Consider the function $h(x)$ which is conjectured to be $c^\infty$ nowhere analytic, and can be used to generate what we call the base change slog, which is the inverse of the basechange sexp. This is Peter Walker's helper function from his 1991 paper. In http://eretrandre.org/rb/files/Walker1991_111.pdf Peter Walker proves $h(x)$ is $C^{\infty}\;$ infinitely differentiable. How can we prove that $h(x)$ nowhere analytic? $h(x)$ has a surprisingly simple definition, and converges nicely at the real axis.

$$l(x) = \ln(x+1)$$ $$h_n (x) = l^{[n]}\exp^{[n]}(x) $$ $$h(x) = \lim_{n\to \infty} h_n (x) $$

There is an analytic Abel function $\alpha(z)$ for iterating $f(z)=\exp(z)-1,\;\;\alpha(f(z))=\alpha(z)+1$, which has an aymptotic series studied by Ecalle; see https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge. Then Peter Walker's slog function, which has also been referred to the basechange slog, is: $$\text{slog}(x) = \alpha(h(x))\;\;\;\text{slog}(\exp(x))=\text{slog}(x)+1 $$

Peter Walker was aware of the difficulties of defining $h(z)$ in the complex plane, and wrote, "...we cannot identify our function ... with Kneser's function defined by conformal mappings, without an extension of the domain of the function h to include nonreal values. Until both these difficulties have been overcome, the possibility remains that ... two ... distinct generalized logarithms have been constructed."

In fact, we now have more tools to generate Kneser's analytic sexp(z)/slog(z) solution, and there is a 1-cyclic mapping connecting it to Walker's solution. For Walker's solution, the issue is that when $\exp^{[n-1]}(z)=(2m-1)\pi i\;$ then $\exp^{[n]}(z)=-1\;$ so $l(\exp^{[1]}(z))\;$ has a singularity in the definition of $h_n(z)$ and these singularities get arbitrarily close to the real axis as n increases. So even though the iteration of $h_n(x)$ converges superbly at the real axis, it fails to converge in any radius in the complex plane, no matter who small that radius is. As far as I know, it has yet to be proven that Walker's solution doesn't converge to its Taylor series anywhere, so that Walker's solution is nowhere analytic.

Here are some steps I took to understand the $h(x)$ function by understanding the $h_n(x)$ sequence. I continue to use the shorthand $l(x)=\ln(x+1)\;$ along with $l^{[n]}(x)$ for the iterated $l(x)$. I also make use of the shorthand $\chi(x)=\exp(x)\;\;\;\chi^{[n]}(x)=\exp^{[n]}(x);$ and the shorthand $\Delta_n(x)=h_n(x)-h_{n-1}(x)$.

$$ h_0(x)=x;\;\;\; h_1(x) = h_0(x) + l\left(\frac{1}{\exp(x)}\right) $$

$$ \Delta_1(x) = l \left(\frac{1}{ \chi^{[1]}(x) }\right)$$

$$ \Delta_2(x) = l \left( l \left(\frac{1}{ \chi^{[2]}(x) }\right)\frac{1}{ (\chi^{[1]}(x))+1 } \right)$$

$$ \Delta_3(x) = l \left(l \left(l \left(\frac{1}{\chi^{[3]}(x)}\right)\frac{1}{ \chi^{[2]}(x)+1 } \right)\frac{1}{ l^{[1]}(\chi^{[2]}(x))+1 }\right)$$

$$ \Delta_4(x) = l \left(l \left(l \left(l \left(\frac{1}{\chi^{[4]}(x)}\right)\frac{1}{ \chi^{[3]}(x)+1 } \right)\frac{1}{ l^{[1]}(\chi^{[3]}(x))+1 }\right)\frac{1}{ l^{[2]}(\chi^{[3]}(x))+1 }\right)$$ $$ \Delta_4(x) \approx \frac{1}{\chi^{[4]}(x)}\cdot\frac{1}{ \chi^{[3]}(x)+1 } \cdot \frac{1}{ l^{[1]}(\chi^{[3]}(x))+1 }\cdot\frac{1}{ l^{[2]}(\chi^{[3]}(x))+1 }\cdot\left(1-\frac{O}{2\chi^{[4]}(x)}\right)$$

The above equation can be extended arbitrarily. One can also expland these equations using the Tayor series $l(x)=x+\frac{-x^2}{2}+\frac{x^3}{3}+\frac{-x^4}{4}+...$, which is what I think the next step is which leads to an expansion whose first term is the approximation above. But then ultimately, we need to show that sum of these $\Delta_n(x)$ is nowhere analytic by studying the Taylor series expansion of $\Delta_n(x)$. Then one hopes to show that as n increases, for low enough terms, the Taylor series for $\Delta_n(x)$ is very small compared with $\Delta_{n-1}(x)$ but for large enough terms, the Taylor terms grow faster than any $r^n$, so that the series radius of convergence gets arbitrarily small.


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