Convolution Theorem in Physics I'm getting ready for my classes to start next semester in Grad school, and I'm reading over Fourier Transforms and their applications. I came across the Convolution Theorem, namely, that if we have a convolution $$ \int_{-\infty}^\infty f(x)g(z-x)\, dx = h(z) ,$$ then the Fourier transform $${\hat f}[f(x)g(x)] = \left[{1 \over {\sqrt{2\pi}}}\right]{\hat f(k)}\ast{\hat g(k)}$$ of the product of $f(x)$ and $g(x)$ essentially equals the individual convolutions of the same functions in their "frequency" interpretation. The actual convolution theorem states $$\widehat {h(k)} = \sqrt{2\pi}{\hat f(k)}{\hat g(k)}$$ My question is: 
 What's so important about these results? What's their significance when I am actually in a laboratory, taking measurements?
 A: *

*Fourier transforms occur very often in most fields of physics.

*Products of functions occur very often in most fields of physics. 


As a consequence of points 1 and 2, it is common to encounter Fourier transforms of products when manipulating an algebraic expression. To move forward algebraically, you would need to apply the convolution rule.
A: A measurement typically involves the convolution of the thing being measured with the response function of the instrument. Now if the Fourier Transform of your response function has zeros in it, the convolution theorem tells you that information at the corresponding frequencies will be destroyed by the measurement process
There are many possible examples of this - but I will give just one. If you have an optical system (lens), the diffraction limit is often given as $\frac{1.22 \lambda}{d}$; this is actually the location of the first null in the response function, which is actually a Bessel function (Airy's disk). If your original image has a repeating pattern at exactly the frequency corresponding to that null, it will be destroyed. Higher and lower frequencies will be present (but attenuated to different extent).
So the convolution theorem helps you see how the frequency content of a measurement is affected by the measurement system. This can be quite useful in gaining a deeper understanding of your setup.
A: A process that is linear and shift invariant can be described by a convolution integral. As an example, consider the scalar diffraction of light, which can be computed with the Fresnel diffraction integral. Scalar diffraction is a linear shift invariant process. That's why the Fresnel diffraction integral is a convolution integral.
A: You ask about laboratory experiments involving convolution. A common example is measuring the time profile of some event when the process by which the event is triggered has a time profile similar to the duration of the event itself. For example measuring the decay of an exited species when excited by a laser pulse. A common experiment in biophysics and chemical physics is measuring fluorescence decays by time correlated photon counting. 
The measured decay is distorted by the convolution of the exciting pulse with the true decay. The two can be untangled by re-convolution and so fitting by non-linear least squares to a model of the decay process with the measured excitation pulse profile.(The alternative, which is deconvolution, seldom works well in practice).  When using Fourier transforms to do the convolution it is important to have equal (say zero) signal at the start and end of the data set since the Fourier transform assumes a repeating signal and any discontinuity here distorts the data.
Another example is the distortion of spectral lines by the finite width of slits in a spectrograph.
(Note that although convolution is fast using digital fast fourier transform algorithms, convolution can still be performed numerically using the integral over the finite range of the experiment.)
