Prove $\frac{H(x^2)}{H(x)}$ increases.

Question

For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.

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Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.

The numerator of the derivative of the sought fraction is

\begin{align} &H(x)^2\frac{dh(x)}{dx} \\ =&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\ =& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\ =& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1 \end{align} All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get \begin{align} g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\ &=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\ &= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt \end{align} for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$. $$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$ Let $p(x):=2x^3+3x^2-2x-7$. $p(x)\le p(1)=-4, \forall x\in[0,1]$. This is true since $p(x)$ is convex as $p''(x)=12(x+\frac12)>0$ on that interval and $p(0)=-7<-4=p(1)$. We can take $a=\frac13$ since we can show, with a bit of work, $g''(\frac13)>0$.

(to be continued)

Answer 1

I thought about using the general properties $H(0)=H(1)=1$, $H(x)=H(1-x)$, $H''(x)<0$ and $H'(0)=+\infty$.

However, if you plug in $f(x)=\sin(\pi\sqrt{x})$ and $g(x)=f(x)+f(1-x)$ on a graphing calculator,

$\dfrac{g(x^2)}{g(x)}$ seems to increase from $x=0$ to around $x=0.997$, and then it decreases ever so slightly.

Therefore, you probably need to use the precise expression of $H(x)$, and I don't expect a proof much more elegant than the one you sketched.

Answer 2

Let $$u = \ln x, \quad v = \ln(1-x), \quad w = \ln(1 + x).$$ We have \begin{align*} &[H(x)]^2h'(x)\\[6pt] =\,& [(1-x)^2v - (1+x^2)u]w + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2\\[6pt] \ge\,& [(1-x)^2v - (1+x^2)u]\cdot (x - x^2/2)\\[6pt] &\qquad + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2 \tag{1}\\[6pt] =\,& 2u^2x^2 - [x(1-x/2)(1+x^2) + (1-3x)(1-x)v]u\\[6pt] &\qquad + v(1-x)^2(v + x - x^2/2)\\[6pt] \ge\,& 0. \tag{2} \end{align*} Explanations:
(1): $\ln(1+x) \ge x - x^2/2$ for all $x \ge 0$;
Using $(1-x)\ln(1-x) > - 1$, we have \begin{align*} (1-x)^2v - (1+x^2)u &= (1-x)^2\ln(1-x) - (1+x^2)\ln x\\ &\ge -(1-x) - \ln x\\ &\ge 0. \end{align*}

(2): $\ln(1-x) + x - x^2/2\le 0$ for all $x\in [0, 1)$;
If $1/3 \le x < 1$, clearly $$x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x) \ge 0$$ and if $0 \le x < 1/3$, using $(1-x)\ln(1-x) \ge -x + x^2/2$ for all $0 \le x < 1$, we have \begin{align*} &x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x)\\ \ge\,& x(1-x/2)(1+x^2) + (1-3x)(-x + x^2/2)\\ =\,& x^2(3+x)(1-x/2)\\ \ge\,& 0. \end{align*}

We are done.

Answer 3

Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.

The numerator of the derivative of the sought fraction is

$$\begin{align} &H(x)^2\frac{dh(x)}{dx} \\ =&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\ =& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\ =& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1 \end{align}$$ All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get $$\begin{align} g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\ &=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\ &= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt \end{align}$$ for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$. $$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$ Let $p(x):=2x^3+3x^2-2x-7$. $p(x)\le p(1)=-4, \forall x\in[0,1]$. This is true since $p(x)$ is convex as $p''(x)=12(x+\frac12)>0$ on that interval and $p(0)=-7<-4=p(1)$. We can take $a=\frac13$ since we can show, with a bit of work, $g''(\frac13)>0$.

(to be continued)

Answer 4

Partial Hints :

We broke the fraction into two part :

Show that :

$$f(x)=x^{-1}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)$$

Is increasing on $[0,2/5]$

And :

$$g(x)=x^{-1}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$

Is decreasing on $(0,1)$

Obviously a increasing function divided by a decreasing function both positive is increasing .

We can make the same thing on $x\in[2/5,3/5]$ with :

$$\frac{1}{1+x}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right),\frac{1}{1+x}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$

Next we use symmetry to get :

$$r\left(x\right)=\frac{\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)}{x},\frac{r\left(1-x^{2}\right)}{x},\frac{t\left(1-x^{2}\right)}{x},t\left(x\right)=\frac{-x\ln\left(x^{2}\right)-\left(1-x\right)\ln\left(\left(1-x\right)^{2}\right)}{x}$$

Currently the function $\frac{r\left(1-x^{2}\right)}{x}$ is decreasing on $[0.5,1)$ and $\frac{t\left(1-x^{2}\right)}{x}$ is increasing both positive so the fraction is decreasing remains to substitute $y=1-x^2$.

For a end see RiverLi's answer I stop here .

Answer 5

Not quite what you asked, but a simple proof of a very related inequality, by Boppana: https://arxiv.org/abs/2301.09664

Theorem. The binary entropy function $h$ satisfies $$ \frac{h(x^2)}{h(x)} \geq \phi\cdot x, \quad \quad \forall p\in[0,1] $$ where $\phi=\frac{1+\sqrt{5}}{2}$ denotes the golden ratio.