Let $\varphi$ be an harmonic function such that $D\varphi \in L^q(\mathbb R^n)$ for $q \in (1, +\infty)$. I read in Partial Differential Equations of Quin Han and Fanghua Lin that for $q = 2$, $\varphi$ has to be constant. My professor told me that it is possible to generalize this result to any $q \in (1,+\infty)$ but I couldn't find a reference of this anywhere. Would someone please tell me where I could find a proof ?
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4$\begingroup$ A tempered harmonic distribution is a polynomial (so in particular a $\varphi$ as above is a polynomial, but the only $L^p$ polynomials are the constant ones). See Theorem 4.18 of my lecture notes on distributions here (or many other places): math.ou.edu/~cremling/teaching/lecturenotes/ln-dist.pdf $\endgroup$– Christian RemlingCommented Nov 29, 2020 at 22:27
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$\begingroup$ I believe it is only true for simply connected domain, but it is difficult to check without fourier when not in $\mathbb{R}^n, \mathbb{T}^n$. $\varphi$ is a tempered harmonic distribution, so $\forall f \in S\left(\mathbb{R}^{n}\right), \int \varphi \Delta f=0,$ integral by part, $\int \nabla \varphi \nabla f=0,$ by poincare lemma $\forall g \in S\left(\mathbb{R}^{n}\right), \exists f \in S\left(\mathbb{R}^{n}\right), \nabla f=g,$(This is not true but there should have the same phenom) and smooth fuction is dense in $S\left(\mathbb{R}^{n}\right),$ so $D \varphi=0$ so $\varphi=0 .$ $\endgroup$– katagoCommented Nov 30, 2020 at 5:55
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$\begingroup$ Do not need consider approximate general function in $L^q(R^n)$ use function in $S(R^N)$ only need to consider there is a good approximation of $\nabla \varphi$ in $S(R^N)$, and this could be done by cut off $\nabla \varphi$ to $(\nabla \varphi)_c$ use a ball with big radius and consider $\eta_{\epsilon}*(\nabla \varphi)_c$, then use approximation and young inequality. But in general I think tempered function space can not approximate $L^p(\Omega)$ for $\pi(\Omega)\neq 0$, so there is no reason to believe the phenomenon is true in general non-simple connected domain. $\endgroup$– katagoCommented Nov 30, 2020 at 6:08
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1 Answer
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If $\phi$ is harmonic over $\mathbb{R}^n$, then all its partial derivatives $\partial_i \phi$ are harmonic. As a consequence, all we are left to prove is that any harmonic function that belongs to $L^p$ is zero. $\phi$ will have all its partial derivatives vanish so it is constant.
Let $\psi$ be an harmonic function that belongs to $L^p$ for some $p \in [1, \infty)$. By the mean value property, we have, for any $x \in \mathbb{R}^n$, that $$ \psi(x) = \frac{1}{\mathrm{Vol}(B_r(x))} \int_{B_r(x)} \psi(y) dy^n. $$ We now use Hölder's inequality to get $$ |\psi(x)| \leq \frac{1}{\mathrm{Vol}(B_r(x))} \|\psi\|_{L^p(B_r(x))} \mathrm{Vol}(B_r(x))^{1-\frac{1}{p}} \leq \mathrm{Vol}(B_r(x))^{-\frac{1}{p}} \|\psi\|_{L^p(\mathbb{R}^n)}. $$ Letting $r$ tend to $+\infty$, we conclude that $\psi(x) = 0$.
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$\begingroup$ If you insist on having a referece,I don't know any but the mean value property is proven e.g. in Evans' book "Partial differential equations" $\endgroup$ Commented Dec 2, 2020 at 9:39
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$\begingroup$ Nice answer! Just how do you know that $\partial_i \phi$ is harmonic? $\endgroup$– FalconCommented Dec 2, 2020 at 20:31
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1$\begingroup$ Because you can commute derivatives by Schwartz lemma. Harmonic functions are $C^\infty$ by elliptic regularity. $\endgroup$ Commented Dec 2, 2020 at 22:15