Fredholm integral with functions constrained to [0;1] I am trying to feed information about the solution when solving an inverse problem given by a Fredholm integral of the form
$$
g(t)=\int_{a}^{b}K(t,s)f(s)ds.
$$
Say I know $g(t)$ and $K(t,s)$, and want to know $f(s)$. But I have some additional knowledge: $g(t)$ and $f(s)$ are both constrained to the interval $[0;1]$. 
Since this leads to an ill-posed problem, I wonder if the extra knowledge about the solution interval may help in finding a solution without struggling too much with inverse problem methods.
Any help appreciated!
 A: There is not enough information for a thorough answer. An a priori bound on the solution may indeed help theoretically and practically. As usual with measured data you may not want to solve the equation $Af=g$ (where $A$ denotes the integral operator, i.e $Af(t) = \int k(t,s) f(s) ds$) but a "least squares"  type problem, i. e. 
$$\min_f \|Af-g\|. $$
You shall expect that minimizers do not exist (lack of coercivity). An a priori bound $0\leq f\leq 1$ can help as, depending on the function spaces, the problem
$$\min_{0\leq f\leq 1} \|Af-g\|$$
may have minimizers. 
From a practical point of view: Solving the constrained problem may indeed be easier but this depends on the specific situation. 
As a first try you could give the projected gradient method a shot. Assuming that you work with the $L^2$ norm, this amounts to the iteration
$$
f^{n+1} = P_{[0,1]}(f^n - \tau A^*(Af^n-g)) 
$$
with the adjoint $A^*$, a stepsize $0<\tau<2/\|A\|^2$ and the projection $P_{[0,1]} $ onto the interval, i. e.  you apply pointwise
$$
x\mapsto \min(\max(x, 0),1).
$$
