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 $Ag=f$ but a "least squares" type problem, i. e. $$\min_g \|Ag-f\|. $$ 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 $$\min_{0\leq g\leq 1} \|Ag-f\|$$ 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 $$ g^{n+1} = P_{[0,1]}(g^n - \tau A^*(Ag^n-f)) $$ 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 $$ x\mapsto \min(\max(x, 0),1). $$

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 $$ x\mapsto \min(\max(x, 0),1). $$

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 $Ag=f$ but a "least squares" type problem, i. e. $$\min_g \|Ag-f\|. $$ 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 $$\min_{0\leq g\leq 1} \|Ag-f\|$$ may have minimizers.

From a practical point of view: Solving the constrained problem may indeed be easier but this depends on the specific situation.

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 $Ag=f$ but a "least squares" type problem, i. e. $$\min_g \|Ag-f\|. $$ 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 $$\min_{0\leq g\leq 1} \|Ag-f\|$$ 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 $$ g^{n+1} = P_{[0,1]}(g^n - \tau A^*(Ag^n-f)) $$ 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 $$ x\mapsto \min(\max(x, 0),1). $$