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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.