Differentiable entropy inverse branches and shifted contracts

Every point of $(0,\mathrm{\infty })$ is a limit point of this domain, and every $a$ in it is nonzero. Apply F8 to the assertion $P(n)$ that the derivative formula below holds at order $k=n+1$. Its base case $n=0$, namely $(\log )\mathrm{\prime }(a)=a^{-1}$, is F2. Induction using F5 and the scalar-multiple rule in F6 then gives $(\log {)}^{(k)}(a)=(-1{)}^{k-1}(k-1)!a^{-k}$ for every integer $k\ge 1$: differentiating the formula at order $k$ multiplies its coefficient by $-k$ and replaces $a^{-k}$ by $a^{-k-1}$, yielding the formula at order $k+1$. Each displayed derivative is itself differentiable by F5 and F6, hence continuous by F7; the logarithm itself is continuous by F2 and F7. Since $I\mathrm{\prime }=\log$ by F1, $I$ is continuous by F7 and every positive-order derivative of $I$ exists and is continuous. In dimension one the word derivatives in F4 are precisely ordinary iterated derivatives, so $I$ is smooth in that definition.

Let $t_1<t_2$ lie in the range of the same branch. On the lower branch, if $a_{-}(t_1)\le a_{-}(t_2)$, then strict decrease of $I$ in F1 would give $t_1=I(a_{-}(t_1))\ge I(a_{-}(t_2))=t_2$, a contradiction. Hence $a_{-}(t_1)>a_{-}(t_2)$. On the upper branch, if $a_{+}(t_1)\ge a_{+}(t_2)$, strict increase of $I$ would similarly give $t_1\ge t_2$. Thus $a_{+}(t_1)<a_{+}(t_2)$, proving both asserted monotonicities directly from branch order.

On either open branch, $I\mathrm{\prime }(a)=\log a\mathrm{\ne }0$ by F1, so its one-dimensional derivative is invertible, with inverse multiplication by $1\mathrm{/}\log a$. Step 1.1 and F3 therefore supply a smooth inverse locally at every point of the branch range. Branch uniqueness in F1 identifies every such local inverse with $a_{\pm }$ on its neighbourhood. Thus $a_{\pm }$ is smooth on each stated open range, with $a_{\pm }\mathrm{\prime }(t)=1\mathrm{/}\log a_{\pm }(t)$. Since F9 gives $\log a<\log 1=0$ for $0<a<1$ and $\log a>0$ for $a>1$, this derivative is negative on the lower branch and positive on the upper branch.

For each real $s$, substitution of $a=1+c+s$ is a bijection between admissible real $c$ and positive $a$. F1 gives $I(1)=0$ and positive values on both open branches, so $I(a)\ge 0$, with equality exactly at $a=1$. Hence there is no solution for $t<0$, and for $t=0$ the unique solution is $c=-s$, which is admissible because $1+c+s=1$. For $t>0$, a solution cannot have $a=1$, so F1 gives exactly $c=a_{\pm }(t)-1-s$ for each defined inverse value: two solutions for $0<t<1$, and just the upper-branch solution for $t\ge 1$. In particular, shifted-root comparisons require both the chosen branch and the right-hand side; monotonicity of an unspecified root is insufficient.