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 $\|v-u\|_{L^2(U)} \leq \epsilon$.


2 Answers 2


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).

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    $\begingroup$ But if you remove a thin slice of $U$ so that it becomes simply connected, you are back in business. $\endgroup$
    – username
    Feb 15, 2022 at 21:58
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    $\begingroup$ 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 $B'$. $\endgroup$ Feb 15, 2022 at 23:23
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    $\begingroup$ @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. $\endgroup$
    – RBega2
    Feb 16, 2022 at 1:12
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    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Feb 16, 2022 at 1:21
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    $\begingroup$ @TerryTao: I believe du Plessis proved this in 1969, see doi.org/10.1112/jlms/s2-1.1.404 $\endgroup$ 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 $0$.


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