Can any analytic function be written as the difference of two monotonically increasing analytic functions?

Question

This question is not a duplicate of this one or this one, since the solutions shown there contain jumps.

Let $f(x)$ be an analytic function on $\mathbb{R}$. We can take its Taylor series and group the terms with a positive coefficient in one function $f_+(x)$ and terms with a negative coefficient in a function $f_-(x)$. Then $f(x) = f_+(x) - f_-(x)$, both components are monotonically rising on $\mathbb{R}_+$, and both are analytic. However, at $0$ the components would have to swap to maintain monotonicity, creating a discontinuity. Can this be avoided?

For an easier question, consider specifically the case of $f(x) = x^2$. I can't even find a simple decomposition for this basic case. I'm not convinced it exists -- this is not homework, I have no idea which way the truth lies.

Answer 1

Define two new analytic functions $$ g(x) = \int_0^x (1+f'(t)^2)\,dt \quad\text{and}\quad h(x) = g(x) + f(x). $$ Then certainly $f(x) = h(x)-g(x)$; and $$ g'(x) = 1+f'(x)^2 > 0 \quad\text{and}\quad h'(x) = 1+f'(x)^2 + f'(x) = \tfrac34 + (f'(x)+\tfrac12)^2 > 0, $$ so both functions are increasing.

Answer 2

One way to solve this is to construct an increasing analytic function $g$ so that $f+g$ is also increasing. For that to hold, we need $g'(x)\ge\max(0, -f'(x))$.

Let $$ g(x) = \int_0^x h\left(f'(t)\right)\,dt $$ where $h$ is an analytical function with $h(u)>\max(0,-u)$: eg $h(u)=\sqrt{1+u^2}$. This makes $g$ analytical and increasing, and $f'(x)+g'(x)>0$ so that $f+g$ is too.