Let $w$ be a positive continuous function for which $\int_0^1 w(x)dx = \int_0^1 x^2w(x)dx = 1$. Prove that $\int_0^1 xw(x)dx < 1$.

Question

Let $w$ be a positive continuous function for which $$\int_0^1 w(x)dx = \int_0^1 x^2w(x)dx = 1.$$ Prove that $\int_0^1 xw(x)dx < 1$.

I was thinking of using the Reisz Representation Theorem for this but my intuitive understanding of it is lacking. I would love some guidance on this proof, thanks!

Answer 1

Let $\mu (E)=\int_E w(x)dx$. Then $\mu$ is a probability measure and $\int x^{2}d\mu(x)=1$. Hence $\int xd\mu (x)\leq \sqrt {\int x^{2} d\mu (x)}=1$. Equality can hold only when the integrand is a constant a.e. $[\mu]$ which is obviously false. Hence $\int xd\mu (x) <1$.

Alternatively $\int xw(x)dx=\int (x\sqrt {w(x)})(\sqrt {w(x)} dx \leq \sqrt {\int x^{2}w(x)dx} \sqrt {\int w(x)dx}=1$ and equality is ruled out as above.