# Thread: Harmonic Functions

1. ## Harmonic Functions

Let $\displaystyle f: \Omega \subset \mathbb{R} ^n \longrightarrow \mathbb{R}$ such that $\displaystyle f \in L^1 ( \Omega )$ and for all $\displaystyle x \in \Omega$ and $\displaystyle R_x$ such that $\displaystyle \overline{B_{R_x} (x)} \subset \Omega$ we have $\displaystyle f(x)=\frac{1}{ \vert B_{R_x}(x) \vert } \int_{B_{R_x}(x)} \ f$ . Is $\displaystyle f$ harmonic in $\displaystyle \Omega$ ?

I know the result is true if $\displaystyle f \in C^2( \Omega )$, but I don't know how to argue that if $\displaystyle f \in L^1 ( \Omega )$ then it's twice differentiable.

2. Originally Posted by Jose27
Let $\displaystyle f: \Omega \subset \mathbb{R} ^n \longrightarrow \mathbb{R}$ such that $\displaystyle f \in L^1 ( \Omega )$ and for all $\displaystyle x \in \Omega$ and $\displaystyle R_x$ such that $\displaystyle \overline{B_{R_x} (x)} \subset \Omega$ we have $\displaystyle f(x)=\frac{1}{ \vert B_{R_x}(x) \vert } \int_{B_{R_x}(x)} \ f$ . Is $\displaystyle f$ harmonic in $\displaystyle \Omega$ ?

I know the result is true if $\displaystyle f \in C^2( \Omega )$, but I don't know how to argue that if $\displaystyle f \in L^1 ( \Omega )$ then it's twice differentiable.
Okay, hadn't given this one much thought. Here's one proof for continous functions under slightly different assumptions; and although it is an improvement from the $\displaystyle C^2$ case I'm still interested in the general case.

Let $\displaystyle \Omega \subseteq \mathbb{R} ^n$ open and $\displaystyle f: \Omega \rightarrow \mathbb{R}$ be such that for all $\displaystyle x\in \Omega$ and $\displaystyle r>0$ such that $\displaystyle \overline{B} _r(x) \subset \Omega$ we have $\displaystyle f(x)= \frac{1}{|\partial B_r(x)|} \int_{\partial B_r(x)} f(y)dy$.

Let $\displaystyle \eta : \mathbb{R} ^n \rightarrow \mathbb{R}$ be given by $\displaystyle \eta (x)= ce^{\frac{1}{\| x \| ^2-1}}$ where $\displaystyle c>0$ is such that $\displaystyle \int_{\mathbb{R} ^n} \eta =1$. Define $\displaystyle \eta_{ \varepsilon } (x)= \varepsilon ^{-n} \eta \left( \frac{x}{\varepsilon } \right)$. It is a standard result that if $\displaystyle f\in L_{loc}^1 (\Omega )$ then $\displaystyle f_{\varepsilon }:= f\ast \eta _{\varepsilon } \in C^{\infty} ( \Omega _{\varepsilon } )$ where $\displaystyle \Omega_{ \varepsilon } := \{ x\in \Omega : d(x,\partial \Omega )> \varepsilon \}$.

With this in mind, let $\displaystyle f: \Omega \rightarrow \mathbb{R}$ be continous then:

$\displaystyle f_{\varepsilon } (x)= \int_{B_{\varepsilon }(0)} f(x-y) \eta_{\varepsilon }(y)dy=\int_{0}^{\varepsilon } \eta _{\varepsilon }(r) \int_{ \partial B_{\varepsilon }(x)} f(y)dSydr$$\displaystyle = \int_{0}^{\varepsilon } \eta _{\varepsilon } (r) f(x) |\partial B_{\varepsilon }(x)|dr=f(x) \int_{B_{\varepsilon }(x) }\eta _{\varepsilon } (y)dy=f(x)$

This proves that $\displaystyle f$ is harmonic in $\displaystyle \Omega$. Now the issue here, if we wanted to generalize this procedure to $\displaystyle f$ only integrable (or locally so), is that $\displaystyle \partial B_r(x)$ has measure zero so our change into "polar coordinates" can not be applied as we did here.

Any comments or suggestions are appreciated.