Convolution can smooth an input function, is there an operation which bunches it up? An easy to remember description of what the convolution of two functions is, is to say that one is a weight function and the result is a weighted average of the other function. 
The canonical example would be a lowpass filter, which smoothes the input function by means of the filter (weight) function.
A highpass filter, on the other hand does roughly the opposite by removing the smoothness.
What I am looking for: is there an operation where a function is "bunched up" by another one, where "bunching up" should roughly mean:


*

*keep the value of the integral,

*reduce the support, or at least reduce function values which are already small, while increasing the larger ones,

*keep the symmetry between the two input functions (despite the fact that we call one the filter or weight).


My hunch is that I could Fourier transform the input, shift all frequencies towards zero somehow making use of the second function and then reverse transform and normalize, but then the two input functions are not used symmetrically. And I don't want to invent this. I would think that this concept exists already.
EDIT: If I understand the first answer correctly, @josh suggests that if a convolution kernel $\phi$ had a convolutional inverse $\phi^{-1}$, we could reverse a smoothing. Smoothing an $f$ by $f\ast\phi = g$ could then be reverted to $f\ast\phi\ast\phi^{-1} = g\ast\phi^{-1} = f$.
As mentioned in that answer, the "convolutional inverse" does not seem to exist always as a well behaved function. So I would like to invite more answers, not bound to convolution at all, maybe even non-linear operations.
 A: Convolution need not smooth the input if the integral kernel against which the convolution is performed is not a regular function.  Convolution against, e.g. the Gaussian $\phi_{\mu, \sigma}(x)$ with mean $\mu$ and standard deviation $\sigma$, will always smooth a function: $\phi_{\mu, \sigma} \ast f$ is "less lumpy" than $f$ itself (unless $f$ is constant or unbounded and linear).
Now consider the convolutional inverse of the Gaussian, i.e. $\psi$ such that $\psi \ast \phi_{\mu, \sigma} = \delta(x)$, where the RHS is the Dirac delta function localized at $x = 0$.  Note that
$$
\psi \ast (\phi_{\mu, \sigma} \ast f) = (\psi \ast \phi_{\mu, \sigma}) \ast f = \delta \ast f = f\,,
$$
where we use the facts that the convolution product is associative and that $\delta$ is the unit of this product.  So, convolution with $\psi$ transforms $\phi_{\mu, \sigma} \ast f$ (less lumpy) into $f$ (more lumpy).  This means that convolution has not smoothed the function!
The above is an example of "deconvolution" as linked above.  It's still a convolution, however it is against an integral kernel which maybe isn't a well behaved function at all.  Try to work out the form that $\psi$ takes.
Also note there are lots of technical issues related to deconvolution.  You can only convolve our $\psi$ against a function which is "smooth enough" or else you singular behavior.  By construction, we are guaranteed that $\phi_{\mu, \sigma} \ast f$ is smooth enough.  But $f = \exp(-|x|)$ is not, e.g.
