/questions/2530213/when-can-we-interchange-integration-and-differentiation – Jeff Jul 06 '18 at 00:15

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