# Density of restrictions of harmonic functions inside a ball

Let $$B$$ be the closed unit ball in $$\mathbb R^3$$ centered at the origin and let $$U= \{x\in \mathbb R^3\,:\, \frac{1}{2}\leq |x| \leq 1\}.$$ Let $$S_U= \{u \in C^{\infty}(U)\,:\, \Delta u =0 \quad\text{on U^{\textrm{int}}}\},$$ and $$S_B= \{u \in C^{\infty}(B)\,:\, \Delta u =0 \quad\text{on B^{\textrm{int}}}\}.$$

Is the following statement true? Given any $$\epsilon>0$$ and any $$u \in S_U$$, there exists an element $$v \in S_B$$ such that $$\|v-u\|_{L^2(U)} \leq \epsilon$$.

No. If $$\varphi \in C^\infty_c(B)$$ is a bump function equal to $$1$$ in $$\lvert x\rvert \leq 1/2$$ then from Green's theorem we have $$\int_B u \Delta \varphi = 0$$ for all $$u \in S_B$$, but the same is not true in general for typical $$u \in S_U$$, which by the Cauchy–Schwarz inequality implies that $$u$$ is a positive distance away from $$S_B$$ in the $$L^2(U)$$ norm. For instance, if we take $$u = K\rvert_U \in S_U$$ where $$K(x) = \frac{-1}{4\pi \lvert x\rvert}$$ is the Newton potential (the fundamental solution to $$\Delta K = \delta$$) then $$\int_B u \Delta \varphi = \int (\Delta K) \varphi = 1 \neq 0$$ and hence $$u$$ is a positive distance from $$S_B$$.

One can create similar obstructions using functions $$\varphi$$ which behave like a specified spherical harmonic in the angular variable (instead of being constant in the angular variable, which is basically what is being done here).

• But if you remove a thin slice of $U$ so that it becomes simply connected, you are back in business. Feb 15, 2022 at 21:58
• If $u$ can be approximated then it admits an harmonic extension to $B$. In fact one can replace the $L^2$ norm of $u-v$ by the sup-norm in a smaller annulus $U'$. If $v_n$ correspond to $\epsilon=1/n$, then $(v_n)$ is Cauchy in $U'$ in the sup-norm and, by the maximum principle, in a ball $B'$. Feb 15, 2022 at 23:23
• @username Since the ambient space is $\mathbb{R}^3$, $U$ is already simply connected. Even in $\mathbb{R}^2$ what I think you are proposing would make the situation even worse, e.g., one could have something like the real part of $\sqrt{z}$ being harmonic. Feb 16, 2022 at 1:12
• In 2D I think Runge's theorem (or Mergelyan's theorem) gives the required density in the simply connected case. For higher dimensions I think an analogous theorem should hold for harmonic functions on a contractible domain (with a connected exterior), though I don't have a reference at hand for this. Feb 16, 2022 at 1:21
• @TerryTao: I believe du Plessis proved this in 1969, see doi.org/10.1112/jlms/s2-1.1.404 Feb 16, 2022 at 8:42

Suppose that $$u\in S_{U}$$. Then I claim that $$\inf\{\|u-v\|_{L^{2}(U)}:v\in S_{B}\}$$ is bounded below by the standard deviation of the spherical symmetrization $$u^{\sharp}$$ of $$u$$.

For this post, the $$L^{2}(U)$$ norm will be with respect to the normalized area probability measure on $$U$$. Let $$\mu$$ be the Haar probability measure on the group of all $$3\times 3$$ orthogonal matrices. Let $$\nu$$ be the normalized area probability measure on $$S^{2}$$.

Define the spherical symmetrization $$w^{\sharp}$$ of a function $$w$$ by letting $$w^{\sharp}(x)=\int_{A\in O(3)}(w\circ A)(x)d\mu(A).$$ Observe that $$w^{\sharp}(x)=\int_{y\in S^{2}}w(\|x\|\cdot y)d\nu(y).$$

Observe that if $$w$$ is harmonic, then $$w^{\sharp}$$ is also harmonic, and there are constants $$\alpha,\beta$$ such that $$w^{\sharp}(x)=\frac{\alpha}{\|x\|}+\beta$$ (this fact generalizes to all dimensions $$n\geq 2$$).

By Jensen's inequality, if $$r\in[0,1)$$ and $$x\in S^{2}$$, then $$(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}=(\int_{y\in S^{2}}f(ry)-g(ry)d\nu(y))^{2}\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$ Therefore, by integrating, we obtain $$\int_{x\in S^{2}}(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}dx\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$

Therefore, if $$f,g:U\rightarrow\mathbb{R}$$ are continuous, then $$\|f^{\sharp}-g^{\sharp}\|_{L^{2}(U)}\leq\|f-g\|_{L^{2}(U)}$$

Suppose $$u\in S_{U},v\in S_{B}$$. Since $$v$$ is harmonic on $$B$$, the function $$v$$ satisfies the mean-value property, so the function $$v^{\sharp}$$ is constant.

Therefore, $$\text{Var}(u^{\sharp})\leq\|u^{\sharp}-v^{\sharp}\|_{L_{2}(U)}^{2}\leq\|u-v\|_{L^{2}(U)}^{2}.$$

There are plenty of functions $$u$$ that are harmonic on $$U$$ but where $$u^{\sharp}$$ is non-constant on $$U$$ (such as the Newtonian potential), and for each such function, we have $$\text{Var}(u^{\sharp})>0.$$ This proof generalizes to any dimension $$n\geq 2$$ where the balls $$B,B\setminus U$$ have any radii but are still centered at $$0$$.