Interpolation between iterated logarithms I am investigating the family of functions $$\log_{(n)}(x):=\log\circ \cdots \circ \log(x)$$
Is there a known smooth interpolation function $H(\alpha, x)$ such that $H(n,x)=\log_{(n)}(x)$ for $n\in\mathbb{N}$?
 A: Your function $H(\alpha,x)$ is the same as $\exp^{-\alpha}(x)$ in the notation of wikipedia, and as such it has been widely studied. Two good sources are the Citizendium page and also this great paper by Dmitri Kouznetsov, whose results I have managed to reproduce.

D. Kouznetsov (July 2009). "Solution of $F(z+1)=\exp(F(z))$ in complex z-plane". Mathematics of Computation 78 (267): 1647–1670.

The function he studies is $F(-\alpha)=H(\alpha,1)$ in your notation. Moreover, $H(\alpha,x)$ for different $x$ are just shifted versions of each other, in the sense that $H(\alpha,y)=H(\beta+\alpha,1)$ for some $\beta$ dependent on $y$. This is because if there is some $\beta$ such that $H(0,y)=H(\beta,1)$, then if I define $H'(\alpha)=H(\beta+\alpha,1)$, we have $H'(0)=H(0,y)$ and $H'(\alpha+1)=\log H'(\alpha)$, so $H'$ satisfies the same functional equations as $H$ and thus is the same function (assuming the additional translation-invariant regularity properties, which means analyticity and an asymptotic form as $\alpha\to i\infty$, see Kouznetsov's paper for details).
Over the reals, the fact that $H(0,y)=H(\beta,1)$ has a solution for any $y$ follows from the fact that the range of $H(\beta,1)$ for $\beta\in(-2,\infty)$ is all of ${\Bbb R}$, since $\lim_{\beta\to-2}H(\beta,1)=-\infty$ and $\lim_{\beta\to\infty}H(\beta,1)=\infty$. Over the complexes, it follows from Picard's little theorem, since $H$ is non-constant.
