Can I apply the convolution theorem with an adaptive kernel?

Question

Consider a function $f(x)$ and a Gaussian kernel function $g(x)$

The convolution $h(x)=(f\ast g)(x)$ results

and as observed, there is some smoothing around $x=5$. I want to prevent this smoothing and only have $f$ smoothed at it's boundaries ($x=-5,10$). This can be done in real space by just setting $h(x)=f(x)$ from (say) $x\in[-4,9]$ after the convolution is done.

However, I intend to perform this convolution in Fourier space using the convolution theorem $H(\kappa)=F(\kappa)G(\kappa)$. So my only option is to transform back $H(\kappa)$ to $h(x)$ and perform the above operation. If I want to avoid this last step, how can I do that in Fourier space? Maybe an adaptive kernel function can work, eg is Gaussian close to the boundaries and is a Dirac delta function in the interior part. Playing with the Gaussian standard deviation can provide something like this.

I think it is easy enough to implement this in real space in order to avoid that crude operation, but I can't think of a way to do it in Fourier space. How can I change the kernel shape when it's only a point-wise multiplication (and not a slide of one function on top of the other one)?

Answer 1

1. No convolution kernel can perform the operation you suggest. There is a quick way of seeing this: for the smoothing at $-5$ and $10$ to take place, the convolution kernel $g$ must be smooth $^{[1]}$, but then $f\ast g$ will also be smooth at all points, because $$ \frac{d^k}{dx^k}( f\ast g)= f\ast \frac{d^kg}{dx^k}.$$

As mentioned in comments, this is related to a standard limitation of Fourier analysis, connected with the uncertainty principle of Fourier analysis.

2. As for the adaptive kernel. You suggest the use of a kernel that "changes shape as it slides". This operation is called an integral transform and is given by $$ T_K f(x)=\int_{-\infty}^\infty K(x, y)f(y)\, dy.$$ The shape of the kernel at each point $y\in\mathbb R$ is given by the function $K(\cdot, y)$. Convolutions are the integral transforms in which the kernel is of the form $K(x, y)=g(x-y)$, which corresponds to a fixed shape that "slides" along the given function $f$.

To sum up, adaptive kernels are not convolutions and so they do not turn into pointwise multiplication in Fourier space.

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[1] $g$ must also satisfy some appropriate assumptions at infinity.