I want to prove, assuming I can, that an error function has unique global max.
For $x \geq 0$ consider the function $f(x) = \tanh(x)$, such a function is always positive, strictly increasing and convex. Let $h$ be a real number strictly positive, and let also $0 \leq y < + \infty$, in the interval $Y = [y,y+h]$ the function $f$ can be approximated by the function
$$ l(x;y) = f(y) + \frac{f(y+h)-f(y)}{h}(x-y) $$
The error function in $Y$ is given by
$$ e(x;y) = f(x) -l(x;y) = f(x) - f(y) - \frac{f(y+h)-f(y)}{h}(x-y) $$
Such a function has the property that $e(y;y) = e(y+h;y) = 0$, it belongs also to $C^{\infty}(Y)$ so by the Rolle theorem there's a valye $\xi_y \in (y,y+h)$ such that $e^{(1)}(\xi_y;y)=0$. Also because $e(x;y)$ is both convex and positive, the stationary point is unique, and the point $\xi_y$ is the unique global max. Explicit computation of such $\xi_y$ provides
$$ \xi_y = \tanh^{-1} \left(\sqrt{1 - \frac{f(y+h)-f(y)}{h}} \right). $$
Because for $y \in [0,+\infty)$ the function
$$ g(y) = -\frac{f(y+h)-f(y)}{h} $$
is strictly increasing e because $\xi_y$, as a function of $y$, can be defined as composition of strictly increasing function then $\xi_y$ is strictly increasing as well.
Substituing $\xi_y$ instead of $x$ I have
$$ e(\xi_y;y) = f(\xi_y) -l(\xi_y;y) = f(\xi_y) - f(y) - \frac{f(y+h)-f(y)}{h}(\xi_y-y) $$
I now define the following function for $0 \leq y < +\infty$
$$ u(y) = e(\xi_y;y) $$
I have done some plots of such function, like the one below (please see the function $u(x)$, and I do believe that there must be a simple way to state that the max of $u$ is unique, but I can't figure out how to prove it.
Could anyone give me an hint?
If it is possible I would like to avoid explicit computation of derivatives, since They would probably be quite complicated. Otherwise if there's no theorem I can exploit to prove the statement the derivative is fine.
Update : Some other properties, if it is of any help, I have 1. $$ 0 \leq u(y) \leq f(y+h) - f(y) $$ 2. $$ y < \tanh^{-1} \left(\sqrt{1 - \frac{f(y+h)-f(y)}{h}} \right) < y + h $$ 3. $$ f(y) < \sqrt{1 - \frac{f(y+h)-f(y)}{h}} < f(y + h) $$
Update 2:
I made a small progress
By studying the difference
$$ d(y) = u(y) - u(0)e^{-y} $$
I can easily prove that there's an $y_0$ such that $u^{(1)}(y_0) < 0$, if I define now
$$ \hat{y} = \inf \left\{y : u^{(1)}(y) < 0 \right\} $$
And assuming that there's a real $h > 0$ such that $\hat{y} = 0$ I can probably get a contradiction, the issue is how?
