Harmonic Functions: Weyl Lemma

In this post, we prove a generalization of the result that any (clasically) harmonic function {u} defined on a bounded, open set {\Omega\subset\mathbb{R}^{n}} is analytic. If {u\in L_{loc}^{1}(\Omega)} is weakly harmonic, i.e. {\int(\Delta u)\varphi=\int u\Delta\varphi=0} for any compactly supported {C^{\infty}} function {\varphi} (the space of such functions is denoted {C_{0}^{\infty}(\mathbb{R}^{n})}, then {u} is strongly or classically harmonic and therefore equal a.e. to an analytic function. This result is due to Hermann Weyl and appears to be known in the literature as the Weyl Lemma.

The proof I will give is based on an outline in a problem set I found online.

Let {\eta:\mathbb{R}^{n}\rightarrow\mathbb{R}} be the radial mollifier defined by

\displaystyle \eta(x):=\begin{cases} C\exp\left(\dfrac{1}{\left|x\right|^{2}-1}\right) & {\left|x\right|<1}\\ 0 & {\left|x\right|\geq 1} \end{cases}

where {C} is the normalizing constant. Define an approximate identity {\eta_{\varepsilon}=\varepsilon^{-n}\eta(-/\varepsilon)}. I leave it to the reader to verify that {\eta} is indeed {C^{\infty}} (this is a standard calculus problem) and that {\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}} is an approximate identity.

For {\varepsilon>0}, define an open subset of {\Omega} by {U_{\varepsilon}:=\left\{x\in\Omega:d(x,\partial\Omega)>\varepsilon\right\}} (i.e. the set of points in {\Omega} which are at least {\varepsilon} distance from the boundary of {\Omega}). As {\eta_{\varepsilon}} is supported on {B(0,\varepsilon)}, the convolution {u_{\varepsilon}:=u\ast\eta_{\varepsilon}} is well-defined. Furthermore, {u_{\varepsilon}} is {C^{\infty}}. Iif {x\in U_{\varepsilon}}, then {B(x,\varepsilon)\subset\Omega}, so {u_{\varepsilon}} is integrable on {U_{\varepsilon}}. By Young’s inequality, we have the estimate

\displaystyle \left\|u_{\varepsilon}\right\|_{L^{1}(U_{\varepsilon})}\leq\left\|u\right\|_{L^{1}(\Omega)}\left\|\eta_{\varepsilon}\right\|_{L^{1}(\Omega)}=\left\|u\right\|_{L^{1}(\Omega)}

I claim that {u_{\varepsilon}} is (classicaly) harmonic on {U_{\varepsilon}}. Indeed, as {\eta_{\varepsilon}\in C_{0}^{\infty}}, we can use the Lebesgue dominated convergence theorem to differentiate inside the integral in the definition of {u_{\varepsilon}} (look at the Taylor expansion of {\eta_{\varepsilon}}) to obtain

\displaystyle \Delta u_{\varepsilon}(x)=\int_{\Omega}[\Delta\eta_{\varepsilon}](x-y)u(y)dy=0,

since {\eta_{\varepsilon}(x-\cdot)\in C_{0}^{\infty}} and {u} is weakly harmonic. Consequently, {u_{\varepsilon}} has the mean value property.

We now want to extract a sequence of the {u_{\varepsilon}} which converges to some continuous function {v} and then argue that this function {v} must be harmonic. To accomplish the first task, we use the Arzelà-Ascoli theorem. First, we have that {\left\{u_{\varepsilon}\right\}} is equicontinuous and uniformly bounded. Fix an R>0, and set V:=U_{R}.

Lemma 1 The family {\left\{u_{\varepsilon}\right\}_{\varepsilon>0}} is uniformly bounded on {\overline{V}} for {0<\varepsilon<R/2}.

Proof: For {\varepsilon>0}, {u_{\varepsilon}} satisfies the mean value property. For {B(x,r)\subset\Omega},

\displaystyle \left|u_{\varepsilon}(x)\right|=\dfrac{1}{\left|B(x,r)\right|}\left|\int_{B(x,r)}u_{\varepsilon}(y)dy\right|\leq\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}\left|u_{\varepsilon}(y)\right|dy

If {x\in\overline{V}}, then {B(x,R/2)\subset\Omega}, so we obtain the estimate

\displaystyle \left|u_{\varepsilon}(x)\right|\leq\dfrac{1}{\left|B(x,R/2)\right|}\int_{B(x,R/2)}\left|u_{\varepsilon}(y)\right|dy=\dfrac{2^{n}}{\alpha(n)R^{n}}\leq\dfrac{2^{n}}{\alpha(n)R^{n}}\left\|u\right\|_{L^{1}(\Omega)}

for all {x\in\overline{V}}. \Box

Lemma 2 The family {\left\{u_{\varepsilon}\right\}_{\varepsilon>0}} is equicontinuous on {\overline{V}} for {0<\varepsilon<R/2}.

Proof: For any {x,y\in\overline{V}} and {0<\varepsilon<R/2}.

\displaystyle \begin{array}{lcl} \displaystyle\left|u_{\varepsilon}(x)-u_{\varepsilon}(y)\right|&=&\displaystyle\left|\dfrac{1}{\left|B(x,s)\right|}\int_{B(x,s)}u_{\varepsilon}(z)dz-\dfrac{1}{\left|B(y,s)\right|}\int_{B(y,s)}u_{\varepsilon}(z)dz\right|\\ [2 em]&=&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\int_{B(x,s)}u_{\varepsilon}(z)dz-\int_{B(y,s)}\left|u_{\varepsilon}(z)dz\right|\\ [2 em]&\leq&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\left[\int_{B(x,s)\setminus B(y,s)}\left|u_{\varepsilon}(z)\right|dz+\int_{B(y,s)\setminus B(x,s)}\left|u_{\varepsilon}(z)\right|dz\right] \end{array}

As {u_{\varepsilon}} is uniformly bounded on {\overline{V}} for all {0<\varepsilon<R/2} (with constant independent of {\varepsilon}) and the Lebesgue measure of the two domains depends only on {\left|x-y\right|} and goes to zero as {\left|x-y\right|\rightarrow 0}, we obtain equicontinuity. \Box

By the Arzelà-Ascoli theorem, there exists a sequence {u_{\varepsilon_{k}}} which converges uniformly to a continuous function {v\in C(\overline{V})}. Replacing {\left\{\eta_{\varepsilon}\right\}_{\varepsilon}} with a countable subset with indices tending to zero, we may assume that {\varepsilon_{k}\rightarrow 0}.

To prove that {v} is harmonic, we need the following lemma.

Lemma 3 If {v\in L_{loc}^{1}(\Omega)} has the mean value property, then {v} is (classically) harmonic.

Proof: Set {\chi_{r}(x):=\left|B(0,r)\right|^{-1}\chi_{B(0,r)}(x)}. Observe that by the Lebesgue dominated convergence theorem {v\ast\chi_{r}} is continuous. By the mean value property,

\displaystyle v\ast\chi_{r}(x)=\dfrac{1}{\left|B(0,r)\right|}\int_{\Omega}v(x-y)\chi_{B(0,r)}(y)dy=\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}v(z)dz=v(x),

where we use the translation invariance of the Lebesgue measure to obtain the penultimate equality. Hence, {v\in C(\Omega)}.

We now show that {v} is equal to a smooth function a.e. on {\Omega}. Set {v_{\varepsilon}:=v\ast\eta_{\varepsilon}}. Then

\displaystyle \begin{array}{lcl} v_{\varepsilon}(x)&=&\displaystyle\int_{B(0,\varepsilon)}v(x-y)\eta_{\varepsilon}(y)dy\\ [2 em]&=&\displaystyle\varepsilon^{-n}\int_{B(0,\varepsilon)}v(x-y)\eta(y/\varepsilon)dy\\ [2 em]&=&\displaystyle\int_{B(0,1)}v(x-\varepsilon y)\eta(y)dy\\ [2 em]&=&\displaystyle\int_{0}^{1}\int_{\partial B(0,s)}v(x-\varepsilon z)\eta(z)dS(z)ds\\ [2 em]&=&\displaystyle\int_{0}^{1}v(x)n\alpha(n)s^{n-1}\eta(s)ds\\ [2 em]&=&\displaystyle v(x)\cdot\int_{B(0,\varepsilon)}\eta_{\varepsilon}(y)dy=v(x) \end{array}

In particular, {v\in C^{2}(\Omega)}, so by the converse to the mean value property (see [Evans] Theorem 3, pp. 26), {v} is harmonic. \Box

As {\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}} is an approximate identity, {\left\|u_{\varepsilon}-u\right\|_{L^{1}}\rightarrow 0} as {\varepsilon\rightarrow 0}. Hence, {v=u} a.e. As we proved last time, u is therefore equal to an analytic function a.e.

Next time, we will extend the Weyl lemma to distributions that satisfy the Laplace equation.

  1. L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.
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4 Responses to Harmonic Functions: Weyl Lemma

  1. BEatrice Dulvo says:

    Matt,
    it seems to me that u_\epsilon being smooth harmonic, u_\epsilon\star \eta_\alpha coincides with u_\epsilon by the mean value property, since we are doing averages on circles. As a result u_{\epsilon}\star\eta_{\alpha}=(u\star\eta_{\epsilon})\star\eta_{\alpha}=(u\star\eta_{\alpha})\star\eta_{\epsilon}=u_{\alpha}. So on growing (as \epsilon goes to 0) subsets of \Omega, the u_{\epsilon} coincide. Since as \epsilon goes to 0, u_{\epsilon} converges to u\star\delta=u, this means u is smooth in the interior of \Omega. Whether u is mooth up to the boundary is another story.

    • Matt R. says:

      Beatrice,

      Thank you for your comment. We could have indeed proceeded more succinctly as you suggested. In fact, your outline generalizes nicely to the case where u is an arbitrary distribution.

      Regards,
      Matt

  2. Anonymous says:

    What is the set V here and what is R?

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