Showing that an implicitly defined function is analytic on $(0,\infty)$

Question

I would like to know some suggestions as to how to approach this problem:

Let $h\geq 0$, $\beta \in \mathbb{R}$, and denote by $f(h)$ the largest solution of $$\tanh (2\beta x + h) = x. $$ Prove that $f(h)$ is analytic on $(0,\infty)$.

Thanks!

Answer 1

Note that $x\in (-1,1)$ here, since $|\tanh|<1$ always. Consider the function $$g(x) = \tanh^{-1} x -2\beta x,\quad -1<x<1$$ which is real analytic. (Indeed, $\tanh x$ is the quotient of two functions given by power series, and the inverse of a real analytic function is real analytic). Note that $g$ is strictly convex on $(0,1)$, since $\tanh$ is strictly concave on $(0,\infty)$. Therefore, either $g$ strictly increases from $g(0)=0$ onward, or it initially dips into negative territory and then strictly increases.

Let $a$ be the largest value of $x$ for which $g(x)=0$. Restrict attention to the interval $(a,1)$, on which $g$ is strictly increasing by the above. This interval is mapped bijectively onto $(0,\infty)$. It remains to observe that $g^{-1}$ is your function; and as mentioned above, taking the inverse of a strictly monotone function preserves real-analyticity.