I am trying to show that the following matrix is positive semi-definite (PSD):
$$H=\left[\begin{array}{cc} \int v(x)\,dx & \int xv(x)\,dx\\ \int xv(x)\,dx & \int x^2v(x)\,dx \end{array}\right]$$
where $v(x)$ is a differentiable function of unknown sign.
The definition of PSD requires that for all $w,z \in \mathbb{R}$, $$w^2 \int v(x)dx + 2wz\left(\int xv(x)\,dx\right)^2+z^2\int x^2v(x)\,dx \geq 0$$
Even if I assume $v(x) \geq 0$ (I have $x\geq0$), it doesn't look like Holder's inequality is useful to get to the desired result.
Alternatively, assuming that $\int v(x) dx \geq 0$, I tried proving that the determinant of $H$ is positive, but that did not lead anywhere either.
Are there any other avenues?
A way to verify that a matrix is PSD is the to test that the determinants of all upper-left sub-matrices are non negative.
This is true for the first sub-matrix if you assume $\int v(x) \ dx\ge0$.
For the second one you need to have $$\left(\int v(x) \ dx\right)\left(\int x^2v(x) \ dx\right)\ge \left(\int xv(x) \ dx\right)^2$$ which is Cauchy-Schwarz inequality for the inner product $\langle f,g \rangle = \int fg$ with $f(x) = \sqrt{v(x)}$ and $g(x)=x\sqrt{v(x)}$ providing those functions are well defined. Which is the case if $v$ is supposed non negative.