📂Partial Differential Equations

Smoothing Effect of Harmonic Functions

Theorem

Mean Value Property

$$\begin{align*} u(x) = -\!\!\!\!\!\! \int_{\partial B(x,r)} udS = -\!\!\!\!\!\! \int _{B(x,r)} udy \end{align*}$$

Assuming $u \in C(\Omega)$ satisfies the mean value property in each open ball $B(x,r)\subset \Omega$, then the following holds.

$$u \in C^{\infty}(\Omega)$$

Description

If it’s Harmonic, it means it’s smooth inside. It is important to note that smoothness or continuity is not guaranteed at the boundary $\partial \Omega$.

Proof

Assuming $\epsilon>0$ is given. The $\epsilon$-Mollification of $u$ is as follows.

$$u^\epsilon=\eta_\epsilon *u \in C^\infty(\Omega_{>\epsilon})$$

At this time, $\Omega_{>\epsilon} := \left\{ x \in \Omega : \mathrm{dist}(x, \partial \Omega) > \epsilon \right\}$ applies. Let’s say $x \in \Omega_{>\epsilon}$. Then, the following holds.

$$\begin{align*} u^\epsilon (x) &= \int_\Omega \eta_\epsilon (x-y)u(y)dy \\ &= \dfrac{1}{\epsilon^n}\int_{\Omega} \eta \left( \frac{x-y}{\epsilon} \right) u(y)dy \\ &= \dfrac{1}{\epsilon^n}\int_{B(x,\epsilon)}\eta \left( \frac{|x-y|}{\epsilon} \right) u(y) dy \end{align*}$$ The third equality holds because, by definition of the Mollifier $\eta$, the value outside the ball is $0$. Separating the surface integral and the integral over the radius yields the following.

$$\dfrac{1}{\epsilon^n} \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) \left( \int _{\partial B(x,r)} udS \right) dr$$

Using the mean value property leads to the following.

$$\begin{align*} &\quad \dfrac{1}{\epsilon^n} \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) \left( \int _{\partial B(x,r)} udS \right) dr \\ &= \dfrac{1}{\epsilon^n} \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) \left( \dfrac{n\alpha (n)r^{n-1}}{n\alpha (n)r^{n-1}}\int _{\partial B(x,r)}u dS \right) dr \\ &= \dfrac{1}{\epsilon^n} \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) n\alpha (n)r^{n-1} -\!\!\!\!\!\! \int _{\partial B(x,r)}u(y) dS(y)dr \\ &= \dfrac{1}{\epsilon^n} \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) n\alpha (n)r^{n-1} u(x) dr \\ &= \dfrac{1}{\epsilon^n}u(x) \int_{0}^\epsilon \eta \left( \frac{r}{\epsilon} \right) n\alpha (n)r^{n-1}dr \end{align*}$$

Here, $n\alpha (n)r^{n-1}$ is the surface area of the ball with radius $r$, so the above integral can be rewritten as follows.

$$\dfrac{1}{\epsilon^n}u(x) \int_{B(0,\epsilon)}\eta \left( \frac{r}{\epsilon} \right) dr=u(x) \int_{B(0,\epsilon)} \eta_\epsilon (y)dy=u(x)$$

The last equality holds because, by the definition of $\eta_\epsilon$, the integral within the ball $B(r,\epsilon)$ is $1$. Therefore, for all $\epsilon$, $u=u^\epsilon \ \mathrm{in}\ \Omega_{>\epsilon}$ and $u^\epsilon \in C^\infty(\Omega)$, thus $u \in C^\infty (\Omega)$

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