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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 $C^{-1}$, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Inverse Fourier transformation, with $k_n=i\partial/\partial x_n$, then gives Equation (1), $$(\mu_{C}*f)(x) = \exp\left(\tfrac{1}{2}\sum_{n,m}\frac{\partial}{\partial x_n} C_{nm} \frac{\partial}{\partial x_m} \right)f(x).$$ This should hold irrespective of whether $f$ is polynomial or not.

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.

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 $C^{-1}$, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Inverse Fourier transformation, with $k_n=i\partial/\partial x_n$, then gives Equation (1). This should hold irrespective of whether $f$ is polynomial or not.

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 $C^{-1}$, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Inverse Fourier transformation, with $k_n=i\partial/\partial x_n$, then gives Equation (1), $$(\mu_{C}*f)(x) = \exp\left(\tfrac{1}{2}\sum_{n,m}\frac{\partial}{\partial x_n} C_{nm} \frac{\partial}{\partial x_m} \right)f(x).$$ This should hold irrespective of whether $f$ is polynomial or not.

Upon Fourier transformation the convolution becomes a product, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Inverse Fourier transformation, with $k_n=i\partial/\partial x_n$, then gives Equation (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 $C^{-1}$, $${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$ Inverse Fourier transformation, with $k_n=i\partial/\partial x_n$, then gives Equation (1). This should hold irrespective of whether $f$ is polynomial or not.