subharmonic functions and the volume average over a ball is nodecreasing

Question

Let $u \in C(\Omega)$, $\Omega \subset \mathbb{R}^n$. Using the notations of Sect. 3.4.1, prove the following statements:

The spherical average $S(u; x, r)$ is defined by: $ S(u; x, r) = \frac{1}{\omega_n r^{n-1}} \int_{\partial B_r(x)} u \, d\sigma, $ where $\omega_n$ is the surface area of the unit sphere in $\mathbb{R}^n$.

The average over the volume $A(u; x, r)$ is defined by: $ A(u; x, r) = \frac{n}{\omega_n r^n} \int_{B_r(x)} u(y) \, dy. $ :If $u \in C(\Omega)$ is subharmonic in $\Omega$, then $r \mapsto S(u; x, r)$ and $r \mapsto A(u; x, r)$ are nondecreasing functions.

i can proof that $S$ is nondecreasing .

Let $g(r) = \frac{1}{\omega_n r^{n-1}} \int_{\partial B_r(x)} u(\sigma) \, d\sigma$.

Perform the change of variables $\sigma = x + r\sigma'$. Then $\sigma \in \partial B_1(0)$, $d\sigma = r^{n-1} \, d\sigma'$, and

$ g(r) = \frac{1}{\omega_n} \int_{\partial B_1(0)} u(x + r\sigma') \, d\sigma'. $

Let $v(y) = u(x + ry)$ and observe that

$ \nabla v(y) = r \nabla u(x + ry), \quad \Delta v(y) = r^2 \Delta u(x + ry). $

Then we have

\begin{align*} g´(r) &= \frac{1}{\omega_n} \int_{\partial B_1(0)} \frac{d}{dr} u(x + r\sigma') \, d\sigma' \\ &= \frac{1}{\omega_n} \int_{\partial B_1(0)} \nabla u(x + r\sigma') \cdot \sigma' \, d\sigma' \\ &= \frac{1}{\omega_n r} \int_{\partial B_1(0)} \nabla v(\sigma') \cdot \sigma' \, d\sigma' \quad (\text{divergence theorem}) \\ &= \frac{1}{\omega_n r} \int_{B_1(0)} \Delta v(y) \, dy \\ &= \frac{r}{\omega_n} \int_{B_1(0)} \Delta u(x + ry) \, dy \\ &\geq 0. \end{align*} because $u$ is subharmonic, so $\Delta(u) \geq 0$. Then $S$ is nodecreasing, but for $A$ im can´t the similary argument

Answer 1

I will leave my other answer up, even though it is incorrect, so that I can reference the notation used in it. The work we have done has been sufficient to show that $v'(0)\geq 0$, but not $v'(r)\geq 0$ for all $r\geq 0$.

To improve this, you can compute (exercise) $$v'(r)=\frac{n}{r}\big(a(r)-v(r)\big) \\ \forall r > 0$$

And so it suffices to show $a(r)\geq v(r)$. To do this we just compute $v(r)$ and apply an estimate:

$$v(r)=\frac{1}{V_nr^n}\int_0^r A_{n}{r'}^{n-1}~a(r')\mathrm dr'$$ However, since $a$ is non-decreasing, we know that $a(r')\leq a(r)$ for all $r'\leq r$, and hence $$v(r)=\frac{1}{V_nr^n}\int_0^r A_{n}{r'}^{n-1}~a(r')\mathrm dr'\leq \frac{a(r)}{V_nr^n}\int_0^r A_n {r'}^{n-1}\mathrm dr'$$ Doing the integration and using the fact that $A_n=nV_n$, we get $$v(r)\leq a(r)$$ Thus $v'(r)\geq 0$ for all $r>0$ so $v$ is non-decreasing.

Answer 2

EDIT: THIS IS INCORRECT, but it does show that $v'(0)\geq 0$ which we need.

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Letting $$v(r)=\boldsymbol{-\kern-11.2pt}\int_{\mathbb B(x,r)}\phi~\mathrm d\mu^n=\frac{1}{V_nr^n}\int_{\mathbb B(x,r)}\phi~\mathrm d\mu^n$$ And $$a(r)=\boldsymbol{-\kern-11.2pt}\int_{\partial\mathbb B(x,r)}\phi~\mathrm d\mu^{n-1}=\frac{1}{A_nr^{n-1}}\int_{\partial\mathbb B(x,r)}\phi~\mathrm d\mu^{n-1}$$ You can see that $$v(r)=\frac{1}{V_nr^n}\int_0^r\int_{\partial\mathbb B(x,r)}\phi~\mathrm d\mu^{n-1}~\mathrm dr \\ =\frac{1}{V_nr^n}\int_0^r A_{n}{r'}^{n-1}~a(r')\mathrm dr'$$ Since $a$ is non-decreasing (you showed this) $$v(r)\geq \frac{1}{V_nr^n}\int_0^r~A_n {r'}^{n-1}~a(0)\mathrm dr'$$ But, $a(0)=\phi(x)=v(0)$, and thus $$v(r)\geq v(0)~\frac{1}{V_nr^n}\int_0^r A_n {r'}^{n-1}\mathrm dr' \\ =v(0)\frac{1}{V_n r^n}~\frac{A_nr^n}{n} \\ =v(0)\frac{1}{V_nr^n}V_nr^n \\ =v(0)$$ So $v(r)\geq v(0)$ for all $r\geq 0$ thus (*) $v$ is non-decreasing.

(*): This is not true!!