Identity on convolution with Gaussian measure

Question

I've came across an identity once (I don't remember where) concerning convolutions of Gaussian measures. If I'm not mistaken, this identity was \begin{eqnarray} (\mu_{C}*f)(y) = \exp\bigg{[}\frac{1}{2}\bigg{(}\frac{\partial}{\partial x}, C\frac{\partial}{\partial x}\bigg{)}\bigg{]}f(x)\bigg{|}_{x=y} \tag{1}\label{1} \end{eqnarray} where $C=(C_{ij})$ is a $n\times n$ Gaussian covariance, i.e. a positive-definite matrix and $$\bigg{(}\frac{\partial}{\partial x},C\frac{\partial}{\partial x}\bigg{)} := \sum_{i,j}\frac{\partial}{\partial x_{i}}C_{ij}\frac{\partial}{\partial x_{j}}$$ is a differential operator.

I recently discovered that D. Brydges states a very similar result in his notes on functional integrals (page 34, exercise 4.3.3). He mentions that this identity is related to Wick's theorem for Gaussian measures. Note, however, that Brydges states (\ref{1}) in the case $f$ is a polynomial.

I checked some references I know to look for this version of Wick's Theorem but I have found nothing and I'd like to know more about this identity. If $f$ is assumed to be a polynomial, it seems to me that the natural way to prove it is to consider the Fourier transform of $\mu_{C}$ in order to obtain its moments. But does this identity hold in more general cases? If this is the case, how to address it and prove it? Also, any reference is welcome here. Thanks!

Answer 1

Upon Fourier transformation the convolution becomes a product of the Fourier transform ${\cal F}[f]$ of the function $f$ and the Fourier transformed Gaussian measure, which is again a Gaussian with covariance matrix $C^{-1}$, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Upon inverse Fourier transformation $k_n\mapsto i\partial/\partial x_n$, hence $$(\mu_{C}*f)(x) = \exp\left(\tfrac{1}{2}\sum_{n,m}\frac{\partial}{\partial x_n} C_{nm} \frac{\partial}{\partial x_m} \right)f(x).$$ This holds irrespective of whether $f$ is polynomial or not.

Answer 2

(Not a complete answer really, but too long for a comment.)

I believe the answer depends on the way you define the operator $$A = \exp(\tfrac12 \sum_{n,m} \tfrac\partial{\partial x_n}C_{nm}\tfrac\partial{\partial x_m}).$$

• In the sense of functional calculus, $A$ is the exponential of the elliptic operator $$L = \tfrac12 \sum_{n,m} \tfrac\partial{\partial x_n}C_{nm}\tfrac\partial{\partial x_m}$$ (that is, $A = \exp(L)$). Since $L$ generates a strongly continuous semigroup of operators $\exp(t L)$ on various function spaces (for example, on $L^p(\mathbb{R}^n)$), $A$ is well-defined on these function spaces, and it is a classical result that $A$ is the convolution with the Gauss–Weierstrass kernel (up to change of variables $x \mapsto C^{-1/2} x$). Thus, the identity (1) works for $f$ in these function spaces. (This is essentially what Carlo Beenakker wrote in his answer.)

• If one understands $A$ formally as a series of differential operators, then convergence is an issue. For polynomials, however, this series becomes a finite sum, and it is not difficult to verify this formally, either by appealing to the Fourier transformation (again as in Carlo's answer), or by integration by parts (as in Ander Aguirre's answer, if I understand it correctly). The same should be true if $f$ is a real-analytic function with power series coefficients decaying sufficiently fast, but I did not attempt to work out the exact condition.

Answer 3

Chapter 2 of these notes by Brydges might help. In particular exercise 2.1.3 in page 30.