Let $f:\mathbb{R}\to\mathbb{R}$ be a Baire one function. I need to prove $f^n$ is Baire one for $n=1,2,\dots,k$ by using $\epsilon-\delta$ characterization of Baire one functions:
A function $f:\mathbb{R}\to\mathbb{R}$ is said to be a Baire one function iff for every $\epsilon>0$, there is a positive function $\delta$ on $\mathbb{R}$ such that for any $x,y\in\mathbb{R}$, $$|x-y|<\min\{\delta(x),\delta(y)\}\Rightarrow |f(x)-f(y)|<\epsilon.$$
where $f^n=f\circ f^{n-1}$ and $f^1=f$. In this article is shown that composition of two Baire one functions is not necessarily Baire one. But, I think there is still a possibility to prove $f^n$, $n=1,2,\dots,k$ is Baire one. Here is my attempt.
For $n=1$, it is clear that $f$ is Baire one.
For $n=2,\dots, k$, I try to prove it by using mathematical induction.
For the base step, I need to show that $f^n$ is Baire one for $n=2$.
Let $\epsilon>0$. Since $f$ is Baire 1, then there exists a positive function $\delta_1:\mathbb{R}\to\mathbb{R}$, such that for any $x,y\in\mathbb{R}$, $$|x-y|<\min\{\delta_1(x),\delta_1(y)\}\Rightarrow |f(x)-f(y)|<\epsilon.$$
Let $\eta=\frac{1}{2}\min\{\delta_1(f(x)),\delta_1(f(y)),\epsilon\}>0$. Since $f$ is Baire one, there exists a positive function $\delta_2:\mathbb{R}\to\mathbb{R}$, such that for any $x,y\in\mathbb{R}$, $$|x-y|<\min\{\delta_2(x),\delta_2(y)\}\Rightarrow |f(x)-f(y)|<\eta<\min\{\delta_1(f(x)),\delta_1(f(y))\}.$$
Therefore, for any $x,y\in\mathbb{R}$,
$$|x-y|<\min\{\delta_2(x),\delta_2(y)\}\Rightarrow |f(f(x))-f(f(y))|=|f^2(x)-f^2(y)|<\epsilon.$$
Thus, $f^2$ is Baire one.
For induction step, assume that $f^n$ is Baire one for $n=m\le k-1$ and I need to prove that $f^n$ is Baire one for $n=m+1$.
Note that $f^{m+1}=f\circ f^m$, where $f$ and $f^m$ are Baire one. From the base step, we have that $f^2=f\circ f$ is Baire one where $f$ is Baire one. Therefore $f^{m+1}=f\circ f^m$ is Baire one.
Then, I conclude that $f^n$ is Baire one for $n=1,2,\dots,k$.
However, I have some questions:
- For the base step, is it allow to prove like that?
I feel strange in the italic sentences since it appears twice but with different positive function, i.e. $\delta_1$ and $\delta_2$. Also for the first, I use $\epsilon>0$ and for the second, I use $\eta>0$. - Is the induction step correct?
Sorry for the long question, I will appreciate for any help. Thank you in advance.
The proof "For the base step" seems circular, as $\eta$ depends on $x,y$, causing $\delta_2$ depends on $x,y$ as well.
More precisely, if we want to prove $f^2$ is Baire-$1$ using the $\epsilon$-$\delta$ characterization, we need to prove the following.
Given any $\epsilon>0$, there is a positive function $\delta:\mathbb{R}\to\mathbb{R}^+$ such that for any $x,y\in\mathbb{R}$, we have $$|x-y|<\min\{\delta(x),\delta(y)\}\Rightarrow |f^2(x)-f^2(y)|<\epsilon.$$
Note that $\delta$ can only depend on $\epsilon$.
Let $\epsilon>0$ be given. In your proof, you let $\delta=\delta_2$, which depends on $\eta$. But the choice of $\eta$ depends on $x,y$. This means that for different $x,y$, we would get different $\eta$, and thus different $\delta$, which does not fulfill the positive function $\delta$ in the characterization.