"anti-gaussian" 2D convolution kernel What is the 2D kernel, k, that when convoluted with a 2D signal, f, that is convoluted again with a gaussian 2D kernel, g, produces a result that is closest to the original signal, f'.
Something like this, I think:
For
f' = f $\star$ k $\star$ g (where $\star$ denotes convolution)
Find k, such that |f - f'| is minimal.
 A: For concreteness, let us define the Gaussian kernel $g$ as 
$$g(x,y) = \frac{1}{2\pi \sigma_x \sigma_y} \exp\left[-\frac{x^2}{2 \sigma_x^2}-
\frac{y^2}{2\sigma_y^2}\right].$$ Its Fourier transform is again a Gaussian
$$\hat g(k_x,k_y) = \int\!dx\, dy\, e^{i kx x + i ky y}\, k(x,y)=
 \exp\left[-\frac{\sigma_x^2 k_x^2}{2}-
\frac{\sigma_y^2 k_y^2}{2}\right]$$
is again a Gaussian. 
Under Fourier transform the convolution becomes a product, i.e.,
$\widehat{k\star g} = \hat{k} \times \hat{g}$. 
The difference is given by $$\begin{align}\int\!dx\,dy\,|f(x,y)-f'(x,y)|^2
&=\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,|\hat{f}(k_x,k_y)-\hat{f'}(k_x,k_y)|^2\\
&=\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,|\hat{f}|^2|1-\hat{k} \hat{g}|^2
\end{align}.$$ In order to minimize the difference, $\hat{k} \hat{g}$ should be as close as possible to 1 (and of course $k$ still normalized).
Therefore, we have to minimize
$$\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,[|1-\hat{k} \hat{g}|^2+\lambda|\hat{k}|^2],$$ where $\lambda$ is a Lagrange multiplier to keep the norm of $\hat k$ at 1. We can variate this equation, and obtain
$$\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,[2 \hat{g}(1- \hat{k} \hat{g})+2 \lambda\hat{k}] \delta \hat k,$$
with the solution
$$\hat{k} = \frac{\hat{g}}{\hat{g}^2+\lambda}=
\frac{\exp\left[-\frac{\sigma_x^2 k_x^2}{2}-
\frac{\sigma_y^2 k_y^2}{2}\right]}{\exp\left[-\sigma_x^2 k_x^2-
\sigma_y^2 k_y^2\right]+\lambda}.$$ $\lambda$ is determined from the normalization condition.
