# A moment inequality

Let $$\chi(s)=\int_{0}^{1}x(t)^{s}f(t)dt$$, where $$x(t)$$ and $$f(t)$$ are real valued continuous functions for $$t\in[0,1]$$, and $$f(t)\geq0$$.

Is it possible to show that

$$\left(\chi(0)\chi(2)-\chi(1)^{2}\right)\left(\chi(4)\chi(2)-\chi(3)^{2}\right)-\left(\chi(3)\chi(1)-\chi(2)^{2}\right)^{2}\geq0$$

Note: I believe that

$$\chi(0)\chi(2)-\chi(1)^{2}\geq0$$

$$\chi(4)\chi(2)-\chi(3)^{2}\geq0$$

follows from the Cauchy-Schwarz inequality. (correct me if I am wrong about this).

• Editing to give a more informative title would be useful. – YCor Nov 20 '18 at 15:13

This inequality is false in general, by homogeneity considerations. Indeed, it can be rewritten as $$L(x)\ge R(x)$$, where $$L(x):=\left(\chi(0)\chi(2)-\chi(1)^{2}\right)\left(\chi(4)\chi(2)-\chi(3)^{2}\right)$$ and $$R(x):=\chi(3)\chi(1)-\chi(2)^{2}$$. Take any non-constant positive $$x$$, so that, by the Cauchy--Schwarz inequality, $$R(x)>0$$. Then for small enough $$t>0$$ we have $$L(tx)=t^8L(x).
Added in response to the OP fixing the typo: The inequality is now true. Indeed, let $$m_j:=\chi(j)$$. Without loss of generality, $$m_0 m_2-m_1^2>0$$, by the Cauchy--Schwarz inequality. The matrix $$M:=[m_{i+j}]_{i,j=0}^2$$ is positive semidefinite. This follows because $$$$0\le\int_0^1\Big(\sum_{i=0}^2 a_i\, x(t)^i\Big)^2f(t)\,dt=\sum_{i,j=0}^2 m_{i+j}a_i a_j$$$$ for all real $$a_0,a_1,a_2$$.
So, the determinant of $$M$$ is $$\ge0$$, which is equivalent to $$$$m_4\ge m_4^*:=\frac{m_2^3-2 m_1 m_3 m_2+m_0 m_3^2}{m_0 m_2-m_1^2}.$$$$ On the other hand, the difference between the left and right sides of your inequality (with $$m_j$$ in place of $$\chi(j)$$) is increasing in $$m_4$$, and the value of this difference is $$0$$ at $$m_4=m_4^*$$. Thus, your inequality follows.