Is this enough to show this is the delta distribution?

Question

Let $f$ be a probability distribution with $$\int xf(x) = 0, \quad \int x^2f(x) = 0$$ Clearly, Dirac's delta distribution is a candidate for $f(x)$. Is the previous constrain enough to prove that it is the only possibility?

Answer 1

I'll use a measure theoretic approach (as you are clearly not interested in densities): Let $\mu$ be a probability measure on $\mathbb{R}$ and assume that $$\int x^2 d\mu(x)=0$$ This implies that the integrand $x^2=0$ $\mu$-almost everywhere. But because $x^2\neq 0$ on $\mathbb{R}\setminus \{0\}$ we must have $\mu(\mathbb{R}\setminus \{0\})=0$. Thus, all mass has to be concentrated at $0$ and you obtain the dirac measure.

Or, more compactly: your second assumption is enough to conclude the dirac distribution.