Power towers: to infinity and all the way back

Solution 1:

I fix some $a > e$ and use subscripts for $n$.

Define $f_n(x) = \log^n(\operatorname{pow}_a^n(x))$, so that $f_0(x) = x$, and $f_{n+1}(x) = \log(f_n(a^x))$.

Since $\log$ and $pow_a$ are increasing, it's easy to check that the operator $T : f \mapsto \log \circ f \circ pow_a$ is "increasing" in several ways : it takes increasing functions to increasing functions, and if $f \ge g$ on $\Bbb R^+$, then $T(f) \ge T(g)$ on $\Bbb R$. Hence, since $f_0(x) = x$ is increasing, and $f_1(x) = \log(a^x) \ge \log(e^x) = f_0(x)$ for $x \in \Bbb R^+$, all the $f_n$ are increasing functions of $x$, and $(f_n)$ is an increasing sequence of functions : $f_{n+1}(x) \ge f_n(x)$ except for $x < 0$ and $n = 0$.

Hence, $\forall x \in \Bbb R, f_n(x) \ge f_1(x) = (\log a) x$.

Next, by looking at what $T$ does on affine functions ($L_{A,B}(x) = Ax+B$), we can find an affine upper bound for all the $f_n$ for large enough $x$ : $T(L_{A,B})(x) = \log(Aa^x+B)$, and $\log(Aa^x) = \log A + x \log a = L_{\log a,\log A}(x)$. Finally, a simple comparisons shows that $L_{\log a, \log A}(x) \le T(L_{A,B})(x) \le L_{\log a, \log A}(x) + B/(Aa^x)$

Let $A = \log a > 1 $, and let $B = (\log A)/(1-1/A)$.
Then for $x \ge 0$, $L_{\log a, \log A}(x) + B/(Aa^x) \ge L_{\log a, \log A}(x) + B/A = L_{A, B}(x)$.

Now we can show by induction that from $f_1 = L_{A,0}$, we get $L_{A,\log A} \le f_n \le L_{A,B}$ forall $x \ge 0$ and $n \ge 2$.
As for $x \le 0$, we have$L_{A,\log A} \le f_n$ for $n \ge 2$, and $f_n(x) \le L_{A,\log A} + B/(Aa^x)$ for $n \ge 3$.

This gives a uniform bound on all the $f_n$, which proves that $f(x)$ exists forall $x \in \Bbb R$. We get that $f$ is increasing, $f(x) \ge Ax$, $\lim_{x \to +\infty} f(x)/x = A$, and $f$ satisfies the functional equation $f = T(f)$, or also $e^{f(x)} = f(a^x)$.

As for negative $x$, $\lim_{x \to - \infty} f(x) = \lim_{x \to - \infty} \log(f(a^x)) = \lim_{x \to 0} \log(f(0)) \ge \log(f(0))$, with equality if $f$ is continuous at $0$.