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


Note: I believe that



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

  • $\begingroup$ Editing to give a more informative title would be useful. $\endgroup$ – 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 $L(tx)=t^8L(x)<t^4R(x)=R(tx)$.

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 \begin{equation} 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 \end{equation} for all real $a_0,a_1,a_2$.

So, the determinant of $M$ is $\ge0$, which is equivalent to \begin{equation} 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}. \end{equation} 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.

  • $\begingroup$ Yes, sorry! There should have been a square on the second term. Sorry about this, I have edited the question. Thank you $\endgroup$ – hopeless Nov 20 '18 at 13:28

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