Double integrals of this type sometimes appear when using differentiation under the integral sign with respect to two variables. Therefore, I am interested in reducing them to (simpler) single integrals. If $f$ is Hölder continuous, i.e. $f \in C^{0,\alpha}([0,1])$ for some $\alpha \in (0,1]$, the integrals in the following calculation exist and all manipulations are easy to justify ($F$ is an arbitrary antiderivative of $f$): \begin{align} D(f) &\equiv \int \limits_0^1 \int \limits_0^1 \frac{f(x) - f(y)}{x-y} \, \text{d} x \, \text{d} y = 2 \int \limits_0^1 \int \limits_0^y \frac{f(x) - f(y)}{x-y} \, \text{d} x \, \text{d} y \\ &= 2 \int \limits_0^1 \int \limits_0^y \frac{f(y) - f(y-t)}{t} \, \text{d} t \, \text{d} y = 2 \int \limits_0^1 \int \limits_t^1 \frac{f(y) - f(y-t)}{t} \, \text{d} y \, \text{d} t \\ &= 2 \int \limits_0^1 \frac{F(1) - F(1-t) - F(t) + F(0)}{t} \, \mathrm{d} t = 2 \int \limits_0^1 \log(t) [f(t) - f(1-t)] \, \mathrm{d} t \\ &= 2 \int \limits_0^1 \log\left(\frac{t}{1-t}\right) f(t) \, \mathrm{d} t \equiv S(f) \, . \end{align} However, this identity is also true for many less regular functions. The integral on the right-hand side exists for all $f$ in the weighted $L^1$ space $$A \equiv L^1 \left([0,1], \left\lvert\log \left(\frac{\cdot}{1-\cdot}\right)\right\rvert \lambda^1\right)$$ ($\lambda^1$ is the one-dimensional Lebesgue measure). It contains the function $g \equiv t \mapsto \left(t-\frac{1}{2}\right)^{-1}$, for which we have $S(g) = \pi^2$, while $D(g)$ does not exist. Therefore, I would like to find the set $$B = \{f \in A : D(f) \text{ exists and } D(f) = S(f)\} \subsetneq A$$ on which the above equation holds.
In a first attempt, I have tried to work with $$ C \equiv L^1 \left([0,1], - \log \left[\cdot (1-\cdot)\right] \lambda^1\right) = L^1 ([0,1],\lambda^1) \cap A \, .$$ $f \in C$ has an absolutely continuous antiderivative, so we can work backwards to show $$ S(f) = 2 \int \limits_0^1 \int \limits_t^1 \frac{f(y) - f(y-t)}{t} \, \text{d} y \, \text{d} t \stackrel{?}{=} 2 \int \limits_0^1 \int \limits_0^y \frac{f(y) - f(y-t)}{t} \, \text{d} t \, \text{d} y \, ,$$ but I do not know how to justify changing the order of integration here. Tonelli's theorem does the trick for monotone functions, but according to numerical calculations $S(f) = D(f)$ holds for other $f \in C$ as well, which leaves me with the following questions:
- Can we prove that the order of integration may be changed or is there an alternative way to show that $C \subset B$ holds? (No! See edit.)
- What is the largest subset of $B$ (maybe even $B$ itself?) we can find?
Some examples ($\mathrm{G}$ is Catalan's constant): \begin{align} D(\arctan) \stackrel{\text{Hölder cont.}}{=} S(\arctan) &= 2 \mathrm{G} - \frac{\pi^2}{16} - \frac{\pi}{4}\log(2) + \frac{1}{4} \log^2(2) \\ D(\log^3) \stackrel{\text{monotone}}{=} S(\log^3) &= 2 \pi^2 + \frac{2\pi^4}{15} + 12 \zeta(3)\\ D\left(t \mapsto \frac{\log(1-t)}{t^{3/2}}\right) \stackrel{?}{=} S\left(t \mapsto \frac{\log(1-t)}{t^{3/2}}\right) &= \frac{2\pi^2}{3} - 16 \log(2) (1 - \log(2)) \end{align} The last function is neither Hölder continuous nor monotone, so it is unclear whether the integrals are equal. Mathematica gives approximately $3.1754$ for the double integral as opposed to $3.1766$ for the single integral, so there is numerical evidence, but a proof would be better, obviously.
Edit 26 April 2020
The answers to this question show that there are measurable sets $M \subset [0,1]$ whose indicator function $1_M$ lies in $C$ (obvious) but not in $B$, since $$\int \limits_0^1 \int \limits_0^1 \frac{\lvert 1_M(x) - 1_M(y) \rvert}{\lvert x-y \rvert} \, \mathrm{d} x \, \mathrm{d} y = 2 \int \limits_{M \times M^\text{c}} \frac{\mathrm{d} x \, \mathrm{d} y}{\lvert x-y \rvert} = \infty \, .$$ This confirms fedja's comment and disproves $C \subset B$, so the first question is obsolete.