$ f(-f(x)) = -x , f(0) = 0$ Im looking for real-analytic functions such that
For all real $x$
A) $ f ' (x) > 0 $
B) $ f '' (x) > 0 $
C) $ f(0) = 0$
D) $ f( - f(x) ) = -x$
I prefer entire functions.
If I am not mistaken When $f$ is a solution , So is $f(f)$. That is nice for dynamics.
Also , I wonder about Its zero's in the complex plane , the regions where it is univalent and Its Riemann surface.
Notice that $D) $ is like holding a straight mirror at $ y = -x $ and asking $f$ to be its OWN reflection.
Some plots would be nice too.
 A: If we take the graph of $f$ and rotate it by $45^\circ$ around the origin, then we get a curve that's symmetric around the vertical axis. Its direction must always be within $45^\circ$ of horizontal (or $f$ would not be a function), and it cannot have a $45^\circ$ asymptote (or $f$ would not be defined on all of $\mathbb R$). It must also be convex, like $f$ is.
On the other hand, whenever we have a curve that satisfies this and passes through $(0,0)$, we can rotate it back to produce an $f$ that satisfies your A-D. The only question is whether $f$ can be real analytic. The analytic implicit function theorem says it can, if only the starting curve is analytic, but it may be more convincing with a concrete example:
The simplest curve with these properties is one branch of a hyperbola. We can easily write a degree-2 equation for such a hyperbola in the rotated coordinates
$$\tag{a} y = \frac12(\sqrt{1+x^2}-1) $$
that is
$$ 4y^2+4y+1 = 1+x^2 $$
This remains a second-degree polynomial in $x$ and $y$ of the original coordinate system, so we can solve for $y$ using the quadratic formula to get an explicit expression for a possible $f$ involving a square root.
If we scale the hyperbola by $\sqrt2/2$ before rotating it (which gives slightly nicer numbers) we get this concrete solution for $f$:
$$ f(x) = \frac{5x-2+\sqrt{(4x-1)^2+3}}3 $$
This produces a real analytic solution, but not an entire one. It's not clear to me that there are entire solutions that extend convex real functions.

Showing that the above $f(x)$ is a solution is left as an exercise for the reader, as is keeping everything straight while parameterizing the solution for varying top angles of the hyperbola and/or scaling factors.
In any case, such parameterization would not produce a complete solution to the problem. Many other curves than hyperbolas could be used instead of (a), such as for example
$$ \tag b y = \frac12 \log(\cosh x) $$
(which would yield a nightmarish formula after rotating it by $45^\circ$ -- but that's the nice case; asymptotic slopes other than $\frac12$ would not yield a closed formula at all).
