# Multiply and divide by the same function in convolution

I am calculating the convolution of two functions $$F(x), G(x)$$ in $$\mathbb{R}^{n}$$, n-dimensional space. I have another function $$h(x)$$ that is a Gaussian.

What effect does multiplying $$F(x)$$ by $$h(x)$$ and concurrently dividing $$G(x)$$ by $$h(x)$$ have, on the resulting convolution?

Specifically, I am interested in the quantity: $$\frac{ (F(x) h(x)) * (G(x) / h(x)) }{ F(x) * G(x) }$$

Both $$F(x), G(x)$$ are results of Fourier transforms of other functions.

• What do you mean by "what effect"? Are you asking whether the result can be simplified, or whether anything cancels? Then the answer is no. Jul 10 at 3:58
• I imagine that such a transformation is commonplace in practice. Do you have references that I can read up about the above expression? Jul 10 at 4:03
• Since $F$ and $G$ are Fourier transforms of other functions, why do you suppose that they are in $\mathbb{R}^n$? Jul 10 at 11:26

### Can it be simplified?

$$\frac{(fh) * (g/h)}{f * g} \Leftrightarrow \frac{(F * H)(G * 1/H)}{F G}$$

where $$G=\mathcal{F}(g)$$, by convolution theorem. The numerator on right side cannot be simplified any further: nothing cancels. That $$h$$ is Gaussian doesn't change this.

Also, in practice, $$g/h$$ is a bad idea with Gaussian $$h$$: Gaussian decays to sub-machine epsilon very quickly, yielding numeric instabilities. Whether one can do g / (h + epsilon) depends on the purpose of g/h.

### What does it "do"?

• $$fh$$: windows $$f$$. That is, tapers off $$f$$'s endpoints into negligibility.
• $$g/h$$: opposite effect, inflates $$g$$'s endpoints, and ~zeros the center.
• $$/ (f * g)$$: no general-purpose description. However, if $$f$$ is close to $$f/h$$, and $$g$$ is close to $$gh$$, then it has the effect of making the output closer to $$1$$ at each point. Since $$h$$ is Gaussian, this is basically never exactly true, but it can be true to varying degrees of approximation, and one could make descriptions on regularity of output (e.g. expected flat regions).

So the numerator's reverse-windowings effectively convolve the center of $$f$$ with edges of $$g$$. Dividing by the denominator can be seen as a pointwise measure of similarity of numerator with denominator, which also measures closeness of $$f$$ to $$f/h$$ and of $$g$$ to $$gh$$. But there are much better ways to measure these similarities, so this can't be its primary purpose.

Without knowing the context, I doubt much else can be said, but it's worth seeing others' responses.

• Your answer is great. I added a little more context, can you take a look and consider those to see if they are helpful? Jul 10 at 4:29
• @hsiaomichiu Since $H$ is just $h$ with different width, the expression is now effectively same as the right equation instead of left, and the "what does it do" needs to reinterpret accordingly. This greatly changes the interpretation and effectively the question itself, which is discouraged on StackExchange. In short, the numerator is now a product of lowpassing of $f$ and highpassing of $g$, which might also interpret as similarity with $fg$. Jul 10 at 4:41