Asymptotic inverses of asymptotic functions

Question

Under what conditions are the inverses of asymptotic functions themselves asymptotic? (A simple example where this is not the case is the pair of functions $\log (x)$ and $\log (2x)$.)

Answer 1

This isn't meant to be comprehensive, but here is a fairly straightforward result that is also general enough to handle your question about $\pi(x)$. I'll state it in a way that might feel a bit backwards at first — but this allows us to give a concise definition by making use of the existing concept of uniform continuity.

Let $f: \mathbb R^+ \to \mathbb R^+$ be a strictly increasing function with $\lim_{x\to\infty} f(x) = \infty$, and define the conjugate function $g(x) := \log f(\exp(x))$ (which also strictly increases to infinity). If $g^{-1}$ is uniformly continuous on some interval $[a,\infty)$, then $y \sim f(x)$ implies $x \sim f^{-1}(y)$ .

If $f(x) = x^n$, then the conjugate is just the linear function $nx$, which is of course uniformly continuous. A more interesting case is $f(x) = x /\log x$: its conjugate is $x - \log x$, whose derivative is bounded away from $0$ on $[2,\infty)$, and thus it has uniformly continuous inverse. In fact it's not hard to see that this property is preserved by multiplication and division by slowly-growing functions like $\log x$ or $\log \log x$ or even $\exp(\sqrt{\log x})$, so it applies to most of the counting functions you'll encounter in Hardy and Wright.

This result might be considered "obvious" for a fixed function $f$, so it could easily be taken for granted. But the general principle is not so hard to prove, either: assume that the conditions hold and $y \sim f(x)$, so $y / f(x) \to 1$ as $x\to\infty$. Taking logs, we have $\log y - \log f(x) \to 0$, so letting $X = \log x$ and $Y = \log y$ we get $Y - g(X) \to 0$. Now applying uniform continuity of $g^{-1}$ gives $g^{-1}(Y) - X \to 0$, which unravels to $\log f^{-1}(y) - \log x \to 0$, in other words $x \sim f^{-1}(y)$.

Answer 2

In fact, it's a problem about asymptotic solution of transcendental equations.

For equations with real variables, we have the following conclusion.

$\mathbf{Theorem\ 1.}$ Let $f(\xi)$ be continuous and strictly increasing in an interval $a<\xi<\infty$, and $$f(\xi)\sim \xi, \ \xi \to \infty.$$ Denote by $\xi(u)$ the root of the equation $f(\xi)=u$ which lies in $(a,\infty)$ when $u>f(a)$.Then $$\xi(u)\sim u,\ u\to \infty.$$

It's fortunate to generalize this conclusion to complex variables, but there're several constraints.

$\mathbf{Theorem\ 2.}$ Suppose that $f(z)$ is an analytic function of complex $z$, which is holomorphic in a region containing a closed annular sector $\mathbf{S}$ with vertex at the origin and angle less that $2\pi$. $$\ $$ Let $\mathbf{S}_1$ and $\mathbf{S}_2$ be closed annular sectors with vertices at the origin, $\mathbf{S}_1$ being properly interior to $\mathbf{S}$, and $\mathbf{S}_2$ being proper interior to $\mathbf{S}_1$.Then $$z(u)\sim u, as\ u\to \infty\ in\ \mathbf{S}_2,$$ where $z(u)$ is a root of the equation $u=f(z)$.