2 Preliminaries

Here, we define what is to be understood as to solve the Dirichlet problem at infinity. First, we define a compactification for M. To this end, define the set $\overline {M}$ :

$$\begin{align*}\overline{M}=N\times \left[0,\infty\right]/\left(N\times\left\{0\right\}\right), \end{align*}$$

where $\left [0,\infty \right ]$ is a compactification of $\left [0,\infty \right )$ which is homeomorphic to $\left [0,1\right ]$ . The subspace $\partial _{\infty }M:=N\times \left \{\infty \right \}$ plays the role of a boundary, and in fact, when N is homeomorphic to $\mathbb {S}^{n-1}$ , $\overline {M}$ has the structure of a topological manifold with boundary. Given the previous definition, a way of defining that the Dirichlet problem is solvable at infinity is as follows: given $f\in C\left (N\right )$ , there exists a function $u:\overline {M}\rightarrow \mathbb {R}$ which is in $C\left (\overline {M}\right )$ , is harmonic in M, and such that its restriction to $\partial _{\infty }M$ is f. In the case of Cartan–Hadamard manifolds, this definition of solvability coincides with the definition of solvability given by Choi, who uses the cone topology as defined by Eberlein and O’Neill [Reference Eberlein and O’Neill8], and which is equivalent, in the sense of homeomorphism, to the one defined for the compactification above: the resulting spaces in both cases are homeomorphic to the closed n-ball.

A stronger definition of solvability was used in the statement of Theorem 1; let us recall it. We shall say that the Dirichlet problem is uniformly solvable at infinity if there is a harmonic function $u:M\longrightarrow \mathbb {R}$ such that

$$\begin{align*}\lim_{r\rightarrow \infty}u\left(\omega, r\right)=f\left(\omega\right) \end{align*}$$

uniformly on $\omega \in N$ . Notice that using this definition, if $f\in C\left (N\right )$ , then we can extend u to $\overline {M}$ continuously and hence this definition of solvability implies the one given above. In the case of $N=\mathbb {S}^{n-1}$ (not necessarily with the round metric), uniform solvability implies solvability in the sense of Choi in [Reference Choi5].

Observe that with the definitions of solvability given in the previous paragraph, if the Dirichlet problem is solvable at infinity for given continuous data, then the corresponding harmonic extension is bounded; thus, solving the Dirichlet problem at infinity gives a method for proving the existence of bounded nonconstant harmonic functions.

Furthermore, in the case of M being a smooth Riemannian manifold, we can define the concept of solvability at infinity for the Dirichlet problem with boundary data in $L^{2}$ . In this case, given $f\in L^{2}\left (\mathbb {S}^{n-1}\right )$ , we say that the Dirichlet problem is $L^2$ -solvable at infinity with boundary data $f\in L^{2}\left (\mathbb {S}^{n-1}\right )$ if there is a function $u:M\longrightarrow \mathbb {R}$ which is harmonic, and such that

$$\begin{align*}\lim_{r\rightarrow \infty} u\left(\omega, r\right)= f\left(\omega\right) \quad \mbox{in} \quad L^{2}\left(\mathbb{S}^{n-1}\right).\end{align*}$$

2.1 The Maximum Principle

To prove uniqueness, we shall make use of the maximum principle. What follows is not new, but we have written it down for the convenience of the reader and for easy reference (see also Choi’s asymptotic maximum principle [Reference Choi5]).

Theorem 3. Let $\overline {M}=N\times \left [0,\infty \right ]$ such that $N\times \left (0,\infty \right )$ is endowed with the metric

$$\begin{align*}g=dr^2+\phi\left(r\right)^2g_N, \end{align*}$$

with $\phi $ smooth and $\phi>0$ on $\left (0,\infty \right )$ . Let $u:\overline {M}\longrightarrow \mathbb {R}$ be continuous function which is harmonic in $N\times \left (0,\infty \right )$ with respect to the metric g. Then, u reaches its maximum and its minimum at $N\times \left \{0\right \}\cup N\times \left \{\infty \right \}$ . If M is a smooth Riemannian manifold (this, of course, requires that g extends smoothly to M at $r=0$ ), and $u:\overline {M}\longrightarrow \mathbb {R}$ is harmonic, then it reaches its maximum and minimum at $\partial _{\infty }M$ .

Proof. Assume that u reaches its maximum at a point in $N\times \left (0,\infty \right )$ , and let the value of this maximum (minimum) be a. Consider the set

$$\begin{align*}A=\left\{\left(\omega,r\right)\in N\times\left(0,\infty\right)\,:\, u\left(\omega,r\right)=a\right\}\hspace{-1.5pt}. \end{align*}$$

By the strong maximum principle for elliptic operators (Theorem 3.5 in [Reference Gilbarg and Trudinger9]), there is a small ball around any point in A where u is constant and equal to a. Thus, A is open in $N\times \left (0,\infty \right )$ . However, A is also closed by the continuity of u, and as $N\times \left (0,\infty \right )$ is connected, we must have that $A=N\times \left (0,\infty \right )$ , and thus u is constant in $N\times \left (0,\infty \right )$ and, as a consequence, by continuity, also in $N\times \left [0,\infty \right ]$ : this implies the first part of the statement of the theorem. In the case of M being a smooth Riemannian manifold, if u is harmonic, it is smooth in M, and thus if it reaches its maximum in M, similar arguments as above show that u must be constant; the same holds if u reaches its minimum in M. Thus, we conclude that the maximum and the minimum are reached at $\partial _{\infty }M$ .

As is well known, an important consequence of the maximum principle is a uniqueness result for the Dirichlet problem, and which we shall use in what follows.

Corollary 3. Let g be as in the previous theorem. Let $u_1, u_2:N\times \left [0,\infty \right ]\longrightarrow \mathbb {R}$ be continuous functions which are harmonic in $N\times \left (0,\infty \right )$ . If $u_1=u_2$ on $N\times \left \{0\right \}\cup N\times \left \{\infty \right \}$ , then $u_1\equiv u_2$ in $N\times \left [0,\infty \right ]$ . If M is a smooth Riemannian manifold, and $u_1,u_2:\overline {M}\longrightarrow \mathbb {R}$ are harmonic in M, then if $u_1=u_2$ in $\partial _{\infty }M$ , then $u_1\equiv u_2$ .

3 A proof of Theorem 1

We first solve the following problem in $N\times \left [0,\infty \right ]$ . Given $f\in C^{\infty }\left (N\right )$ , we shall construct a harmonic function u in $N\times \left (0,\infty \right )$ such that as $r\rightarrow \infty $ , $u\left (\omega , r\right )\rightarrow f\left (\omega \right )$ uniformly, and such that

$$\begin{align*}u\left(\omega, r\right)\rightarrow \overline{f}\colon =\frac{1}{\mbox{Vol}\left(N\right)}\int_N f\,dV_{g_N} \end{align*}$$

also uniformly as $r\rightarrow 0$ , so that u passes to the quotient $\overline {M}$ as a continuous function which is harmonic, as it has been defined in the introduction, and such that it also solves the Dirichlet problem at infinity. To prove this, we use separation of variables.

We let $f_{m,k}$ , $k=0,1,2,\dots , k_m$ , be eigenfunctions of the mth eigenvalue, $\lambda _m^2$ , $m=0,1,2,\dots $ , such that the set $\left \{f_{m,k}\right \}_{m,k}$ is an orthogonal basis for $L^{2}\left (N\right )$ with respect to the inner product induced by (6). We use the convention that $\lambda _m>0$ , if $m>0$ , and that $\lambda _0=0$ is the trivial eigenvalue of the Laplacian of N whose eigenfunctions are the constant functions; we shall also fix the convention $f_{0,0}\equiv 1$ . Next, we must find functions $\varphi _m$ so that the product $\varphi _m\left (r\right ) f_{m,k}\left (\Omega \right )$ satisfies $\Delta _g \left (\varphi _m f_{m,k}\right )=0$ ; this time we use the convention $\varphi _0\equiv 1$ . Then, it is elementary to prove that the equation to be satisfied by $\varphi _m$ , $m>0$ , is

(8) $$ \begin{align} \mathcal{L}_m\varphi_m:=\varphi_m"+\left(n-1\right)\frac{\phi'}{\phi}\varphi_m'-\frac{\lambda_m^2}{\phi^2}\varphi_m=0. \end{align} $$

First, we show that there is a solution to (8) such that $\varphi _m\left (0\right )=0$ , if $m\neq 0$ , and that we can choose $\varphi _m>0$ near 0 (for $m=0$ , as said before, we just choose the constant function 1). Indeed, since $\phi \left (r\right )\sim r$ , $r=0$ is a regular singular point of the equation, and thus, near $r=0$ , a solution can be written as

$$\begin{align*}q\left(r\right)=r^{s}p\left(r\right), \end{align*}$$

where $s=\dfrac {-\left (n-2\right )+\sqrt {\left (n-2\right )^2+4\lambda _m^2}}{2}>0$ satisfies the indicial equation

$$\begin{align*}s\left(s-1\right)+\left(n-1\right)s-\lambda_m^2=0, \end{align*}$$

p is smooth function, and $p\left (0\right )=1$ (see page 45 in [Reference Bôcher3], a classical paper by Bôcher, and Chapters 4 and 5 in [Reference Coddington and Levinson6]). From this, our assertion follows.

Next, we show that $\varphi _m$ is nondecreasing. From its general form, it is clear that $\varphi _m'\left (r\right )> 0$ near $0$ . Assume then that at some point $r_0>0$ , $\varphi _m'\left (r_0\right )=0$ occurs for the first time. Then, since for $0<r<r_0$ $\varphi _m(r)> 0$ , using equation (8) shows that $\varphi _m"\left (r_0\right )> 0$ , which in turn implies that $\varphi '\left (r\right )>0$ a little beyond $r_0$ . This shows that

$$ \begin{align*} \sup{} \left\{ r:\, \varphi_{m}^{\prime} \left(\rho\right)\geq 0 \quad \mbox{for all}\quad \rho\in \left[0,r\right]\right\}=\infty, \end{align*} $$

and thus our claim follows.

In order to show that $\varphi _m$ is bounded, define the function

$$\begin{align*}\eta_m\left(r\right)=\exp\left[-\int_{r}^{\infty}\frac{\lambda_m}{\phi\left(s\right)}\,ds\right]. \end{align*}$$

Of course, here is where we need the fundamental assumption that

$$\begin{align*}\int_1^{\infty}\frac{1}{\phi\left(s\right)}\,ds<\infty, \end{align*}$$

which implies that $\eta _m$ is bounded on $\left [R_0,\infty \right )$ for any $R_0>0$ (the bound depends on $R_0$ ).

A straightforward computation shows that if $r\geq R_0$ , $R_0>0$ as in the statement of Theorem 1, then

(9) $$ \begin{align} \mathcal{L}_m \eta_m =\left(n-2\right)\frac{\phi'}{\phi}\lambda_m \eta_m \geq 0. \end{align} $$

This is why we need the hypothesis $\phi '\left (r\right )\geq 0$ for r large enough. We note in passing that when $\mbox {dim}\left (N\right )=n-1=1$ , the $\eta _m$ ’s thus defined give explicit solutions to (8) (this was pointed out to me by J. E. Bravo): this is the reason why in dimension 2 the assumption $\phi '\geq 0$ is not required.

Next, we are going to use inequality (9) to show that for $r\geq R_0>0$ , it holds that $0\leq \varphi _m\left (r\right )\leq A_m\eta _m\left (r\right )$ for an appropriate constant $A_m$ . So, choose $A_m>0$ large enough so that

$$\begin{align*}\varphi\left(R_0\right)<A_m\eta_m\left(R_0\right),\quad \mbox{and}\quad \varphi_m'\left(R_0\right)<A_m\eta_m'\left(R_0\right). \end{align*}$$

We now claim that $\left (A_m\eta _m-\varphi _m\right )'\left (r\right )\geq 0$ must also hold for all $r\geq R_0$ . In order to prove our claim, let

$$\begin{align*}r_0= \sup{} \left\{r:\, \left(A_m\eta_m-\varphi_m\right)'\left(r\right)\geq 0 \quad \mbox{in}\quad \left[R_0,r\right]\right\}, \end{align*}$$

and let us show that $r_0=\infty $ . Indeed, assume that $r_0<\infty $ . By continuity, it is clear that $\left (A_m\eta _m-\varphi _m\right )'\left (r\right )\geq 0$ up to $r_0$ and equality holds at $r=r_0$ . Hence, by our choice of $A_m$ , it also holds that $h_m:=A_m\eta _m-\varphi _m> 0$ up to $r_0$ . Then, using (9), we have that

$$\begin{align*}h_m"\left(r_0\right)\geq \frac{\lambda_m^2}{\phi^2}h_m\left(r_0\right)> 0, \end{align*}$$

and hence a bit beyond $r_0$ we would still have $h_m'>0$ , which leads to a contradiction, and thus we must have $r_0=\infty $ . This shows that for all $r\in \left [R_0,\infty \right )$ , $\left (A_m\eta _m-\varphi _m\right )'\left (r\right )\geq 0$ , which together with the fact that $\varphi \left (R_0\right )<A_m\eta _m\left (R_0\right )$ implies that $A_m\eta _m\geq \varphi _m$ for all $r\in \left [R_0,\infty \right )$ , and this shows that $\varphi _m$ is bounded above.

Notice that in the case that the sectional curvature satisfies $K\leq -1$ everywhere, by the Bishop–Gromov theorem (which in the case of rotational symmetry reduces to a simple ODE comparison argument), $\phi \left (r\right )\geq \sinh r$ , and hence we have that $0\leq \varphi _m \leq A_m\tanh ^{\lambda _m} r$ , and the claimed estimate (3) holds. In any case, the previous estimates show that, by multiplying by appropriate constants, we may assume that $\lim _{r\rightarrow \infty } \varphi _m\left (r\right )=1$ .

Given $f\in C^{\infty }\left (N\right )$ , we can represent it as

$$\begin{align*}f\left(\omega\right)=\sum_{m\geq 0} \sum_k c_{m,k}f_{m,k}\left(\omega\right), \end{align*}$$

where the $f_{m,k}$ ’s are eigenfunctions of the Laplacian as defined above. Notice that the coefficient $c_{0,0}$ of $\varphi _0 f_{0,0}$ is the average value of f:

$$\begin{align*}c_{0,0}=\frac{1}{\mbox{Vol}\left(N\right)}\int_N f\,dV_{g_N}. \end{align*}$$

From the Fourier representation of f, we get a harmonic extension

(10) $$ \begin{align} u\left(\omega, r\right)=\sum_{m\geq 0} \varphi_m\left(r\right)\sum_k c_{m,k}f_{m,k}\left(\omega\right). \end{align} $$

Since $f\in C^{\infty }\left (N\right )$ , it is not difficult to show that u is twice differentiable and that it satisfies $\Delta _g u=0$ in $N\times \left (0,\infty \right )$ . All we need to prove next is that u satisfies the boundary conditions at infinity and that it is continuous at $r=0$ . We start by proving that the boundary condition at infinity is satisfied. Let $\epsilon>0$ , we estimate as follows: By the triangular inequality,

$$ \begin{align*} \left|f\left(\omega\right)-u\left(\omega, r\right)\right| \leq \sum_{m\geq 0} \left(1-\varphi_{m}\left(r\right)\right) \left|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right|. \end{align*} $$

Pick H such that

$$\begin{align*}\sum_{m\geq H} \left|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right|\leq \frac{\epsilon}{2}. \end{align*}$$

This can be done by Peetre’s result on the absolute convergence of a Fourier series as soon as f is smooth enough [Reference Peetre17]. Let $R>0$ be such that if $r\geq R$ , for $m=0,1,\dots , H$ ,

$$\begin{align*}1-\varphi_m\left(r\right)\leq \frac{\epsilon}{2L}, \end{align*}$$

where L bounds

$$\begin{align*}\sum_{m\leq H} \left|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right| \end{align*}$$

from above. Under this considerations, we obtain that, for $r\geq R$ ,

$$\begin{align*}\left|f\left(\omega\right)-u\left(\omega, r\right)\right|< \epsilon, \end{align*}$$

and our claim is now proved: the boundary condition at infinity is satisfied.

Notice that $u=\overline {f}$ at $r=0$ , so, to show continuity at $r=0$ , we must show that $u\left (\omega ,r\right )\rightarrow \overline {f}$ as $r\rightarrow 0$ . To do so, just observe that

$$\begin{align*}\left|u\left(\omega, r\right)-\overline{f}\right|\leq \sum_{m> 0} \varphi_{m}\left(r\right) \left|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right|, \end{align*}$$

and as for $m>0$ , $\varphi _m\left (r\right )\rightarrow 0$ when $r\rightarrow 0$ , by the dominated convergence theorem our assertion follows.

Hence, the function u defined by (10) passes to the quotient and its value at $r=0$ is given by $\overline {f}$ . The uniqueness statement of Theorem 1 when M is a smooth Riemannian manifold follows from Corollary 3. So all is left to show to finish the proof of Theorem 1 is to prove that there is uniqueness of the solution constructed above in the case of M not being necessarily a manifold.

To prove uniqueness of the harmonic extension of f when M is not a manifold, we first show that the only possible value that a solution u to the Dirichlet problem at infinity with boundary condition (at infinity) equals to f can take at the vertex of the cone is $\overline {f}$ , the average of f over N. To proceed, we let

$$\begin{align*}\overline{u}\left(r\right)=\frac{1}{\mbox{Vol}\left(N\right)}\int_N u\left(\omega, r\right) \,dV_{g_N}\left(\omega\right). \end{align*}$$

We claim that $\overline {u}$ is constant. Indeed, let $0<r<r_1<\infty $ ; then, by Green’s theorem, we have that

$$\begin{align*}\phi\left(r_1\right)^{n-1}\frac{d \overline{u}}{d r}\left(r_1\right)- \phi\left(r\right)^{n-1}\frac{d \overline{u}}{d r}\left(r\right) =\frac{1}{\mbox{Vol}\left(N\right)}\int_{N\times \left[r,r_1\right]} \Delta u\left(\omega, r\right)\, dV_g\left(\omega, r\right)=0, \end{align*}$$

that is,

$$\begin{align*}\frac{d \overline{u}}{d r}\left(r\right)= \frac{\phi\left(r_1\right)^{n-1}}{\phi\left(r\right)^{n-1}}\frac{d \overline{u}}{d r}\left(r_1\right)= \frac{A}{\phi\left(r\right)^{n-1}}. \end{align*}$$

Therefore, if $d\overline {u}\left (r_1\right )/d r\neq 0$ , we would have that $\lim _{r\rightarrow 0^+}\overline {u}\left (r\right )=\pm \infty $ . Indeed, by the Fundamental Theorem of Calculus,

$$\begin{align*}\overline{u}\left(r\right)=\overline{u}\left(r_1\right)+\int_{r_1}^r\frac{A}{\phi\left(\rho\right)^{n-1}}\,d\rho. \end{align*}$$

However, near $0$ , $\phi \left (r\right )^{n-1}\sim r^{n-1}$ , $n\geq 2$ , and thus the previous identity shows our claim because, depending on the sign of A, which in turn depends on the sign of $\overline {u}'\left (r_1\right )$ , we then would have that

$$\begin{align*}\lim_{r\rightarrow 0^+}\int_{r_1}^r\frac{A}{\phi\left(\rho\right)^{n-1}}\,d\rho=\pm\infty. \end{align*}$$

However, by the continuity of u, we must have $\lim _{r\rightarrow 0^+}\overline {u}\left (r\right )=u\left (0\right )$ . Therefore,

$$ \begin{align*} d\overline{u}\left(r_1\right)/dr=0 \end{align*} $$

must hold. Since $r_1$ is arbitrary, this shows that $\overline {u}$ is constant, and in fact, since we must have that $\lim _{r\rightarrow \infty }\overline {u}=\overline {f}$ , that $u\left (0\right )=\overline {f}$ . The maximum principle (Corollary 3) implies the uniqueness statement even in the case when the cone is not a smooth manifold.

4 $L^2$ -solvability

If we only require $f\in L^{2}\left (\mathbb {S}^{n-1}\right )$ , the harmonic extension u constructed above is locally integrable and harmonic in the sense of distributions, and hence almost everywhere equal to a smooth harmonic function, and thus harmonic, and we can also show that $u\left (\omega , r\right )\rightarrow f\left (\omega \right )$ as $r\rightarrow \infty $ in $L^{2}\left (\mathbb {S}^{n-1}\right )$ , and in consequence that the Dirichlet problem is $L^2$ -solvable. Let us give a proof of these claims.

We let

$$\begin{align*}u_l\left(\omega, r\right)=\sum_{m\leq l} \varphi_m\left(r\right)h_m\left(\omega\right), \end{align*}$$

where

$$\begin{align*}h_m\left(\omega\right)=\sum_k c_{m,k}f_{m,k}\left(\omega\right). \end{align*}$$

First of all, the sequence $\left \{u_l\right \}_{l=1,2,3,\dots }$ converges in $L^2_{loc}\left (M\right )$ . Indeed, for $R>0$ , we can compute as follows (we employ the notation $B_R\left (0\right )$ for the geodesic ball of radius $R>0$ centered at the pole of the manifold):

$$ \begin{align*} &\qquad\qquad\int_{B_R\left(0\right)} \left|\sum_{m_1<m\leq m_2} \varphi_m\left(r\right)h_m\left(\omega\right)\right|{}^2\,dV_g \\&\qquad\qquad\qquad\qquad\qquad\quad\hspace{5pt} =\\&\int_0^R\int_{\mathbb{S}^{n-1}} \sum_{m_1<m,m'\leq m_2} \varphi_m\left(r\right)\varphi_{m'} \left(r\right) h_{m'}\left(\omega\right)h_m\left(\omega\right) \,dV_{g_{\mathbb{S}^{n-1}}}\left(\omega\right) \phi\left(r\right)\, dr\\&\qquad\qquad\qquad\qquad\qquad\quad\hspace{5pt} = \\&\qquad\int_0^R \sum_{m_1<m\leq m_2} \varphi_m\left(r\right)^2 \left\|h_m\right\|_{L^{2}\left(\mathbb{S}^{n-1}\right)}^2 \phi\left(r\right)\, dr \\&\qquad\qquad\qquad\qquad\qquad\quad\hspace{5pt} \leq \\&C_R\sum_{m_1<m\leq m_2}\left\|h_m\right\|^2_{L^{2}\left(\mathbb{S}^{n-1}\right)} = C_R\left\|\sum_{m_1<m\leq m_2}h_m\right\|^2_{L^{2}\left(\mathbb{S}^{n-1}\right)}, \end{align*} $$

where we have used the orthogonality in $L^2\left (\mathbb {S}^{n-1}\right )$ of eigenfunctions of the Laplacian with different eigenvalues. Since $\sum _{m\leq l} h_m\rightarrow f$ in $L^2\left (\mathbb {S}^{n-1}\right )$ , given $\epsilon>0$ , there is an L such that whenever $m_2,m_1\geq L$ , then

$$\begin{align*}C_R\left\|\sum_{m_1<m\leq m_2}h_m\right\|_{L^{2}\left(\mathbb{S}^{n-1}\right)}<\epsilon, \end{align*}$$

which shows that $\left \{u_l\right \}_l$ is a Cauchy sequence in $L^{2}\left (B_R\left (0\right )\right )$ , which in turn implies that the sequence converges in $L^{2}\left (B_R\left (0\right )\right )$ . This proves our claim. We call u the limit of the sequence $u_l$ in $L^{2}_{loc}\left (M\right )$ . Notice that $u\in L^1_{loc}\left (M\right )$ .

Next, we show that u is weakly harmonic. This is standard: given $\varphi \in C^{\infty }_0\left (\mathbb {S}^{n-1}\right )$ as each $u_l$ is harmonic, it then holds that

$$\begin{align*}\int_M u_l\left(x\right)\Delta_g\varphi\left(x\right)\,dV_g=0. \end{align*}$$

However, $u_l\rightarrow u$ in $L^2_{loc}\left (M\right )$ , and so, since $\Delta _g \varphi \in C^{\infty }_0\left (M\right )$ , we must have that

$$\begin{align*}\int_M u\left(x\right)\Delta_g \varphi\left(x\right)\,dV_g=0, \end{align*}$$

which is what we wanted to show. Since u is weakly harmonic and locally integrable, Weyl’s lemma applies and we can conclude that u is almost everywhere equal to a proper harmonic function on M.

The following computation, which was originally suggested in [Reference Bravo, Cortissoz and Peters-Stein4], and that we reproduce for the convenience of the reader, shows that the boundary data are satisfied in an $L^2$ -sense as described above. Again, using the orthogonality of the eigenfunctions of the Laplacian with different eigenvalues, we can estimate

$$ \begin{align*} \left\|f\left(\omega\right)-u\left(\omega,r\right)\right\|^2_{L^2\left(\mathbb{S}^{n-1}\right)}&= \left\|\sum_m \left(1-\varphi_{m}\left(r\right)\right) \sum_k c_{m,k} f_{k,m}\left(\omega\right)\right\|^2_{L^2\left(\mathbb{S}^{n-1}\right)}\\ &= \sum_m \left|1-\varphi_{m}\left(r\right)\right|{}^2 \left\|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right\|^2_{L^2\left(\mathbb{S}^{n-1}\right)}. \end{align*} $$

Pick H such that

$$\begin{align*}\sum_{m\geq H} \left\|\sum_k c_{m,k} f_{m,k}\left(\omega\right)\right\|^2_{L^2\left(\mathbb{S}^{n-1}\right)}\leq \frac{\epsilon^2}{8}, \end{align*}$$

and let $R>0$ such that if $r\geq R$ , for $m=0,1,\dots , M$ , we have that

$$\begin{align*}\left|1-\varphi_m\left(r\right)\right|\leq \frac{\epsilon}{2K}, \end{align*}$$

where K is such that

$$\begin{align*}\sum_{m\leq H} \left\|\sum_k c_{m,k} f_{m, k}\left(\omega\right)\right\|^2_{L^2\left(\mathbb{S}^{n-1}\right)}\leq K. \end{align*}$$

Notice that $\left |1-\varphi _m\left (r\right )\right |\leq 1$ . Therefore, putting all this together, we find that given any $\epsilon>0$ for r large enough,

$$\begin{align*}\left\|f\left(\cdot\right)-u\left(\cdot, r\right)\right\|_{L^2\left(\mathbb{S}^{n-1}\right)}<\epsilon, \end{align*}$$

which is what we wanted to prove.

5 Some remarks

In the case when N is a Lie group, Taylor in [Reference Taylor19] gave sufficient conditions for the eigenfunction expansion of f to converge uniformly to f. For instance, if $N=\mathbb {S}^3$ with the round metric, then if $f\in C^{\frac {1}{2}}\left (\mathbb {S}^3\right )$ , that is, if f is Hölder continuous with exponent $\frac {1}{2}$ , the Dirichlet problem is not just solvable, but uniformly solvable at infinity in $\mathbb {R}^4$ endowed with a rotationally symmetric metric such that the factor $\phi $ satisfies (2) and that $\phi '\geq 0$ , or if $\mathbb {R}^4$ with the given metric is a Cartan–Hadamard manifold. On the other hand, if N is a Lie group of dimension $4k$ , it is only required that $f\in H^{2k}\left (N\right )$ for the limit $\lim _{r\rightarrow \infty } u\left (\omega ,r\right )=f\left (\omega \right )$ to be uniform.