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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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $p>0$, $p$ is even and unimodal, and $\int p=1$, whereas $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{49}{40}>1=H(1)=H(c).\quad\Box$$

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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $p>0$, $p$ is even and unimodal, and $\int p=1$, whereas $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{31}{20}>1=H(1)=H(c).\quad\Box$$

The idea here is to make $p$ "just barely" unimodal.

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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $p>0$ and $\int p=1$, whereas $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{49}{40}>1=H(1)=H(c).\quad\Box$$

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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $p>0$, $p$ is even and unimodal, and $\int p=1$, whereas $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{49}{40}>1=H(1)=H(c).\quad\Box$$

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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{49}{40}>1=H(1)=H(c).\quad\Box$$

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 is not true. E.g., suppose that $c=1$ and $$p(x)=p_t(x):=(1-t)q(x)+t\frac12\,e^{-|x|}$$ for real $x$, where $t\downarrow0$ and $$q(x):=\frac{1(|x|<1/2)}2+\frac{1(1/2\le|x|<5/2)}8.$$ Then $p>0$ and $\int p=1$, whereas $$H(2)\to\int_{-5/2}^{5/2}\frac{2q(x-2)}{q(x-1)+q(x+1)}\,q(x)\,dx \\ =\frac{49}{40}>1=H(1)=H(c).\quad\Box$$