# Is convolution area-preserving?

Consider two functions f(x) and g(x) with indefinite integrals ("area under the curve") Af and Ag. Does the convolution f * g preserve the area, i.e. is

Af *g = Af * Ag

i.e.

∫ ( ∫f(x)g(t−x)dx ) dt = (∫f(x)dx) (∫g(x)dx) ?

I've learned that convolution in one space is multiplication in the dual space, or more precisely, that direct multiplication in one space corresponds to polynomial multiplication in the dual space. There is also a nice, short video where this intuition is explained, but I'm having difficulty transferring this to the question regarding areas and drawing the right conclusions.

To provide some practical context: Image formation in an optical system can be described as

Image = Source * PSF

with PSF being the point spread function describing the system's impulse response. The PSF is often approximated by a Gaussian and the convolution results in a Gaussian blur.

If the Gaussian is normalized to unit area (i.e. a normal distribution), does this equate to the fact that the integrated intensities of source and image are preserved? Meaning that the photons are merely "redistributed" (blurring process), but the total photon number is preserved (= energy conservation in an ideal, lossless system)?

Conversely, can I add e.g. a 10% loss of photons due to absorption by simply changing the normalization of the PSF from 1.0 to 0.9 (rescaling the Gaussian's normalization factor)?

• Are you asking whether $\int\left(\int f(x)g(t-x)dx\right)dt = \left(\int f(x)dx\right)\left(\int g(x)dx\right)$ ? Nov 24, 2020 at 9:43
• @lisyarus Hi, I think...yes! :) Nov 24, 2020 at 9:44
• I have checked for the case of the convolution of a box function with a normal distribution in Wolfram Alpha and, indeed, the resulting area seems to be preserved. I'm wondering if this holds in general for all kinds of functions f(x) and g(x) Nov 24, 2020 at 9:48
• @lisyarus Actually, this is precisely my question. I will add it to the post Nov 24, 2020 at 9:53

If by convolution you mean the standard

$$(f * g)(t) = \int f(x)g(t-x)dx$$

and by area

$$A_f = \int f(x)dx$$

then the question is whether the following is true

$$\int\left[\int f(x)g(t-x) dx\right]dt = \left[\int f(x)dx\right]\cdot \left[\int g(x)dx\right]$$

This is indeed true. By a simple change of variables

$$u = x$$ $$v = t - x$$

with Jacobian

$$\det \pmatrix{1 & -1 \\ 0 & 1} = 1$$

we get

$$\int\int f(x)g(t-x) dxdt = \int\int f(u)g(v)dudv = \left[\int f(u)du\right]\cdot \left[\int g(v)dv\right]$$

provided that the conditions of the Fubini theorem are satisfied.