Let $f$ be a quadratically integrable function. Does the inequality
$$ \int_0^1\int_0^1 \min(s,t) f(s) f(t) \,\mathrm{d}s\,\mathrm{d}t \ge 0 $$
hold true? This is obviously related to Mercer's condition. I thought about writing
$$2\int_0^1\int_0^t sf(s)f(t)\,\mathrm{d}s\,\mathrm{d}t=2\int_0^1f(t)\big(\int_0^t sf(s)\,\mathrm{d}s\big)\,\mathrm{d}t.$$
Does partial integration then lead to some conclusion?
Hint. Assume that $f\in C([0,1])$ and let $F(x)=\int_0^xf(t)dt$ then, by integrating repeatedly by parts, the given integral can be written as $$\begin{align} \int_0^1\int_0^1 \min(s,t) f(s) f(t) ds dt&=2\int_0^1 f(s)\left(\int_0^s tf(t) dt\right) ds\\ &=2\int_0^1 f(s)\left(sF(s)-\int_0^s F(t)dt\right)\\ &=2\int_0^1 sF(s)f(s)ds -2\int_0^1 \left(f(s)\int_0^s F(t)dt\right)ds\\ &=F^2(1)-\int_0^1 F^2(s)ds -2\int_0^1 F(t) (F(1)-F(t))dt\\ &=F^2(1)-2F(1)\int_0^1 F(t)dt+\int_0^1 F^2(s)ds\\ &=\int_0^1\left(F(1)-F(t)\right)^2dt\geq 0. \end{align}$$