Let $p$ be a Lebesgue density function with infinite support (i.e. $p(x)>0 \forall x\in \mathbb{R}$ and $\int p(x) dx = 1$). Moreover, assume that $p$ is even (i.e. $p(x) = p(-x)$) and that $p(x) \geq p(x')$ for $0 \leq x \leq x'$ so $p$ is monotonically non-increasing after zero and monotonically non-decreasing before zero. Examples of density that satisfy these properties are the Gaussian, Cauchy and Laplace densities. Now define $$ H(\mu) := \int^{\infty}_{-\infty} \frac{p(x-\mu)}{0.5p(x-c) + 0.5p(x+c)} p(x) d x, $$ for some $c > 0$. One can show that $H(-\mu) = H(\mu)$ (use substitution with $y=-x$). Using this, one can easily show that $2H(c) = H(c) + H(-c) = 2$ by writing the sum as one integral which simplifies a lot. This implies that $H(c) =1.$

I am now trying to show that $H(\mu) \leq 1$ for $|\mu| \geq c$. As $H$ is an even function, it is sufficient to show that $H(\mu) \leq 1$ for $\mu \geq c$.

This can be shown for Gaussians using an additional monotonicity assumption (see this post). Such assumption does not hold for the Cauchy densities. However, computing the integral using Monte Carlo approximations still seems to show that $H(\mu) \leq 1$ for $|\mu| \geq c$ for Cauchy.

I am not a 100% sure whether my current assumptions are sufficient. I at least haven't found a counterexample yet. If anyone knows one, this would be also highly valuable.

Any other insights would also be much appreciated!

Let $p$ be a Lebesgue density function with infinite support (i.e. $p(x)>0 \forall x\in \mathbb{R}$ and $\int p(x) dx = 1$). Moreover, assume that $p$ is even (i.e. $p(x) = p(-x)$) and unimodal: $p(x) \geq p(x')$ for $0 \leq x \leq x'$ so $p$ is monotonically non-increasing after zero and monotonically non-decreasing before zero. Examples of density that satisfy these properties are the Gaussian, Cauchy and Laplace densities. Now define $$ H(\mu) := \int^{\infty}_{-\infty} \frac{p(x-\mu)}{0.5p(x-c) + 0.5p(x+c)} p(x) d x, $$ for some $c > 0$. One can show that $H(-\mu) = H(\mu)$ (use substitution with $y=-x$). Using this, one can easily show that $2H(c) = H(c) + H(-c) = 2$ by writing the sum as one integral which simplifies a lot. This implies that $H(c) =1.$

I am now trying to show that $H(\mu) \leq 1$ for $|\mu| \geq c$. As $H$ is an even function, it is sufficient to show that $H(\mu) \leq 1$ for $\mu \geq c$.

This can be shown for Gaussians using an additional monotonicity assumption (see this post). Such assumption does not hold for the Cauchy densities. However, computing the integral using Monte Carlo approximations still seems to show that $H(\mu) \leq 1$ for $|\mu| \geq c$ for Cauchy.

I am not a 100% sure whether my current assumptions are sufficient. I at least haven't found a counterexample yet. If anyone knows one, this would be also highly valuable.

Any other insights would also be much appreciated!