Help to give/improve a proof on simple problem of Riemann integration

Question

Let $K$ be a nonnegative, symmetric around zero, real function satisfying $$\int_{-1}^1K(u)du=1.$$

I want to show that $0<\int_0^1 K(u)u^2du$.

This is intuitive but I'm struggling to show it. I will sketch what I thought.

By the symmetry, $$\int_{-1}^1K(u)du=1\implies \int_{0}^1K(u)du=1/2.$$ This means that $K$ is not zero everywhere on $[0,1]$. Also $x^2$ is strictly positive almost everywhere on $[0,1]$. Then there is a non-null set $A\subseteq [0,1]:K(u)u^2>0$ for any $u\in A$ implying $\int_0^1 K(u)u^2du>0$.

This sketch is a "intuitive" proof, but the steps are not totally clear to me.

Can you help to improve it or suggest another proof?

Answer 1

If $K$ is non negative then you have $\int_0^1 K(u)u^2 du \ge 0$.

If $\int_0^1 K(u) u^2 du = 0$ then $K(u)u^2 = 0$ ae. and hence $K(u) = 0$ ae. This would imply that $\int_{-1}^1 K(u) du = 0$.

Answer 2

Keeping this in the realm of Riemann integration: As you wrote, we have $\int_0^1K(u)\,du =1/2.$ This implies

$$\lim_{a\to 0^+}\int_a^1 K(u)\,du = 1/2.$$

Thus for small $a>0,$ $\int_a^1 K(u)\,du > 1/4.$ From this it follows that

$$\int_0^1 u^2K(u)\,du \ge \int_a^1 u^2K(u)\,du \ge a^2\int_a^1 K(u)\,du > a^2/4$$

for small $a>0.$