Reference Request - Recovering a function from its definite integrals (inverse problem)

Here is how you make an inverse problem of this problem: Choose a space $X$ for the function $f$ you are looking for (e.g. $L^2(0,1)$ to work in Hilbert spaces, but other spaces may be more suitable, depending on your needs).

I assume that you only have finitely many definite integrals (since I assume that this is a practical problem where the definite integrals come from measurements). Now let us denote your tuples as $(a_1,b_1),\dots (a_N,b_N)$. You forward operator is $$\newcommand{\RR}{\mathbb{R}} K:X\to\RR^N $$ mapping $f$ to the $N$-vector with components $\int_{a_i}^{b_i}f(x)\,dx$. So you are given a vector $g\in\RR^N$ and want some solution to $$ Kf = g. $$ Now you are in business with the standard theory for linear inverse problems.

You have some of the usual problems coming with an inverse problem: Non-uniqueness (the operator is not injective) and probably instability in some sense (depending on you data and values $(a_i,b_i)$). (As far as I see, non-solvability should not be an issue as $K$ should be surjective for meaningful tupels $(a_i,b_i)$).

To deal with non-uniqueness: You may view this as an advantage as you can choose among all solutions of $Kf=g$. To pick one, you can choose regularization functional $R:X\to [0,\infty]$ and define a minimum-$R$-solution as solution of $$ \min\{R(f)\mid f\in X,\ Kf=g\}. $$ From a computational point of view, convex functional $R$ are beneficial and you can choose $R$ to impose some structure on your solution, e.g. $R(f) = \int_0^1 |f'(x)|^2\, dx$ imposes some smoothness (effectively this means that you constrain your solutions to the Sobolev space $H^1$). The most straight-forward choice would be $R(f) = \int_0^1 |f(x)|^2\, dx$ which should produce a linear equality as optimality condition (and you are effectively computing the Moore-Penrose pseudo-inverse). I could say more about regularizing functionals if needed.

If your data vector $g$ is also uncertain, i.e. it may be given by measurement data with an error, you may want to relax your problem and look for solutions of $$ \min\{R(f)\mid d(Kf,g)\leq\delta\} $$ for some discrepancy functional $d$ and some value $\delta>0$. Both should be related to the error in your data. Note that this is in some way equivalent to (generalized) Tikhonov regularization which would be solving $$ \min_f d(Kf,g) + \lambda R(f) $$ for some regularization parameter $\lambda>0$. The most simple case of this would be standard Tikhonov regularization in Hilbert spaces: $$ \min_f \|Kf-g\|_{2}^2 + \lambda\|f\|_{L^2(0,1)}^2 $$ leading to the linear optimality condition $$ K^*(Kf-g) + \lambda f = 0. $$ The adjoint operator $K^*:\RR^N\to L^2(0,1)$ is given by $$ K^*g = \sum_{i=1}^N g_i\chi_{[a_i,b_i]} $$ (where $\chi_{[a_i,b_i]}$ is the characteristic function of $[a_i,b_i]$). So the optimality condition is actually $$ \sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} + \lambda f = 0. $$ This shows that the regularized solution is also a linear combination of the characteristic functions $\chi_{[a_i,b_i]}$ and thus, we still get a finite dimensional linear problem for the coefficients.

If you want some smoothness, try $R(f) = \int_0^1 |f'(x)|^2\, dx$. This would give an optimality conditions like $$ \sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} - \lambda f'' = 0 $$ and thus the solution is piecewise quadratic.

In general, it appears that hardly anything interesting can be said. E.g., let $A=\{(1/5,3/5),(2/5,4/5)\}$; here, it will be convenient to think of $A$ as a set of (say) open intervals, rather than a set of pairs of endpoints of intervals.

However, note first that, without loss of generality, for each open interval in $A$, all the intervals (closed, left-open, right-open) with the same endpoints may be assumed to belong to $A$. Next, let us assume that $\int_0^1|f|<\infty$ and that $A$ is a semi-ring (see measures on semi-rings) and $[0,1]\in A$. Then the formula $\mu(I):=\int_I f$ for $I\in A$ defines a finite signed countably-additive measure $\mu$ on $A$, which can be uniquely extended to a signed measure $\bar\mu$ on the sigma-algebra $\Sigma$ generated by $A$.

The measure $\bar\mu$ determines, and is determined by, the conditional expectation $E(f|\Sigma)$ (of $f$ given $\Sigma$), equal the Radon--Nikodym derivative $\dfrac{d\bar\mu}{d\lambda|_\Sigma}$ (with respect to the underlying Lebesgue measure $\lambda$ over $[0,1]$), and this conditional expectation is then precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$. (One might note that the appearance of the Radon--Nikodym derivative here is in broad agreement with the comment "Recovery of function from its integral is called differentiation" by Alexandre Eremenko.)

E.g., if $A$ consists of all intervals with endpoints in the set $\{j/n\colon j=0,\dots,n\}$, then all that we will know is, in essence, the "histogram" of the average values of $f$ over the intervals $[0,1/n],\dots,[1-1/n,1]$, and this "histogram" is the best approximation to $f$ that we can get in this case.

Extended comment: Dirk suggested an inverse-problem approach. One may note that such an approach will work perfectly well (and, generally, even better) within the above framework of the conditional expectation. Indeed, for a space $X$ of (say) real-valued integrable functions on $[0,1]$ we have the map $X\overset K\to\mathbb R^A$ defined by the formula $Kf:=(\int_I f)_{I\in A}$ for $f\in X$. This map can be factored as follows: \begin{equation} X\overset{E(\cdot|\Sigma)}\longrightarrow X_\Sigma\overset{K_\Sigma}\longrightarrow\mathbb R^A, \end{equation} where $X_\Sigma$ is the set of all integrable $\Sigma$-measurable functions in $X$ and $K_\Sigma$ is the restriction of $K$ to $X_\Sigma$; indeed, by the definition of the conditional expectation/Radon--Nikodym derivative, we have
$K_\Sigma E(f|\Sigma)=Kf$ for all $f\in X$. Thus, instead of $K$, one can deal with its restriction $K_\Sigma$, with the same (or greater) degree of success. In particular, if $A$ is finite, then we have to deal with the finite-dimensional space $X_\Sigma$ instead of the possibly infinite-dimensional space $X$.

This comment may be viewed as an illustration of what was said previously: that the conditional expectation $E(f|\Sigma)$ is precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$.