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 $1$.

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  • $\begingroup$ 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? $\endgroup$ – H17 May 24 at 17:06
  • $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki May 24 at 21:30

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