Is there a function $f:[0;1]\rightarrow \Bbb R ^n$ of class $C^ \infty$ with $f^{(i)}(0)=10^i.f^{(i)}(1) \, \forall \, i \in \Bbb N_0$

Question

I have the function $g:\Bbb R_{>0} \rightarrow [0.1;1)$ given by $$g(x)=10^{-1-\lfloor \log_{10} (x) \rfloor}.x$$ Which kind of "erase" the coma in the decimal expresion of $x$. For example $$g(123,539)=0,123539 \qquad; \qquad g(0,00012)=0,12$$ And I'm trying to find a function $f:[0.1;1]\rightarrow \Bbb R ^n$ of class $C^ \infty$ such that $f \circ g$ is a function of class $C^ \infty$ (because $g$ it's not a pretty function and I want to make it pretty in order to be able to work with it). I've managed to prove $f \circ g$ is $C^ \infty$ if and only if $f$ satisfy the following property $$f^{(i)}(0.1)=10^i.f^{(i)}(1) \, \forall \, i \in \Bbb N_0$$ I've also manage to prove this is equivalent to finding a function $h:[0;1]\rightarrow \Bbb R ^n$ of class $C^ \infty$ with the same property $$h^{(i)}(0)=10^i.h^{(i)}(1) \, \forall \, i \in \Bbb N_0$$ Because we can take $f(x)=h((10x-1)/9)$ and it will satisfy the property we were looking for. In order to avoid trivial solutions such as $h$ beeing constant, I will also ask for the following property $$x\neq y \qquad h(x)=h(y) \Rightarrow x=0 \; , \; y=1$$ So $h$ is injective in $(0;1)$. I haven't been able to find such a function and I don't know how to search for one so I'm kind of lost.

Answer 1

To avoid confusion, the $g$ function from the problem will be called $G_{10}$ $$G_{10}(x)=10^{-1-\lfloor \log_{10}(x) \rfloor}.x \quad \forall \ x \in \Bbb R_{>0}$$

First, let's get some motivation for this problem. Let $f:\Bbb R \rightarrow \Bbb R_{>0}$ be a function, although $G_{10} \circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] \rightarrow S^1$ given by $$g(x)=e^{2\pi i.(10x-1)/9} \tag 1\\$$ Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g \circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h \circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:\Bbb R \rightarrow \Bbb R_{>0}$ is $C^1$ we would like $G_{10} \circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] \rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g \circ G_{10} \circ f \in C^1$ for every $f \in C^1$. It's easy to prove the last condition is equivalent to asking $g \circ G_{10}\in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k \in \Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^\infty$"

So, we are looking for a function $f:[0.1;1] \rightarrow S^1$ inyective over $(0.1;1)$ with $f \circ G_{10} \in C^\infty$. To make it easier to find, we will suppose $f \in C^\infty$. Note $f \circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).

It's "easy" to prove by induction that, if $f:[0.1;1] \rightarrow S^1$ is $C^k$ with $k \in \Bbb N$, then $f \circ G_{10} \in C^k$ if and only if $$f^{(i)}(0.1)=10^i.f^{(i)}(1) \quad \forall \ i \in \{0,1,\ldots,n\}$$ So we want this to happen for all $i \in \Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]\rightarrow S^1$ inyective over $(0.1;1)$ with $f \in C^\infty, \ f(0.1)=f(1)=1$ and satisfying $$f^{(i)}(0.1)=10^i.f^{(i)}(1) \quad \forall \ i \in \Bbb N_0$$

Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 \pi i.h(x)}$ with $h:[0.1;1]\rightarrow \Bbb R$. Observe $f \in C^\infty \Leftrightarrow h \in C^\infty$ and $f(0.1)=f(1)=1 \Leftrightarrow h(0.1),h(1) \in \Bbb Z$ in which case it's "easy" to check $$f^{(i)}(0.1)=10^i.f^{(i)}(1) \quad \forall \ i \in \Bbb N_0 \Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) \quad \forall \ i \in \Bbb N$$ Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.

We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]\rightarrow \Bbb R$ given by $h(x)=\log_{10}(x)$ satisfys $h\in C^\infty, \ h(0.1),h(1) \in \Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check $$h^{(i)}(x)=\frac{1}{\ln(10)} . \frac{(-1)^{i-1}.(i-1)!}{x^i} \quad \forall \ i\geq 1$$ And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) \ \ \forall \ i \in \Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is $$f(x)=e^{2 \pi i.\log_{10}(x)}$$ And we can check $f \circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $\Bbb R_{>0}$ via the same formula and what might seem crazy is $f \circ G_{10}=f$

Interesting Fact: If $g:\Bbb R \rightarrow \Bbb R_{>0}$ is $C^1$, if we know the values of $f \circ g:\Bbb R \rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 \in \Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f \circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $\Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:\Bbb R \rightarrow S^1$ with $h \in C^1$ there's a $g:\Bbb R \rightarrow \Bbb R_{>0}$ with $g \in C^1$ and $f \circ g = h$. If we get anoter $g' \in C^1$ with $f \circ g' = h$ then there's a $k \in \Bbb Z$ with $g'(x)=10^k.g(x) \ \forall x \in \Bbb R$.