Proof. Think of $A,B$ as characteristic functions. Write $A_{<k}$ for $(A_1,\dots ,A_{k-1})$ etc. By the Chain rule it is enough to prove for every $k$ that

$$H[{(A\cup B)}_k|{(A\cup B)}_{<k}]>c(1-p)(H[A_k|A_{<k}]+H[B_k|B_{<k}]).$$

By Submodularity,

$$H[{(A\cup B)}_k|{(A\cup B)}_{<k}]\ge H[{(A\cup B)}_k|A_{<k},B_{<k}].$$

For each $u,v\in {\{0,1\}}^{k-1}$ write $p(u)=\mathbb{P}(A_k=0|A_{<k}=u)$, $q(v)=\mathbb{P}(B_k=0|B_{<k}=v)$.

Then

$$H[{(A\cup B)}_k|A_{<k}=u,B_{<k}=v]=h(p(u)q(v))$$

which by hypothesis is at least

$$c(p(u)h(q(v))+q(v)h(p(u))).$$

So

$$H[{(A\cup B)}_k|{(A\cup B)}_{<k}]\ge c\sum _{u,v}\mathbb{P}(A_{<k}=u)\mathbb{P}(B_{<k}=v)\left(p(u)h(q(v))+q(v)h(p(u))\right).$$

But

$$\sum _u\mathbb{P}(A_{<k}=u)\mathbb{P}(A_k=0|A_{<k}=u)=\mathbb{P}(A_k=0)$$

and

$$\sum _v\mathbb{P}(B_{<k}=v)h(q(v))=\sum _v\mathbb{P}(B_{<k}=v)H[B_k|B_{<k}=v]=H[B_k|B_{<k}].$$

Similarly for the other term, so the RHS equals

$$c(\mathbb{P}(A_k=0)H[B_k|B_{<k}]+\mathbb{P}(B_k=0)H[A_k|A_{<k}]),$$

which by hypothesis is greater than

$$c(1-p)(H[A_k|A_{<k}]+H[B_k|B_{<k}])$$

as required. □

Proof. Write $\psi$ for ${\varphi }^{-1}=\frac{\sqrt{5}-1}{2}$. Then ${\psi }^2=1-\psi$, so $h({\psi }^2)=h(1-\psi )=h(\psi )$ and $\varphi \psi =1$, so $h({\psi }^2)=\varphi \psi h(\psi )$. Equality also when $x=0,1$.

Toolkit:

Let $f(x)=h(x^2)-\varphi xh(x)$. Then

$$\begin{array}{llll}\hfill f^{\prime }(x)& =2x{h}^{\prime }(x^2)-\varphi h(x)-\varphi x{h}^{\prime }(x)& \hfill & \\ \hfill f^{″}(x)& =2h^{\prime }(x^2)+4x^2{h}^{″}(x^2)-2\varphi h^{\prime }(x)-\varphi x{h}^{″}(x)& \hfill & \\ \hfill f^{‴}(x)& =4x{h}^{″}(x^2)+8x{h}^{″}(x^2)+8x^3{h}^{‴}(x^2)-3\varphi h^{″}(x)-\varphi x{h}^{‴}(x)& \hfill & \\ \hfill & =12x{h}^{″}(x^2)+8x^3{h}^{‴}(x^2)-3\varphi h^{″}(x)-\varphi x{h}^{‴}(x)& \hfill & \\ \hfill & & \hfill \end{array}$$

So

$$\begin{array}{llll}\hfill \ln 2f^{<mi>‴</mi>}(x)& =\frac{-12x}{x^2(1-x^2)}+\frac{8x^3(1-2x^2)}{x^4{(1-x^2)}^2}+\frac{3\varphi }{x(1-x)}-\frac{\varphi x(1-2x)}{x^2{(1-x)}^2}& \hfill & \\ \hfill & =\frac{-12}{x(1-x^2)}+\frac{8(1-2x^2)}{x{(1-x^2)}^2}+\frac{3\varphi }{x(1-x)}-\frac{\varphi (1-2x)}{x{(1-x)}^2}& \hfill & \\ \hfill & =\frac{-12(1-x^2)+8(1-2x^2)+3\varphi (1-x){(1+x)}^2-\varphi (1-2x){(1+x)}^2}{x{(1-x)}^2{(1+x)}^2}& \hfill & \end{array}$$

This is zero if and only if

$$-12+12x^2+8-16x^2+3\varphi (1+x-x^2-x^3)-\varphi (1-3x^2-2x^3)=0$$

which simplifies to

$$-\varphi x^3-4x^2+3\varphi x-4+2\varphi =0.$$

Since this is a cubic with negative leading coefficient and constant term, it has a negative root, so it has at most two roots in $(0,1)$. It follows (using Rolle’s theorem) that $f$ has at most five roots in $[0,1]$, up to multiplicity.

But

So $f^{\prime }(0)=0$, so $f$ has a double root at $0$.

We can also calculate (using ${\psi }^2+\psi =1$):

So there’s a double root at $\psi$.

Also, note $f(1)=0$.

So $f$ is either non-negative on all of $[0,1]$ or non-positive on all of $[0,1]$.

If $x$ is small,

so there exists $x$ such that $f(x)>0$. □

Proof. We can extend $f$ continuously to the boundary by setting $f(x,y)=1$ whenever $x$ or $y$ is $0$ or $1$. To see this, note first that it’s valid if neither $x$ nor $y$ is $0$.

If either $x$ or $y$ is small, then

So it tends to $1$ again.

One can check that $f\left(\frac{1}{2},\frac{1}{2}\right)<1$, so $f$ is minimised somewhere in ${(0,1)}^2$.

Let $(x^{\ast },y^{\ast })$ be a minimum with $f(x^{\ast },y^{\ast })=\alpha$.

Let $g(x)=\frac{h(x)}{x}$ and note that

$$f(x,y)=\frac{g(xy)}{g(x)+g(y)}.$$

Also,

$$g(xy)-\alpha (g(x)+g(y))\ge 0$$

with equality at $(x^{\ast },y^{\ast })$. So the partial derivatives of LHS are both $0$ at $(x^{\ast },y^{\ast })$.

$$\begin{array}{llll}\hfill y^{\ast }{g}^{\prime }(x^{\ast }{y}^{\ast })-\alpha g^{\prime }(x^{\ast })& =0& \hfill & \\ \hfill x^{\ast }{g}^{\prime }(x^{\ast }{y}^{\ast })-\alpha g^{\prime }(y^{\ast })& =0& \hfill & \end{array}$$

So $x^{\ast }{g}^{\prime }(x^{\ast })=y^{\ast }{g}^{\prime }(y^{\ast })$. So it’s enough to prove that $x{g}^{\prime }(x)$ is an injection. $g^{\prime }(x)=\frac{h^{\prime }(x)}{x}-\frac{h(x)}{x^2}$, so

Differentiating gives

$$\frac{-1}{x(1-x)}-\frac{\log (x-1)}{x^2}=\frac{-x-(1-x)\log (1-x)}{x^2(1-x)}.$$

The numerator differentiates to $-1+1+\log (1-x)$, which is negative everywhere. Also, it equals $0$ at $0$. So it has a constant sign. □