As the title says, given a differentiable function $f: \ \mathbb R \to \mathbb R$ define
$$E=\{x:\limsup_{y\to x}|f'(y)|<\infty\} $$
and show that $E $ is open and dense in $\mathbb R$.
Here $\limsup_{y\to x}|f'(y)| $ is defined as $\lim_{\epsilon \to 0} (\sup \{|f'(y): y \in B_{\epsilon } (x) \setminus \{x \} \}) $.
I suppose this should be solved with an application of Baire's category theorem but I'm quite at loss on how to proceed!
Any help would be much appreciated!
Best Answer
To show that $E$ is open: Let $F=E^{c}=\{x\mid\limsup_{y\rightarrow x}|f'(y)|=\infty\}$. We go to show that $F$ is closed. Let $(x_{n})$ be a sequence in $F$ and suppose that $x_{n}\rightarrow x$ for some $x\in\mathbb{R}$. We prove that $x\in F$ by contradiction. Suppose the contrary that $x\notin F$. Choose $M>0$ such that $\limsup_{y\rightarrow x}|f'(y)|<M$. There exists $\delta>0$ such that $|f'(y)|<M$ whenever $y\in(x-\delta,x+\delta)\setminus\{x\}$. Since $x_{n}\rightarrow x$ and $x_{n}\neq x$, there exists $n$ such that $x_{n}\in(x-\delta,x+\delta)\setminus\{x\}$. Without loss of generality, we suppose that $x<x_{n}<x+\delta$. Choose $\varepsilon>0$ be sufficiently small such that $x<x_{n}-\varepsilon<x_{n}<x_{n}+\varepsilon<x+\delta$. Since $\lim_{y\rightarrow x_{n}}|f'(y)|=\infty$, there exists $y_{0}\in(x_{n}-\varepsilon,x_{n}+\varepsilon)\setminus\{x_{n}\}$ such that $|f'(y_{0})|>2M$. Note that $y_{0}\in(x-\delta,x+\delta)\setminus\{x\}$, so we also have $|f'(y_{0})|<M$, which is a contradiction.
In the above, we have not used any properties about $f$ nor its derivative $f'$. That is, that $F$ is closed continues to hold if $f'$ is replaced by an arbitrary function $g:\mathbb{R}\rightarrow\mathbb{R}$.
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To show that $E$ is dense: Note that $E^{-}=\mathbb{R}$ iff $\emptyset=E^{-c}=\left(E^{coc}\right)^{c}=F^{o}$. That is, we need to show that $F$ has empty interior. Prove by contradiction. Suppose that there exist $\alpha<\beta$ such that $(\alpha,\beta)\subseteq F$. For each $n\in\mathbb{N}$, let $A_{n}=\{x\in(\alpha,\beta)\mid f'(x)\in(-n,n)\}$, which is a $F_{\sigma}$-subset of the topological space $(\alpha,\beta)$ (See Theorem 2 in the appendix). For each $n$, write $A_{n}=\cup_{k}F_{nk}$, for some closed subsets $F_{nk}$ of $(\alpha,\beta)$. Note that $(\alpha,\beta)=\cup_{n} A_n=\cup\{F_{nk}\mid n,k\in\mathbb{N}\}$. Since $(\alpha,\beta)$ is a Baire space, by Baire Category Theorem, there exist $n,k$ such that $F_{nk}$ has non-empty interior. That is, there exist $\alpha'<\beta'$ such that $(\alpha',\beta')\subseteq F_{nk}\subseteq A_{n}$. Choose $x_{0}\in(\alpha',\beta')$. Note that $x_{0}\in F$, so there exists a sequence $(x_{k})$ with $x_{k}\neq x_{0}$, $x_{k}\rightarrow x_{0}$, and $|f'(x_{k})|\rightarrow\infty$. Observe that $x_{k}\in(\alpha',\beta')$ for large $k$ and hence $|f'(x_{k})|<n$, contradicting to $|f'(x_{k})|\rightarrow\infty$.
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Appendix. We state and prove the following theorems.
Theorem 1: Let $X$ be a topological space. Let $f_{n}:X\rightarrow\mathbb{R}$ and $f:X\rightarrow\mathbb{R}$. Suppose that $f_{n}$ is continuous and $f_{n}(x)\rightarrow f(x)$ for each $x\in X$. Then, for each open subset $O\subseteq\mathbb{R}$, $f^{-1}(O)$ is a $F_{\sigma}$-subset of $X$ (i.e., countable union of closed subsets).
Proof of Theorem 1: See my other post $f_n\rightarrow f$ pointwise, $O$ open subset of $\mathbb{R}$ $\Rightarrow$ $f^{-1}(O)$ is $F_{\sigma}$
Theorem 2: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function. If $O\subseteq\mathbb{R}$ is open, then $f'^{-1}(O)$ is a $F_{\sigma}$-set.
Proof of Theorem 2: For each $n$, let $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f_{n}(x)=n\left[f(x+\frac{1}{n})-f(x)\right]$. Note that $f_{n}$ is continuous and $f_{n}(x)\rightarrow f'(x)$ for each $x\in\mathbb{R}$. Now, the result follows from Theorem 1.