# What penalty function produces optimization-based Gaussian smoothing?

I have just read yet another introduction to image processing, which describes Gaussian smoothing from a convolution perspective and least-squares smoothing from an optimization perspective.

It would help me knit this knowledge together if I could imagine how I could also use optimization to obtain a Gaussian-smoothed signal as well, but I never see this.

Given that a least squares smoothing is obtained from

$$argmin_{x} \| Ax-b \|^{2}_2 + \lambda \| \nabla x \|_2$$

What corresponding penalty function would produce a Gaussian smoothing?

• p.s. someone should create a "regularization" tag for this exchange. – barnhillec Jan 26 '17 at 13:09
• I don't think you can express Gaussian smoothing via regularization. Do you have any evidence that you can? – Wolfgang Bangerth Jan 26 '17 at 20:57
• Very interesting question. First of all I assume that any technique that can be expressed as Ax = b is also solvable by variational means and therefore solvable using regularization. Second I assume that any direct forward technique such as Gaussian smoothing,which can produce a smoothed y in the case of Ax = y where A is a Gaussian smoothing matrix, can also produced a smoothed x in the case of Bx = y, where B is the inverse of A, hence is solved by x = inv(B)y =inv(inv(A))y = Ay . Is this not correct? – barnhillec Jan 27 '17 at 9:15
• I agree that a linear equation can be rewritten as a variational problem, but that doesn't necessarily mean that there will be a regularisation problem attached to it. Since Gaussian smoothing can work point wise through superposition there's no over- or under-fitting needing to be sorted out. – origimbo Jan 27 '17 at 14:22

A mathematically precise way to link various penalty terms to smoothing kernel functions would require the theory of reproducing kernel Hilbert spaces (RKHS). The penalty term for a kernel $K$ is the corresponding RKHS norm $|| \cdot ||_K$. For a Gaussian kernel this is given by: $$||f||_K = \sum_{m=0}^\infty \frac{\sigma^{2m}}{2^m m!} \int_{-\infty}^{\infty} [f^{(m)}(t)]^2 dt$$ where $f^{(m)}$ is the $m^{th}$ derivative and $\sigma$ is the standard deviation of the Gaussian kernel.