For every $\varepsilon>0$ find a piecewise continuous function $q:[0,1]\rightarrow \mathbb{R}$ such that $\int_0^1 q(x)dx=1$ and $$\int_{0}^1 \int_{0}^{s} \left|\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}\right|dt ds<\varepsilon$$ Somehow the question does not look hard, but still I don't see if it is true or not. For the function $1/(x-x^2)$ one has $\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}=0$, but the truncation of this function to $[\varepsilon,1-\varepsilon]$ and normalization do not work.
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$\begingroup$ Is q a nickname for g? $\endgroup$– Michael RenardyCommented Aug 22, 2013 at 18:14
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$\begingroup$ Another way to make the function $1/(x-x^2)$ integrable is to replace it by the function $(x-x^2)^{-\alpha}$ for $\alpha<1$ close to $1$. But (unless my computations are wrong), this does not work either. $\endgroup$– Mikael de la SalleCommented Aug 22, 2013 at 21:17
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$\begingroup$ Good point, Mikael, I did not check this. What does exactly fail? $\endgroup$– Kate JuschenkoCommented Aug 22, 2013 at 21:29
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2$\begingroup$ See my answer below (this was of course not what I had in mind when I wrote my comment, but I found the answer as I was trying to explain in words, rather than by a computation, what fails). $\endgroup$– Mikael de la SalleCommented Aug 22, 2013 at 22:43
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1 Answer
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There does not exist such a function $q$ if $\varepsilon<1/2$.
Indeed, if $q$ is a positive measurable function on $[0,1]$ of integral $1$, pick $a \in [0,1]$ such that $\int_0^a q(x) dx = 1/2$.
Then $$\int_{0}^a \int_{0}^{s} \frac{q(s)q(t/s)}{s} dt ds = 1/2,$$ whereas $$\int_{0}^a \int_{0}^{s} \frac{q(t)q((s-t)/(1-t))}{1-t}dt ds$$ is equal to $$ \int_{0}^a \int_{0}^{\frac{a-t}{1-t}} q(t) q(u) du dt,$$ But since $(a-t)/(1-t) \leq a$ for all $t\leq a$, this quantity is less than $1/4$.
This implies that $$\int_{0}^1 \int_{0}^{s} \left|\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}\right|dt ds \geq 1/2.$$
answered Aug 22, 2013 at 22:26