# I want to disprove an equality involving a double integral

I want to show that the following equality does not hold:

$$\label{at3} \frac{\lambda^2-1}{2}x^2-\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty}K(y_1,y_2,x)\ln g(y_1,y_2)dy_2dy_1=\mbox{constant}\ \ ,\ \ \forall x\in\mathbb{R},$$ where $\lambda>1$, and $$\label{at2} g(y_1,y_2)=\int_{-1}^{1}e^{-\frac{(\lambda^2-1)u^2}{2}+\lambda y_1u}\cosh\left(y_2\sqrt{1-u^2}\right)dF(u),$$ in which $F(u)$ is an unknown cumulative distribution function (CDF) for $U$ with the support $[-1,1]$. It could be discrete or continuous or a mixture. Finally, \begin{align} K(y_1,y_2,x)=e^{-\frac{(\lambda^2-1)}{2}x^2}e^{\frac{-y_1^2-y_2^2+2\lambda y_1x}{2}}\cos \left(y_2\sqrt{x^2-1}\right). \end{align} In order to show that the first equality is not valid, I tried to see the scaling behaviour of the double integral term in $x$. I think that the double integral term is at most $O(x)$, so that it cannot cancel the effect of the first term $\frac{\lambda^2-1}{2}x^2$. However, I was not successful.

• this equality should hold for any $F(u)$? May 10, 2018 at 18:15
• how about $\lambda$? should the equality hold for any $\lambda>1$? --- you may want to give some more background on this question, as a motivation. May 10, 2018 at 20:19
• Yes, for any $\lambda>1$. Actually, $\lambda$ can be viewed as a condition number of a matrix. When it is 1, the equality that I want to disprove holds. I want to show that otherwise, it should not hold. I will send you and email for the further details. May 11, 2018 at 10:21
Quite possibly I am not understanding the question, and you will correct me. You say for $\lambda=1$ the equality holds for any $F(u)$ on $[-1,1]$. Let me try $dF(u)=\delta(u-1)du$, so $$g(y_1,y_2)=e^{y_1},\;\; K(y_1,y_2,x)=\exp\left(\frac{-y_1^2-y_2^2+2 y_1x}{2}\right)\cos \left(y_2\sqrt{x^2-1}\right),$$ and hence $$\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty}K(y_1,y_2,x)\ln g(y_1,y_2)dy_2dy_1=2\pi x\sqrt e\neq\text{constant}.$$
• Sorry if I could not explain it thoroughly. For $\lambda = 1$, there exists a distribution, or a $F(u)$, such that the equality holds. Of course, not for every $F(u)$, but there exists one. However, I want to say if $\lambda >1$, no such distribution exists. So, any distribution that you choose will not make the left hand side a constant. May 11, 2018 at 13:15