# A harmonic function $\varphi$ with $D\varphi \in L^q(\mathbb R^n)$ is constant

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 ?

• 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 Commented Nov 29, 2020 at 22:27
• 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 $\varphi=0 .$ Commented Nov 30, 2020 at 5:55
• 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. Commented Nov 30, 2020 at 6:08

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 $$\psi(x) = 0$$.
• Nice answer! Just how do you know that $\partial_i \phi$ is harmonic? Commented Dec 2, 2020 at 20:31
• Because you can commute derivatives by Schwartz lemma. Harmonic functions are $C^\infty$ by elliptic regularity. Commented Dec 2, 2020 at 22:15