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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!

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  • $\begingroup$ Which formulation of BCT are you using/familiar with? $\endgroup$ – nhmwhhxx Dec 8 '19 at 15:25
  • $\begingroup$ The one for complete metric spaces! $\endgroup$ – MrFranzén Dec 8 '19 at 15:27
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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.

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  • $\begingroup$ 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. $\endgroup$ – Danny Pak-Keung Chan Dec 14 '19 at 2:45

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