## Harmonic Functions: Regularity

I have a confession: I have never actually taken a PDE course. I took an ODE and PDE class offered by the Applied Math department my freshman year, but it wasn’t really a mathematics class–most of the students were physics or engineering sciences concentrators. I have learned a fair amount about PDE in various analysis classes and a mini-course at Princeton’s Analysis and Geometry summer session, but with the exception of the latter, these were not devoted to PDE. So while I impose myself on Georgia Tech’s math department this Fall as a non-degree student, I thought would take their graduate PDE course.

Functions ${u:\Omega\rightarrow\mathbb{R}}$ which arise as solutions of Laplalce’s equation ${\Delta u=0}$ on some open set ${\Omega\subset\mathbb{R}^{n}}$, called harmonic functions have a lot of neat properties, many of them analogous to holomorphic functions on ${\mathbb{C}}$. And indeed, harmonic functions on ${\mathbb{R}^{2}}$ are the real parts of holomorphic functions.

Today, I want to talk about the regularity of harmonic functions that comes as a consequence of their satisfying the mean value property:

$\displaystyle u(x_{0})=\dfrac{1}{S(\partial B(x_{0},r))}\int_{\partial B(x_{0},r)}u(y)dS(y)=\dfrac{1}{\left|B(x_{0},r)\right|}\int_{B(x_{0},r)}u(y)dy$

for all open balls with closure ${\overline{B}(x_{0},r)\subset\Omega}$. (Here, ${S(\cdot)}$ denotes the surface measure and ${\left|\cdot\right|}$ denotes the ${n}$-dimensional Lebesgue measure.)

Some intuition for why this averaging property might give us regularity can be gained from thinking about time series, say for a stock price. The path of stock price over some time period looks differentiable almost nowhere. It may not even be continuous, as stock prices can gap up or down between market closures and opens. However, if we consider a moving average over some window (e.g. 30 days), we get a path that looks `smoother’.

With little effort, we get that a harmonic function, which is only a priori ${C^{2}(\Omega)}$, is in fact smooth (i.e. ${C^{\infty}}$). But as you learned in calculus, smoothness does not imply (real) analyticity (Try to come up with a counterexample). Nevertheless, harmonic functions are analytic. To prove this we will only need three things: any derivative of a harmonic function is harmonic, a local estimate for the ${k^{th}}$ order derivative of a harmonic function in terms of its ${L^{1}}$ norm, and Taylor’s theorem.

Our plan of attack is to prove smoothness is as follows: 1) prove smoothness, 2) prove the local estimate, and 3) prove analyticity. But first, a brief digression on notation. I will use the multi-index notation ${\alpha=(\alpha_{1},\ldots,\alpha_{n})}$, where ${\alpha_{i}\geq 0}$ is integral. We say that ${\alpha}$ is of order ${k}$ if ${\left|\alpha\right|:=\sum_{i}\alpha_{i}=k}$. We define

$\displaystyle \alpha!=\alpha_{1}!\cdots\alpha_{n}!,\indent {\left|\alpha\right|\choose\alpha}=\dfrac{\left|\alpha\right|!}{\alpha!},\indent x^{\alpha}=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}$

and for any sufficiently regular function ${u}$, we define ${D^{\alpha}u=\partial_{1}^{\alpha_{1}}\cdots\partial_{n}^{\alpha_{n}}u}$.

In what follows, ${\Omega\subset\mathbb{R}^{n}}$ will denote an open set.

Recall that if ${u:\Omega\rightarrow\mathbb{R}}$ is harmonic, then it satisfies the mean value property on ${\Omega}$. To prove ${u}$ is smooth, it suffices to prove a more general result: if a locally integrable function ${u}$ satisfies the mean value property for each ball ${B(x,r)\subset\Omega}$, then ${u\in C^{\infty}}$

Proposition 1 If ${u:\Omega\rightarrow\mathbb{R}}$ in ${L_{loc}^{1}(\Omega)}$ satisfies the mean value property, then ${u}$ is ${C^{\infty}}$.

Proof: We will use the mean value property to show that ${u}$ is equal to smooth function on an open neighborhood of every point ${x_{0}\in\Omega}$. Let ${\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}}$ be a ${C^{\infty}}$ compactly supported, radial approximate identity (my survey on Fourier analysis has some notes on these; also, Appendix C.5 [Evans]). Define a ${C^{\infty}}$ function ${u_{\varepsilon}:=\eta_{\varepsilon}\ast u}$ on ${\Omega_{\varepsilon}:=\left\{x\in U|d(x_{0},\partial \Omega)>\varepsilon\right\}}$. For ${x\in\Omega_{\varepsilon}}$,

$\displaystyle \begin{array}{lcl} \displaystyle u_{\varepsilon}(x_{0})&=&\displaystyle\int_{U}\eta_{\varepsilon}(x-y)u(y)dy\\ [1.5 em]&=&\displaystyle\dfrac{1}{\varepsilon^{n}}\int_{B(x,\varepsilon)}\eta\left(\dfrac{\left|x-y\right|}{\varepsilon}\right)u(y)dy\\ [1.5 em]&=&\displaystyle\dfrac{1}{\varepsilon^{n}}\int_{0}^{\varepsilon}\eta\left(\dfrac{r}{\varepsilon}\right)\left(\int_{\partial B(x,r)}udS\right)dr \end{array}$

where the second line is just the definition of ${\eta_{\varepsilon}}$ and the third line is Fubini’s theorem. Observe that for ${0 and ${x\in \Omega_{\varepsilon}}$, ${B(x,\varepsilon)\subset\Omega}$. Using the mean value property, we obtain

$\displaystyle u_{\varepsilon}(x)=\dfrac{u(x)}{\varepsilon^{n}}\int_{0}^{\varepsilon}\eta\left(\dfrac{r}{\varepsilon}\right)n\alpha(n)r^{n-1}dr=u(x)\underbrace{\int_{B(0,\varepsilon)}\eta_{\varepsilon}(y)dy}_{=1}=u(x)$

$\Box$

Lemma 2 For any multi-index ${\alpha=(\alpha_{1},\ldots,\alpha_{n})}$ and harmonic function ${u}$, ${D^{\alpha}u}$ is harmonic (note this differentiation is well-defned since ${u}$ is necessarily smooth).

Proof: Fix a multi-index ${\alpha}$ and set ${v:=D^{\alpha}u}$. Using the equality of mixed partials, we see that

$\displaystyle\partial_{x_{i}}^{2}v=D^{\alpha}\partial_{x_{i}}^{2}u,$

whence by linearity,

$\displaystyle\Delta v=\sum_{i=1}^{n}\partial_{x_{i}}^{2}v=D^{\alpha}\left[\sum_{i=1}^{n}\partial_{x_{i}}^{2}u\right]=D^{\alpha}\left[\Delta u\right]=0$

$\Box$

Proposition 3 If ${u:\Omega\rightarrow\mathbb{R}}$ is harmonic, then for any multi-index ${\alpha}$, with ${\left|\alpha\right|=k}$,

$\displaystyle \left|D^{\alpha}u(x_{0})\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\left\|u\right\|_{L^{\infty}(B(x_{0},r))}$

for all balls ${B(x_{0},r)\subset\Omega}$.

Note that without the result that ${u}$ is ${C^{\infty}}$, it would not make sense to write ${D^{\alpha}u}$ for an arbitrary multi-index.

Proof: The proof will be by induction on the order ${\left|\alpha\right|}$. It is clear from the mean value property that ${C_{0}}$ satisfies Suppose we have a constant ${C_{k}}$ such that

$\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{C_{k}}{r^{k}}\sup_{\overline{B}_{r}(x)}\left|u\right|$

for all balls with ${\overline{B}(x,r)\subset\Omega}$, for all multi-indices ${\alpha}$ of order ${k}$. Suppose ${\left|\alpha\right|=k}$. Then for some ${i}$, we can write ${D^{\alpha}=\partial_{i}D^{\beta}}$, where ${\left|\beta\right|=k}$. As ${D^{\beta}u}$ is harmonic, the mean value property gives

$\displaystyle \begin{array}{lcl}\left|D^{\alpha}u(x)\right|&=&\displaystyle\dfrac{1}{\left|B(x,s)\right|}\left|\int_{B(x,s)}\partial_{i}D^{\beta}u(y)dy\right|\\ &=&\displaystyle\dfrac{1}{\left|B(x,s)\right|}\left|\int_{\partial B(x,s)}D^{\beta}u(y)\cdot\nu_{i}dS(y)\right| \end{array}$

for ${0. If ${y\in \partial B(x,s)}$, then ${B(y,r-s)\subset B(x,r)}$. So by the induction hypothesis,

$\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{1}{\left|B(x,s)\right|}\dfrac{C_{k}}{\left|r-s\right|^{k}}\int_{\partial B(x,s)}\left\|D^{\beta}u\right\|_{L^{\infty}(B(x,r))}dS(y)$

Writing ${s=\theta r}$, where ${0<\theta<1}$, the RHS simplifies to

$\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{nC_{k}}{\theta (1-\theta)^{k}r^{k+1}}\left\|D^{\beta}u\right\|_{L^{\infty}(B(x,r))}$

I leave it to the reader to verify that ${\theta=1/k}$ minimizes the RHS, whence

$\displaystyle \left|D^{\alpha}u(x)\right|\leq n(k+1)\left(\dfrac{k+1}{k}\right)^{k}\dfrac{C_{k}}{r^{k+1}}\leq en(k+1)\dfrac{C_{k}}{r^{k+1}}$

Set ${C_{k+1}:=en(k+1)C_{k}}$. $\Box$

Theorem 4 If ${u:\Omega\rightarrow\mathbb{R}}$ is harmonic, then ${u}$ is real analytic.

Proof: As ${u}$ is smooth, it has a Taylor expansion about ${x\in\Omega}$

$\displaystyle u(x+h)=\sum_{\left|\alpha\right|\leq k-1}\dfrac{D^{\alpha}u(x)}{\alpha!}h^{\alpha}+R_{k}(x,h)$

where ${R_{k}(x,h)=O(\left\|h\right\|^{k})}$. Actually, one form of Taylor’s theorem gives the remainder

$\displaystyle R_{k}(x,h)=\sum_{\left|\alpha\right|=k}\dfrac{D^{\alpha}u(x+th)}{\alpha!}h^{\alpha}$

for ${0. Let ${r>0}$ be such that ${\overline{B}(x,2r)\subset\Omega}$. From our work above, we have the estimate

$\displaystyle \left|D^{\alpha}u(x+th)\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\sup_{\overline{B}(x+th,r)}\left|u(x)\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\underbrace{\sup_{\overline{B}(x,2r)}\left|u(x)\right|}_{:=M}$

Substituting this result into the expression for ${R_{k}(x,h)}$, we obtain the estimate

$\displaystyle \begin{array}{lcl}\displaystyle\left|R_{k}(x,h)\right|&\leq&\displaystyle\dfrac{Mn^{k}e^{k-1}\left\|h\right\|^{k}k!}{r^{k}}\left(\sum_{\left|\alpha\right|=k}\dfrac{1}{\alpha!}\right)\\ [1.5 em]&=&\displaystyle\dfrac{Mn^{k}e^{k-1}\left\|h\right\|^{k}k!}{r^{k}}\dfrac{n^{k}}{k!}\\ [1.5 em]&=&\displaystyle\dfrac{M}{e}\left(\dfrac{n^{2}\left\|h\right\|e}{r}\right)^{k}\\ \end{array}$

If ${\left\|h\right\|^{-1}, then ${R_{k}(x,h)\rightarrow 0}$ as ${k\rightarrow\infty}$. $\Box$

In the next installment on harmonic functions, we will prove the Weyl Lemma which states that any weak solution ${u}$ of the Laplace equation is equal a.e. to a harmonic function. We will also extend the Weyl Lemma to distributions which satisfy the Laplace equation.

L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.

G.B. Folland, Introduction to Partial Differential Equations (Second Edition), Princeton UP, 1995.

J.K. Hunter, PDE Notes Chapter 2, https://www.math.ucdavis.edu/~hunter/pdes/ch2.pdf, Accessed 9/5/14.