# Eigenspace of Gaussian Markov operator

Consider the (one-dimensional) Gaussian distribution $$Q := N(\nu,\tau^2)$$ and the (Gaussian) Markov operator

$$\begin{equation*} \begin{array}{rccc} R : & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) & \to & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) \\ & f & \mapsto & \int f(x)\, N(\cdot,\sigma^2)(\mathrm{d}x). \end{array} \end{equation*}$$

I am interested in the eigenspace $$E_1 := \mathrm{kernel(I-R)},$$ in particular in the dimension of $$E_1.$$

Obviously, the indicator function $$\mathbb{1}_{\mathbb{R}}: x \mapsto 1$$ and the identity $$\mathrm{id}_{\mathbb{R}}: x \mapsto x$$ are both eigenfunctions to the eigenvalue $$1,$$ that is, $$\mathbb{1}_{\mathbb{R}},\ \mathrm{id}_{\mathbb{R}} \in E_1.$$

Are there more linearly independent eigenfunctions?

If I understand correctly, your operator $$R$$ is the convolution operator with the Gauss–Weierstrass kernel. This is a Fourier multiplier with symbol $$\lambda(\xi) = \exp(-\tfrac{1}{2} \sigma^2 |\xi|^2)$$: $$\widehat{R f}(\xi) = \lambda(\xi) \hat f(\xi).$$ If $$f$$ is a tempered distribution, then $$R f = f$$ if and only if $$(\exp(-\tfrac{1}{2} \sigma^2 |\xi|^2) - 1) \hat{f}(\xi) = 0 ,$$ and this is equivalent to $$\hat{f}$$ being supported in $$\{0\}$$. This, in turn, implies that $$f$$ is a polynomial. By inspection, in dimension $$1$$ the eigenspace is indeed spanned by $$f(x) = 1$$ and $$f(x) = x$$. In higher dimensions, however, any harmonic polynomial (i.e. a polynomial $$f$$ such that $$\Delta f = 0$$) will do.
If one goes beyond tempered distributions, there are more solutions, even in dimension one. For example, $$f(x) = e^{z x}$$ is an eigenfunction corresponding to eigenvalue $$\lambda(z) = \exp(-\tfrac{1}{2} \sigma^2 z^2),$$ where $$z$$ is an arbitrary complex number. Choosing, for example, $$z = \sqrt{2 \pi} \sigma^{-1} (1 + i),$$ we get $$\lambda(z) = \exp(-2 \pi i) = 1$$, as desired.
If one insists on real-valued solutions, then $$f(x) = \Re e^{z x} = e^{x \Re z} \cos(\Im z)$$ works (as long as $$\lambda(z)$$ is real). Thus, to give a specific real-valued example, $$f(x) = e^{\sqrt{2 \pi} x / \sigma} \cos(\sqrt{2 \pi} x / \sigma)$$ is another eigenfunction with eigenvalue $$1$$.
• Thank you! Since I am not trained and experienced in distribution theory (and Fourier transforms of distributions), I cannot see that the Fourier transform of $f(x)=x$ for $x \in \mathbb{R},$ that is, essentially $\delta'$ (correct?) is supported in $\{0\}.$ What is the difference to polynomials of higher order? Do you know any good reference for these basics? – H17 May 24 at 17:06
• The Fourier transform of $x$ is indeed $i \delta_0'$, and more generally, the Fourier transform of a polynomial $P(x)$ is $P(-i\partial_x) \delta_0$. I am not sure I have a good reference; Vladimirov's Methods of the Theory of Generalized Functions is one of the standard references, I think. – Mateusz Kwaśnicki May 24 at 21:30