Let $V$ be the real inner product space of continuous functions $[0,1] \to R$ with $\langle f, g \rangle = \int_0^1 fg \ dx$ and let $a \in [0,1]$. Show that there is no non-negative function $f \in V$ such that:
$$\int_0^1 f \ dx = 1, \qquad \int_0^1 xf \ dx = a, \qquad \int_0^1 x^2 f \ dx = a^2. $$
I'm stumped. Here are some thoughts:
The conditions on $f$ are inner products, so we know: $$\langle 1, f \rangle = 1, \qquad \langle x, f \rangle = a, \qquad \langle x^2, f \rangle = a^2.$$
At one point I though this statement might be false so in an attempt to produce a counter example I used these inner products to construct the projection $g$ of $f$ onto the subspace of $V$ spanned by $1,x,x^2$, but for all $a \in [0,1]$ that projection had some $x_0$ for which $g(x_0) <0$.
Using properties of inner products I can arrive at things like:
$$\langle x-a, f \rangle = 0, \qquad \langle x^2 - a^2 , f \rangle = 0, \qquad \langle x^2 - ax, f \rangle = 0,$$ so that $f$ is in the orthogonal complement of the span of $x-a, x^2-a^2, x^2 - ax$, but I cannot make that into anything useful.
I've also thought about considering $f$ as a probability density function for a random variable $X$ where $f$ takes value $0$ when $x \notin [0,1]$. Then $\langle 1,f \rangle = 1$ says that $f$ could possibly be a probability density function. Then $E[X] = a$ and $\text{Var}(X) = 0$ seem like they could lead to some kind of contradiction to $f$ being continuous, but I wasn't quite sure.
I've played around with Cauchy-Schwartz to look for inequalities that must be satisfied by $f$ and $a$, but didn't notice anything useful.
Recently I've been trying to use the Riez Representation Theorem, but I am only familiar with that theorem for finite dimensional vector spaces. I have read a bit about how it applies to Hilbert spaces but wasn't sure if that could be useful here.
What am I missing? Why must there exist some $x_0 \in [0,1]$ such that $f(x_0) < 0$ ?
Consider $$\int_0^1(x-a)^2f(x)\,dx.$$ This equals zero. As the integrand is continuous, and non-negative, it must be identically zero.