If $f(x) = \sum \limits_{n=0}^{\infty} \frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-\frac{x}{2}$ is bounded

Question

For every $x>0$, let $$f(x) = \sum \limits_{n=0}^{\infty} \dfrac{x^n}{2^{n(n-1)/2} n!}.$$ Let $f^{-1}$ be the functional inverse of $f$.

How to show there exists a positive real constant $C$ such that, for all $x$, $$\left(f^{-1}\left(f(x)-f(x-1)\right)-\frac{x}{2}\right)^2 < C $$

Edit : I believe this is true because $f'(x) = f(x/2)$.

Answer 1

Assume that $x\gt1$ then $f'(t)=f\left(\frac{t}2\right)$ for every $t\gt0$ and $f$ is increasing on the interval $\left(\frac{x-1}2,\frac{x}2\right)$ hence $$f(x)-f(x-1)=\int_{x-1}^xf'(t)\mathrm dt=\int_{x-1}^xf\left(\tfrac{t}2\right)\mathrm dt,$$ yields $$f\left(\tfrac{x-1}2\right)\leqslant f(x)-f(x-1)\leqslant f\left(\tfrac{x}2\right),$$ that is, $$\tfrac{x-1}2\leqslant f^{-1}\left(f(x)-f(x-1)\right)\leqslant\tfrac{x}2,$$ in particular, $$\left(f^{-1}\left(f(x)-f(x-1)\right)-\tfrac{x}2\right)^2\leqslant\tfrac14.$$