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 $$\int_{0}^1 \int_{0}^{s} \left\frac{q(s)q(t/s)}{s} \frac{q(t)q((st)/(1t))}{1t}\rightdt ds<\varepsilon$$ Somehow the question does not look hard, but still I don't see if it is true or not. For the function $1/(xx^2)$ one has $\frac{q(s)q(t/s)}{s} \frac{q(t)q((st)/(1t))}{1t}=0$, but the truncation of this function to $[\varepsilon,1\varepsilon]$ and normalization do not work.
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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 $$\int_{0}^a \int_{0}^{s} \frac{q(t)q((st)/(1t))}{1t}dt ds$$ is equal to $$ \int_{0}^a \int_{0}^{\frac{at}{1t}} q(t) q(u) du dt,$$ But since $(at)/(1t) \leq a$ for all $t\leq a$, this quantity is less than $1/4$. This implies that $$\int_{0}^1 \int_{0}^{s} \left\frac{q(s)q(t/s)}{s} \frac{q(t)q((st)/(1t))}{1t}\rightdt ds \geq 1/2.$$ 

