# For a differentiable function $f$ show that $\{x:\limsup_{y\to x}|f'(y)|<\infty\}$ is open and dense in $\mathbb R$

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!

• Which formulation of BCT are you using/familiar with? – mi.f.zh Dec 8 '19 at 15:25
• The one for complete metric spaces! – MrFranzén Dec 8 '19 at 15:27

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 $$\limsup_{y\rightarrow x}|f'(y)|. There exists $$\delta>0$$ such that $$|f'(y)| 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 $$x. Choose $$\varepsilon>0$$ be sufficiently small such that $$x. 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})|, 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 $$|f'(x_{k})|, contradicting to $$|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.

• Here, we have used the fact that non-empty open subset of a Baire space is also a Baire space (with respect to the relative topology). That non-empty open interval is a Baire space also follows from the theorem that every locally compact Hausdorff space is Baire. – Danny Pak-Keung Chan Dec 14 '19 at 2:45