## Mean value property in limit for eigenfunctions of the Laplace-Beltrami operator.(English)Zbl 1466.43005

Let $$X=G/K$$ be a Riemannian symmetric space of noncompact type with rank one. Here $$G$$ is a connected noncompact semisimple Lie group with finite center with real rank one and $$K$$ is a maximal compact subgroup of $$G$$. Let $$o$$ be the origin of $$X$$, let $$B(x, r)$$ be the geodesic ball centered at $$x \in X$$ and of radius $$r$$, let $$\Delta$$ be the Laplace-Beltrami operator on $$X$$, and let $$\rho$$ be the half sum of positive roots. The elementary spherical function $$\varphi_\lambda$$, $$\lambda\in \mathbb C$$, on $$X$$ is the unique radial eigenfunction of $$\Delta$$ with eigenvalue $$-(\lambda^2+\rho^2)$$ and $$\varphi_\lambda(o)=1$$. Let us introduce the convolution $(f * m_r^\lambda)(x) = (V_r^\lambda)^{-1} \, \int_{B(x, r)} \, f(y) \, dy,\ \ {\text {where}}\ \ V_r^\lambda=\int_{B(x, r)} \, \varphi_\lambda (x) \, dx.$ This definition is correct for $$r>0$$ except for the set $$D_0^\lambda=\{r>0 \, | \, V_r^\lambda=0\}$$.
It is known that $$f$$ is an eigenfunction of $$\Delta$$ with eigenvalue $$-(\lambda^2+\rho^2)$$ only if $$f$$ satisfies the generalized mean value property: $$(f * m_r^\lambda)(x)=f(x)$$.
The substance of main results of the paper can be expressed in short as follows: the limit of the convolution with $$m_r^\lambda$$ when $$r\to \infty$$ moves arbitrary functions to $$-(\lambda^2+\rho^2)$$-eigenfunctions of the Laplacian.
Here are exact statements.
(i) Suppose that for a function $$f \in L^1_{\mathrm{loc}}(X)$$ and a fixed $$\lambda\in\mathbb C$$, $(f* m_r^\lambda)(x) \to g(r)$ for every $$x\in X$$ for some function $$g$$ on $$X$$, as $$r$$ tends to $$\infty$$ on $$\mathbb R^+ \setminus D_0^\lambda$$. If there is a positive function $$\psi \in L^1_{\mathrm{loc}}(X)$$ and a positive function $$r_0 \in L^\infty_{\mathrm{loc}}(X)$$ such that for almost every $$x \in X$$, $$|(f *m_r^\lambda)(x)| \leqslant \psi(x)$$ whenever $$r \geqslant r_0(x)$$, then $$\Delta g = -(\lambda^2+\rho^2)g$$.
(ii) Let $$f$$ and $$g$$ be two continuous functions on $$X$$. If for a fixed $$\lambda\in\mathbb C$$, $(f *m_r^\lambda)(x) \to g(x)$ for every $$x \in X$$, uniformly on compact sets, as $$r \to \infty$$ on $$\mathbb R^+ \setminus D_0^\lambda$$, then $$\Delta g = -(\lambda^2+\rho^2)g$$.

### MSC:

 43A85 Harmonic analysis on homogeneous spaces 22E30 Analysis on real and complex Lie groups
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