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It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big)\hspace{-1.25pt} \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $p\ge 3$. It can be shown for $p=3$ and $p=4$, and I believe it is true for $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 2-times monotone, where $k \ge 3$, but this does not apply to the $f(x)$ defined by $(2)$. The inequality does seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.

It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big)\hspace{-1.25pt} \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $f$ defined by $(2)$ when $p\ge 3$. The inequality can be relatively easiliy proved for for $p=3$ and $p=4$, and it seems to be true for all $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 2-times monotone, where $k \ge 3$, but this does not apply to the $f(x)$ defined by $(2)$. The inequality does seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.

It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big) \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $p\ge 3$. It can be shown for $p=3$ and $p=4$, and I believe it is true for $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 2-times monotone, where $k \ge 3$, but this does not apply to the $f(x)$ defined by $(2)$. The inequality does seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.

It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big)\hspace{-1.25pt} \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $p\ge 3$. It can be shown for $p=3$ and $p=4$, and I believe it is true for $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 2-times monotone, where $k \ge 3$, but this does not apply to the $f(x)$ defined by $(2)$. The inequality does seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.

It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big) \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $p\ge 3$. It can be shown for $p=3$ and $p=4$, and I believe it is true for $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 3-times monotone, where $k \ge 2$, but this does not apply to the $f(x)$ in question. The inequality still seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.

It was shown in a previous answer that for: $f(x)=|x|^p$, $\;x\in \mathbb{R}$, $\;p>2$, defining the density:

$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big( \hspace{-1pt}\cos(\phi+\theta)\big) \Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)},\quad 0 \,\le\, \phi,\,\theta \,\le\, \mbox{$\large\frac{\pi}{2}$}$$

we have:

$$ \frac{\partial}{\partial \phi} \frac{\partial}{\partial \theta}\hspace{2pt} \,\log p(\phi;\theta) \,\ge\, 0,\quad 0\, \le\, \phi,\,\theta\,\le\, \mbox{$\large\frac{\pi}{2}$}$$

(It is the case that $\int_0^{\pi/2} p(\phi;\theta)\,d\phi$ is independent of $\theta$ when $f(x)$ is symmetric and increasing on $[0,1]$ This can be shown e.g. using a scale mixture representation of $f$ with a $\text{Heaviside}(x^2-1)$ kernel. The integral is $\int_0^{1}\! \sqrt{1-x}\hspace{1pt}f'(\!\sqrt{x})\, dx$. The derivative inequality doesn't depend on this fact.)

To apply when an independent symmetric random variable is added in the relevant context, it is necessary that the result also hold for:

$$f(x;t) = |x+t|^p + |x-t|^p,\quad t\in\mathbb{R} $$

To prove the inequality, I believe it is sufficient to show that, for $y>x\ge 0$ :

$$\Big(xf'(x) + yf'(y)-f''(x)\big(y^2-x^2\big)\hspace{-01.4pt}\Big)\big(f(y)-f(x)\big) - f'(x)^2\big(y^2-x^2\big) > 0 \tag{1}$$

Since this inequality is scale invariant, it suffices to show that it holds for:

$$f(x) = |x+1|^p + |x-1|^p \tag{2}$$

I conjecture the inequality $(1)$ holds for $p\ge 3$. It can be shown for $p=3$ and $p=4$, and I believe it is true for $p\ge 3$. It apparently does not hold for $p<3$. The derivative expression becomes very complicated in general.

Thanks to an answer to a related question, it can be shown that $(1)$ holds with the reverse inequality when $f''(\!\sqrt{x})$ is 2-times monotone, i.e. $f''(\!\sqrt{x})$ is non-negative, decreasing, and convex on $[0,1]$. It seems likely that a similar result should hold in the greater than case $(1)$, but obvious analogous conditions don't seem to work. I believe $(1)$ holds for functions such that $f^{(k)}(\!\sqrt{x})$ is 2-times monotone, where $k \ge 3$, but this does not apply to the $f(x)$ defined by $(2)$. The inequality does seems to hold though.

This result is important to prove uniqueness of stable optima in the unmixing and deconvolution of linear mixtures of independent random variables with strongly sub- and super-gaussian densities, using variation diminishing property of MLR densities.