the function (f) I want to reconstruct partially could look like this:

The following properties are known:
It consists only of alternating plateau (high/low).
So the first derivation is zero respectively undefined at the edges.
The function was convoluted with a kernel fulfilling the following conditions:
It is a boxcar function
Its center is at x=0
Its integral is 1.
I want to reconstruct only the positions of the edges of the original function (f) from the convolution result (c). So just these positions are of interest to me:

If the convolution kernel width (k) is less than the minimum plateau width (b, 40 in the example above) of f, c looks as follows:
(The width of the box car convolution kernel here is k=31.)
In that case it is easy to reconstruct the edge positions: I look for (possibly broad) extrema, and in between to neighbours [e1_x, e1_y] and [e2_x, e2_y] (one of them is a minimum and one a maximum of course), I search the x0 fulfilling: c(x0) = (e1_y + e2_y) / 2.
The reconstructed edge positions look like that:

But if k > b my approach fails:
(k=57)
Is there a possibility to calculate the original edge positions in f, if g (and so k) and c are known, also for the k>b cases?

This looks like a perfect match for total-variation deconvolution. In a nutshell, you have a model that your given function is $u^0 = h\ast u^\dagger$ with the box-car kernel $h$ and a piecewise constant function $u^\dagger$. To reconstruct $u^\dagger$ from the knowledge of $u^0$ and $h$ you minimize $$ \|u*h - u^0\| + \lambda TV(u) $$ over $u$ for some parameter $\lambda>0$. The first term shall enforce reconstruction while second term shall both regularize the deconvolution and also push the solution toward piecewise constant one. The term $TV$ refers to the total variation and in the discrete and one-dimensional case it is $TV(u) = \sum |u_{i+1}-u_i|$, i.e. the sum of the magnitude of the first differences. The parameter $\lambda$ allows you to balance both effects - since you do not seem to have noise, a very small $\lambda$ should work.