Divergence of a vector field on a sequence of spheres

I'm studying for my exams and I found this problem in the book "Advanced Calculus", written by Friedman:

"Consider a sequence of spheres $S_n$ in $\mathbb{R}^3$ with center $P_n$ and radius $r_n$, such that $P_n \to P, r_n \to 0$ as $ n \to \infty.$ Let $\textbf{F}$ be a continuously differentiable vector field in a neighborhood of $P$. Prove that

$$(\nabla \cdot \textbf{F})(P)=\lim_{n \to \infty}\frac{1}{V_n}\iint\limits_{S_n}\textbf{F}\cdot \textbf{n}\ dS,$$

where $V_n$ is the volume of the ball with boundary $S_n$."

This is what I did:

Observe that, by the Divergence Theorem,

$$\frac{1}{V_n}\iint\limits_{S_n}\textbf{F}\cdot \textbf{n}\ dS = \frac{1}{V_n}\iiint\limits_{D_n}\nabla\cdot\textbf{F} \ dV,$$

where $D_n$ is the region bounded by $S_n$.

Now, notice that $\dfrac{1}{V_n}\displaystyle \iiint\limits_{D_n}\nabla\cdot\textbf{F} \ dV$ is the mean value of $\nabla \cdot \textbf{F}$ on $D_n$. Since F is continuously differentiable, we have that $\nabla \cdot \textbf{F}$ is continuous, so, for each $n$, there is a point $P_n$ such that $$\frac{1}{V_n} \iiint\limits_{D_n}\nabla\cdot\textbf{F} \ dV = (\nabla \cdot \textbf{F})(P_n)$$

So,

$$\lim_{n \to \infty}\frac{1}{V_n}\iint\limits_{S_n}\textbf{F}\cdot \textbf{n} \ dS = \lim_{n \to \infty}\frac{1}{V_n}\iiint\limits_{D_n}\nabla\cdot\textbf{F} \ dV \\ =\lim_{n \to \infty}(\nabla \cdot \textbf{F})(P_n) = (\nabla \cdot \textbf{F})(P),$$

as we wanted.

I have two problems here:

1) I don't know how to prove that because $\nabla \cdot \textbf{F}$ is continuous, we certainly have that there is a point $P_n$ such that $\frac{1}{V_n} \iiint\limits_{D_n}\nabla\cdot\textbf{F} \ dV = (\nabla \cdot \textbf{F})(P_n)$. I know it's true because I saw it in some calculus books, but I haven't found a proof.

2) After I finished, I realize that my points $P_n$ doesn't necessarily have to be the points $P_n$ center of $S_n$. So, because I know that the radius $r_n$ of $S_n$ goes to 0, can I assume that $P$ will be the limit of every sequence $(a_n)$ such that $a_k \in S_k$ $\forall k \in \mathbb{N}$?

I'd be really grateful if someone could help me with my questions above :) Thanks!

Solution 1:

I think that you can find such points by appealing to the first mean value theorem of integration for higher dimensions. If you would like a full proof I would be happy to give one. Then notice that since the balls have radii decreasing to $0$ and since $\nabla\cdot F$ is continuous then limit goes to what you want.

In the following proof we refer to a set, $A$, as Riemann integrable if the characteristic function of $A$ is Riemann integrable.

Lemma:(First Mean Value Theorem of Integration) Let $A\subset\mathbb{R}^{n}$ denote a compact, connected, Riemann integrable set. Let $f:A\to\mathbb{R}$ be a continuous function. Let $g:A\to\mathbb{R}$ be non-negative and Riemann integrable. Then there exists $c\in A$ such that:

$$\int_{A}f(x)g(x)dx=f(c)\int_{A}g(x)dx$$

Proof: If $\int_{A}g(x)=0$ then $g=0$ almost everywhere. Hence, $fg=0$ almost everwhere. Hence, $\int_{A}f(x)g(x)dx=0$. We may choose $c$ arbitrarily in this case to get the above equality. Now we may assume $\int_{A}g(x)dx>0$. Since $A$ is compact and $f$ is continuous then there exist $a\in A$ and $b\in A$ such that:

$$f(a)\le f(x)\le f(b)$$

for all $x\in A$. Since $g$ is non-negative we have:

$$f(a)g(x)\le f(x)g(x)\le f(b)g(x)$$

and finally integrating both over $A$ tells us that:

$$f(a)\int_{A}g(x)dx\le\int_{A}f(x)g(x)dx\le f(b)\int_{A}g(x)dx$$

So $f(a)\le\frac{\int_{A}f(x)g(x)dx}{\int_{A}g(x)dx}\le f(b)$. Since $f$ is continuous and $A$ is connected then $f(A)$ is connected and hence there is $c$ so that

$$f(c)=\frac{\int_{A}f(x)g(x)dx}{\int_{A}g(x)dx}$$

For the case you are interested in take $f=\nabla\cdot F$ and $g=1$.

Solution 2:

Your proof is correct with the proviso that the points $P_n$ occurring in the proof are not the centers of the spheres $S_n$.

The mean value theorem guarantees for any connected body $B$ and continuous function $f:\ B\to{\mathbb R}$ a point $\xi\in B$ with $$\int\nolimits_Bf(x)\ {\rm d}x=f(\xi)\ {\rm vol}(B)\ .$$ You have applied this principle with $f(x):={\rm div}\,{\bf F}(x)\ .$