Smooth compositions with a nonsmooth inner function

TL;DR

This paper explores the application of Baire's category theorem to smooth compositions involving a nonsmooth inner function, providing new insights into their mathematical properties.

Contribution

It introduces a novel approach to analyzing compositions with nonsmooth inner functions using Baire's category theorem.

Findings

Established conditions for smooth compositions with nonsmooth inner functions

Provided new theoretical insights into the structure of such compositions

Extended the application of Baire's category theorem in analysis

Abstract

In this paper, we present an interesting application of Baire's category theorem.

Equations

$${\mathcal{A}}_p=\{f\in C^n(\mathbb{R},\mathbb{R})|f(p(\cdot )+c)\in C^n(\mathbb{R},\mathbb{R})\text{mbox}foranyc\in \mathbb{R}\}.$$

$$\Gamma =\{(x,f(x))\in X\times Y|x\in X\}$$

$$(f{|}_{A_n}{)}^{-1}(F)={\pi }_X(K_n\cap (X\times F))$$

$${\cap }_{n=1}^{\infty }{U}_n\subset {\cup }_{n=1}^{\infty }\mathrm{Int}(A_n).$$

$$\underset{y\to +\infty }{\lim }f(y)=\underset{x\to b^{-}}{\lim }f(p(x)+a)=f(p(b)+a)$$

$$|p(x_n)-p(x_0)|\ge {ϵ}_0$$

$$|p(a_n)-p(x_0)|<{ϵ}_0/2$$

$$p({\tilde{x}}_n)=p(x_0)+{ϵ}_0/2$$

$$p({\tilde{x}}_n)=p(x_0)-{ϵ}_0/2.$$

$$p({\tilde{x}}_n)\to p(x_0)+{ϵ}_0/2$$

$$p({\tilde{x}}_n)\to p(x_0)-{ϵ}_0/2$$

$$\tilde{f}(y)=f(y-p(x_0)+y_0)$$

$${\tilde{f}}^{\prime }(p(x_0))=f^{\prime }(y_0)\ne 0.$$

$$p(x)={\tilde{f}}^{-1}(g(x))$$

$$L=\{d\in\mathbb{R}\ |\ f(y+d)=f(y)\ \mbox{for any $y\in\mathbb{R}$ and any $f\in\mathcal{A}_{p}$}\}.$$

$$\Lambda (x)=\{y\in \mathbb{R}|y=\underset{n\to \infty }{\lim }p(x_n)\text{mbox}forsomesequence{x}_n\to x\}.$$

$$p(x_n)\to y.$$

$$f(p(x)+c)=\underset{n\to \infty }{\lim }f(p(x_n)+c)=f(y+c)$$

$$f=\overline{f}\circ \pi .$$

$$f(p+c)=\overline{f}\circ \pi (p+c)\in C^0(\mathbb{R},\mathbb{R}).$$

$$f=\overline{f}\circ \pi (p+c)\in C^n(\mathbb{R},\mathbb{R}).$$

Full text

1Research partially supported by the Yangfan project from Guangdong Province and NSFC 11571215.

1. Introduction

In this paper, we prove the following interesting result:

Theorem 1.1**.**

Let $n$ be a nonnegative integer or $\infty$, $p:\mathbb{R}\to\mathbb{R}$ be a given function with $p\not\in C^{n}(\mathbb{R},\mathbb{R})$ and

[TABLE]

Then, either $\mathcal{A}_{p}=\mathbb{R}$ or $\mathcal{A}_{p}=C_{d}^{n}(\mathbb{R},\mathbb{R})$ for some nonzero constant $d$. Here $C_{d}^{n}(\mathbb{R},\mathbb{R})$ means the collection of all functions in $C^{n}(\mathbb{R},\mathbb{R})$ with periodicity $d$.

This result is motivated by the work [3] of Christensen and Wu on diffeological vector spaces. A weaker form of Theorem 1.1 is needed in [3]. The method in [3] dealing with the weaker form seems not applicable to prove Theorem 1.1.

It is clear that $\mathcal{A}_{p}$ is a translating invariant subalgebra of $C^{n}(\mathbb{R},\mathbb{R})$. As an example, take $p=\chi_{E}$ with $E$ an abitrary subset of $\mathbb{R}$ ($E\not=\emptyset,\mathbb{R}$). Then, it is clear that $\cos(2\pi x),\sin(2\pi x)\in\mathcal{A}_{p}$. In fact, it is not hard to see that $\mathcal{A}_{p}=C^{n}_{1}(\mathbb{R},\mathbb{R})$ in this specific example.

A key ingredient in the proof of Theorem 1.1 is the following result about the continuity of maps with $\sigma$-compact graph.

Lemma 1.1**.**

Let $X$ be a Hausdorff Baire space, $Y$ be a topological space and $f:X\to Y$ be a map with $\sigma$-compact graph. Then, $f$ is continuous on a dense open subset of $X$.

For a Baire space, we mean a topological space satisfying Baire’s category theorem. That is, a countable intersection of open dense subset is still dense. Typical examples of Baire spaces are complete metric spaces and locally compact Hausdorff spaces. One may compare Lemma 1.1 with Blumberg’s theorem [1] : for any real function on $\mathbb{R}$, there is a dense subset $E\subset\mathbb{R}$, such that $f|_{E}:E\to\mathbb{R}$ is continuous. Blumberg’s theorem was later generalized by Bradford and Goffman [2] for more general spaces (see also [4]).

Acknowledgement. The authors would like to thank the referee for generously sharing ideas that significantly strengthen the results (both Theorem 1.1 and Lemma 1.1) of the previous version of this paper, and to thank their colleague, Prof. Enxin Wu, for many helpful discussions.

2. Proof of Theorem 1.1

We first prove Lemma 1.1.

Proof of Lemma 1.1.

Let

[TABLE]

be the graph of $f$. Suppose that $\Gamma=\cup_{n=1}^{\infty}K_{n}$ where $K_{n}$ is a compact subset of $X\times Y$ for $n=1,2,\cdots$. Then, $A_{n}=\pi_{X}(K_{n})$ is a compact subset of $X$, where $\pi_{X}:X\times Y\to X$ is the natural projection. Note that $f|_{A_{n}}:A_{n}\to Y$ as a map with compact graph is continuous. To see this, let $F$ be any closed subset of $Y$, then

[TABLE]

is compact and hence closed.

Let $F_{n}=A_{n}\setminus{\rm Int}(A_{n})$ where ${\rm Int}(A_{n})$ is the interior of $A_{n}$. Then, $U_{n}=X\setminus F_{n}$ is a dense open subset of $X$. So, $\cap_{n=1}^{\infty}U_{n}$ is a dense subset of $X$ by that $X$ is a Baire space. On the other hand, since $X=\cup_{n=1}^{\infty}A_{n}$,

[TABLE]

Hence $\cup_{n=1}^{\infty}{\rm Int}(A_{n})$ is a dense open subset of $X$. Moreover, $f$ is continuous at the points in $\cup_{n=1}^{\infty}{\rm Int}(A_{n})$ since $f|_{A_{n}}$ is continuous for $n=1,2,\cdots$. This completes the proof of the Lemma. ∎

For clarity, we will separate the proof of Theorem 1.1 into two cases:(i) graph of $p$ is closed and (ii) graph of $p$ is not closed.

Lemma 2.1**.**

Let notations be the same as in Theorem 1.1. Suppose that the graph of $p$ is closed, then $\mathcal{A}_{p}=\mathbb{R}$.

Proof.

We will proceed by contradiction. Assume that $\mathcal{A}_{p}\neq\mathbb{R}$. Let $G\subset\mathbb{R}$ be the largest open subset of $\mathbb{R}$ such that $p$ is continuous on $G$. By Lemma 1.1, $G$ is dense in $\mathbb{R}$. Let $F=\mathbb{R}\setminus G$. For clarity, we divide the proof into several claims.

**Claim 1.By the structure of open subsets in $\mathbb{R}$, $G$ is a disjoint union of countably many open intervals. Let $(a,b)$ be one of such open intervals where $a$ may be $-\infty$ and $b$ may be $+\infty$. Then, $p\in C(\overline{(a,b)})$.

Proof of Claim 1.** We only show the case that $a=-\infty$ and $b$ is finite. The proofs of the other cases are similar. In this case, $\overline{(a,b)}=(-\infty,b]$.

Because the graph of $p$ is closed and $p\in C((-\infty,b)),$ the asymptotic behavior of $p$ as $x$ approaching $b^{-}$ happens in only three cases:

1. (1)

$\lim_{x\to b^{-}}p(x)=p(b)$, 2. (2)

$\lim_{x\to b^{-}}p(x)=+\infty$, and 3. (3)

$\lim_{x\to b^{-}}p(x)=-\infty$.

To prove Claim 1, we only need to exclude (2) and (3). If (2) is true, we have

[TABLE]

for any $a\in\mathbb{R}$ and $f\in\mathcal{A}_{p}$. This means that $\mathcal{A}_{p}=\mathbb{R}$ which is a contradiction. Case (3) can be excluded similarly.

Claim 2. $F$ has no isolated points.

Proof of Claim 2. Suppose that $x_{0}$ is an isolated point of $F$. Then, by Claim 1, $p$ is continuous on a neighborhood of $x_{0}$. So, $x_{0}\in G$ which is a contradiction.

Claim 3. Any continuous point of $p|_{F}:F\to\mathbb{R}$ is a continuous point of $p$.

Proof of Claim 3. Let $x_{0}\in F$ be a continuous point of $p|_{F}$. If $x_{0}$ is not a continuous point of $p$. Then there is a $\epsilon_{0}>0$ and a sequence $\{x_{1},x_{2},\cdots,x_{n},\cdots\}$ of points tending to $x_{0}$ as $n\to\infty$, such that

[TABLE]

for any $n=1,2,\cdots$. Since $x_{0}$ is a continuous point of $p|_{F}$, $x_{n}\not\in F$ for $n$ large enough. Moreover, since $x_{0}$ is not isolated in $F$ (by Claim 2) and by Claim 1, $x_{n}\in(a_{n},b_{n})$ where $(a_{n},b_{n})$ is an open interval in the disjoint open interval decomposition of $G$ for $n$ large enough. It is clear that $a_{n}\to x_{0}$ and $b_{n}\to x_{0}$ as $n\to\infty$. Furthermore, since $x_{0}$ is a continuous point of $p|_{F}$ and $a_{n}\in F$,

[TABLE]

for $n$ large enough. Since $p\in C([a_{n},b_{n}])$ (by Claim 1), and by (2.5) and (2.6), there is a point $\tilde{x}_{n}\in[a_{n},b_{n}]$ such that

[TABLE]

or

[TABLE]

Then there is a sequence $\{\tilde{x}_{1},\tilde{x}_{2},\cdots\}$ of points tending to $x_{0}$ as $n\to\infty$ (taking subsequence if necessary) such that

[TABLE]

or

[TABLE]

as $n\to\infty$. This is a contradiction of that the graph of $p$ is closed Claim 4. $p$ is continuous on $\mathbb{R}$.

Proof of Claim 4. Assume that $F\neq\emptyset$. By Lemma 1.1, $p|_{F}:F\to\mathbb{R}$ is continuous on some dense open subset of $F$. So, there is an open interval $I$ such that $p|_{F}$ is continuous on $I\cap F$ with $I\cap F\neq\emptyset$. By Claim 3, points in $I\cap F$ are continuous points of $p$. So $p$ is continuous on $I$ which implies that $I\subset G$ and contradicts that $I\cap F\neq\emptyset$.

We are now ready to complete the proof. Since $p\not\in C^{n}(\mathbb{R},\mathbb{R})$, we know that $n\geq 1$. Let $f\in\mathcal{A}_{p}$ be a nonconstant function. Then, $f^{\prime}(y_{0})\neq 0$ for some $y_{0}\in\mathbb{R}$. It is clear that

[TABLE]

also belongs to $\mathcal{A}_{p}$ and

[TABLE]

Let $g(x)=\tilde{f}(p(x))$. Then $g\in C^{n}(\mathbb{R},\mathbb{R})$ because $\tilde{f}\in\mathcal{A}_{p}$. Since $p$ is continuous at $x_{0}$,

[TABLE]

for any $x$ in some neighborhood of $x_{0}$. This implies that $p$ is $C^{n}$ in some neighborhood of $x_{0}$. Because $x_{0}$ is arbitrary, $p\in C^{n}(\mathbb{R},\mathbb{R})$ which is a contradiction. ∎

Lemma 2.2**.**

Let notations be the same as in Theorem 1.1. If the graph of $p$ is not closed, then either $\mathcal{A}_{p}=\mathbb{R}$ or $\mathcal{A}_{p}=C_{d}^{n}(\mathbb{R},\mathbb{R})$.

Proof.

Let

[TABLE]

It is clear that $L$ is closed subgroup of $\mathbb{R}$ since $\mathcal{A}_{p}\subset C(\mathbb{R},\mathbb{R})$. Let

[TABLE]

**Claim 1.For any $y\in\Lambda(x)$, $y-p(x)\in L$.

Proof of Claim 1.** Let $\{x_{n}\}$ be a sequence of points tending to $x$ such that

[TABLE]

Then, for any $f\in\mathcal{A}_{p}$,

[TABLE]

for any $c\in\mathbb{R}$ because $f(p(\cdot)+c)$ is continuous. This means that $y-f(x)\in L$.

Because the graph of $p$ is not closed, there is a point $x\in\mathbb{R}$, such that $\Lambda(x)\varsupsetneq\{p(x)\}$. Then, by Claim 1, we know that $L\neq\{0\}$. Therefore, $L=\mathbb{R}$ or $L=d\mathbb{Z}$ for some nonzero constant $d$. For the first case, we know that $\mathcal{A}_{p}=\mathbb{R}$. For the second case, we have $\mathcal{A}_{p}\subset C_{d}^{n}(\mathbb{R},\mathbb{R})$.

Let $\pi:\mathbb{R}\to\mathbb{R}/d\mathbb{Z}$ be the natural projection. Then, by Claim 1, it is clear that $\pi(p+c):\mathbb{R}\to\mathbb{R}/d\mathbb{Z}$ is continuous. Moreover, for any $f\in C_{d}^{n}(\mathbb{R},\mathbb{R})$, $f$ descends to a function $\overline{f}:\mathbb{R}/d\mathbb{Z}\to\mathbb{R}$ such that

[TABLE]

It is clear that $\overline{f}\in C^{n}(\mathbb{R}/d\mathbb{Z},\mathbb{R})$.

When $n=0$, for any $f\in C_{d}^{0}(\mathbb{R},\mathbb{R})$, we know that

[TABLE]

This means that $f\in\mathcal{A}_{p}$. So, we have shown that $\mathcal{A}_{p}=C_{d}^{n}(\mathbb{R},\mathbb{R})$ for the case $n=0$.

When $n\geq 1$, if $\mathcal{A}_{p}\neq\mathbb{R}$, let $f_{0}$ be a nonconstant function in $\mathcal{A}_{p}$. Then a similar argument as in the proof of Lemma 2.1 (after Claim 4 in the proof of Lemma 2.1) using inverse function theorem will show that $\pi(p+c)\in C^{n}(\mathbb{R}/d\mathbb{Z},\mathbb{R})$. Then, for any $f\in C_{d}^{n}(\mathbb{R},\mathbb{R})$,

[TABLE]

This means that $f\in\mathcal{A}_{p}$. So $\mathcal{A}_{p}=C_{d}^{n}(\mathbb{R},\mathbb{R})$. ∎

Proof of Theorem 1.1.

The combination of Lemma 2.1 and Lemma 2.2 gives us Theorem 1.1. ∎