prove that $\forall x \, \, \, h(a+x)=h(a-x) \Longleftrightarrow \mu=0.5$

Question

The question is, prove that

$$\forall x \, \, \, h(a+x)=h(a-x) \Longleftrightarrow \mu=0.5$$

where \begin{eqnarray} \label{eqpdf} h(x) &=&(x^2)\times \left\{ \begin{array}{cc} \int_{0.5}^{1} \frac{1}{(\cos^2(\pi u))^{\frac{k}{2}}} e^{-(\frac{x^2}{2\cos^2(\pi u)})} f(u) du & x<0 \\ \int_{0.5}^1 \frac{1}{(\cos^2(\pi u))^{\frac{k}{2}}} e^{-(\frac{x^2}{2\cos^2(\pi u)})} f(1-u) du & x\geq 0 \end{array} \right. \\ &=& (x^2)\times \int_{0.5}^{1} \frac{1}{(\cos^2(\pi u))^{\frac{k}{2}}} e^{-(\frac{x^2}{2\cos^2(\pi u)})} \left(f(u)1_{\{ x<0 \}} +f(1-u)1_{\{ x\geq 0 \}}\right) du, \end{eqnarray}

where $f(u)=(\frac{u}{1-u})^{2\mu -1}$ , $k>0$ and $\mu \in (0,1)$.

The function $h(x)$ is a function of $(x^2,1_{\{ x<0 \}} )$, it is obvious that for $\mu=0.5$ we have $$f(u) =f(1-u)$$ so $h(x)$ is only a function of $x^2$ and $h(x)=h(-x)$. So $$\mu=0.5 \Longrightarrow h(x)=h(-x)$$. But I want to sure the only condition that $h(x)$ can be symmetric, is for $\mu=0.5$, and $h(x)$ can not be symmetric in any other way.

For $\mu>0.5$, $f(u)$ is increasing and $f(1-u)$ is decreasing. For
$\mu<0.5$, $f(1-u)$ is increasing and $f(u)$ is decreasing. Following code compare $f(u)$ and $f(1-u)$ for various value of $\mu$.

 R code
 > mu=0.75 #An example of the case mu>0.5
 > u<-seq(.55,.99,len=10)
 > (u/(1-u))^(2*mu-1) #f(u)
  [1] 1.105542 1.221914 1.356140 1.515488 1.711763 1.965926 2.320125 2.877212
  [9] 3.997641 9.949874
 > ((1-u)/u)^(2*mu-1) #f(1-u)
  [1] 0.9045340 0.8183879 0.7373868 0.6598535 0.5841932 0.5086662 0.4310113
  [8] 0.3475586 0.2501475 0.1005038
 > 
 > mu=0.25  #An example of the case mu>0.5
 > u<-seq(.55,.99,len=10)
 > (u/(1-u))^(2*mu-1) #f(u)
  [1] 0.9045340 0.8183879 0.7373868 0.6598535 0.5841932 0.5086662 0.4310113
  [8] 0.3475586 0.2501475 0.1005038
 > ((1-u)/u)^(2*mu-1) #f(1-u)
  [1] 1.105542 1.221914 1.356140 1.515488 1.711763 1.965926 2.320125 2.877212
  [9] 3.997641 9.949874

Thanks in advance for any help you are able to provide.

Answer 1

Because the integrand is positive, the integral is always $> 0$. Hence $h(x) > 0$ for $x \ne 0$. But $h(0) = 0$ (technically, when $k \ge 1, h(0) = 0 \cdot \infty$ which is undefined, but always $\lim_{x \to 0} h(x) = 0$). Hence for $a \ne 0, h(a+a) > 0$, while $h(a -a) = 0$. Thus $h$ is not symmetric about $a$.

Note that for $u \in (1/2, 1), u > 1 - u$. Hence for $\alpha = 2\mu -1 > 0$, $$\left(\frac{u}{1-u}\right)^\alpha > \left(\frac{1-u}{u}\right)^\alpha$$ while for $\alpha < 0$, the opposite inequality holds. Because the other factors of the integrand are identical for $x < 0$ and $x > 0$, the same inequalities hold for the integrals. Therefore if $x > 0$ and $\mu < \frac 12, h(-x) < h(x)$, whereas for $\mu > \frac 12, h(-x) > h(x)$, so $h$ is not symmetric about $0$ either, except when $\mu = \frac 12$.