tions are close, the corresponding densities should be close as well. However, in spaces of integrable functions the inverse problem is not well-posed: while the operator I is continuous on L$_{1}$ (or another L$_{p}$ space) the inverse operator ∂ k is not. The example below (from Zinde-Walsh, 2011) illustrates lack of well-posedness. Example. Consider the space D ([0 , 1]) of univariate absolutely continu- ous distribution functions on the interval [0 , 1] in the uniform metric: the dis- tance between two distributions, F$_{1}$, F$_{2}$ is d ( F$_{1}$, F$_{2}$ ) = max x ∈ [0 , 1] | F$_{1}$ ( x ) − F$_{2}$ ( x ) | ; this is the image space of the operator I ( · ) defined on L$_{1}$ ([0 , 1]) . Denote by [ v ] the integer part of v, that is the largest integer that is ≤ v. Let I ( x ∈ A ) denote the indicator function of set A, that equals 1 if x is in A, zero otherwise. With ε = ¯ ε 2 define densities The densities f$_{1}$ and f$_{2}$ have supports that do not intersect, it is easily seen that at each point they differ by 2: | f$_{1}$ ( x ) − f$_{2}$ ( x ) | = 2 ; it follows that the L$_{1}$ ([0 , 1]) difference between them is 2 . The corresponding distributions are F$_{1}$ = I ( f$_{1}$ ) and F$_{2}$ = I ( f$_{2}$ ) . It is easy to establish by integration that max x ∈ [0 , 1] | F$_{1}$ ( x ) − F$_{2}$ ( x ) | ≤ 2 ε = ¯ ε, f$_{2}$ ( x ) = 2 [ ε $_{−}$1 +1 2 ] − 1 ∑ m =0 I ( x ∈ [(2 m + 1) ε, (2 m + 2) ε )) . f$_{1}$ ( x ) = 2 [ ε $_{−}$1 +1 2 ] − 1 ∑ m =0 I ( x ∈ [2 mε, (2 m + 1) ε )); 4