My foundations of analysis is a bit rusty, but here goes:
Problem: Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function on $\mathbb{R}$, and let $A:=\{x\in \mathbb{R}\mid\,\limsup_{y\to x} |f'(y)|<\infty\}$. Show that $A$ is an open and dense subset of $\mathbb{R}$.
I have already shown that $A$ is open in $\mathbb{R}$ by showing that $A^c$ is closed. Now I want to show that $A$ is also dense in $\mathbb{R}$ (i.e. $\bar{A}=\mathbb{R}$).
I use that $\limsup_{y\to x} |f'(y)|=\lim_{\epsilon\to 0}(\sup\{|f'(y)|:y\in B_\epsilon(x)\setminus \{x\}\})$.
What I have tried so far:
Well, I want to show that $\bar{A}=\mathbb{R}$, i.e. that each point of $\mathbb{R}$ is either a point of $A$ or a limit point of $A$. So, I started by trying to show that for any $r\in\mathbb{R}$, every ball $B_\delta(r)$ would contain a point $a\in A, a\not= r$. Suppose not, i.e.
$\neg\Big(\forall\delta>0, \exists a\in B_\delta(r)\setminus \{r\}\text{ s.t. } \lim_{\epsilon\to 0}(\sup\{|f'(y)|:y\in B_\epsilon(a)\setminus\{a\}\})<\infty\Big) \Leftrightarrow \exists\delta>0\text{ s.t. }\forall p\in B_\delta(r)\setminus\{r\}, \lim_{\epsilon\to 0}(\sup\{|f'(y)|:y\in B_\epsilon(p)\setminus\{p\}\})=\infty.$
Now, I felt that I should be able to deduce a contradiction from this somehow. In a sense, we have a set $B_\delta(r)$ of non-zero measure, where for every point in it, $\sup|f'(y)|$ goes off to infinity in shrinking balls around that point. This does not feel right since the function $f$ is differentiable, so $|f'(y)|<\infty, \forall y\in\mathbb{R}$, but we don't know if the derivative is continuous, so we can't use the extreme value theorem, so I got a bit stuck.
Thus I took a step back, and noted that $\bar{A}=A\cup A'$, and since I want to show $\bar{A}=\mathbb{R}$, then perhaps I could show $\bar{A}^c=A^c\cap A'^c=\emptyset$. So, I wanted to show that for no point $r\in\mathbb{R}$ could we have that:
$\begin{cases} \lim_{\epsilon\to 0}\big(\sup\{|f'(y)|:y\in B_\epsilon(r)\setminus\{r\}\}\big)=\infty \\ \exists\delta>0\text{ s.t. }\forall p\in B_\delta(r)\setminus\{r\}, \lim_{\epsilon\to 0}(\sup\{|f'(y)|:y\in B_\epsilon(p)\setminus\{p\}\})=\infty. \end{cases}$
But, it's not that obvious to me that they imply a contradiction.
Where I am right now:
So via the following posts:
Points of continuity of Baire class one functions
How discontinuous can a derivative be?
a differentiable function but its derivative is discontinuous everywhere
I've got:
- $f'(x)$ is Baire Class 1 (simple, but unknown to me before)
- The set of points of continuity of Baire Class 1 functions is comeagre, by e.g. Thm 24.14 in Kechris
- Since we found a set $B_\delta(r)$ with $\delta >0$, where for each point $p\in B_\delta(r)\setminus\{r\}, \lim_{\epsilon\to 0}(\sup\{|f'(y)|:y\in B_\epsilon(p)\setminus\{p\}\})=\infty$, then wouldn't $f'$ be discontinuous everywhere in $B_\delta(r)\setminus \{r\}$? Thus, contradicting 1,2?
UPDATE: Based on the proof of Thm 24.14 in Kechris, I have now tried to rewrite it using things that I feel at least somewhat comfortable with:
Theorem: Let $X,Y$ be metric spaces, $Y$ separable and $f:X\to Y$ of Baire Class 1. Then the set of points $D\subset X$ of discontinuity is of first category, i.e. it can be written as a countable union of nowhere dense sets.
Proof attempt: By the usual definition of continuity in a topological space, we have that $f$ is continuous at $x$ iff for every neighbourhood $W\subset Y$ of $f(x)$, there exists a neighbourhood $U$ of $x$ s.t. $f(U)\subset W$. So, fix a basis $\{V_n\}$ for the topology on $Y$, then $f$ discontinuous at $x$ would mean that there would exist some $n$ s.t. $x\in f^{-1}(V_n)\setminus \mathrm{Int}(f^{-1}(V_n))$, otherwise $x$ would have some open neighbourhood mapped into $V_n$. So $D=\{x\mid f \text{ discontinuous at } x\}=\cup_n f^{-1}(V_n)\setminus \mathrm{Int}(f^{-1}(V_n))$. Now, $f^{-1}(V_n)$ can be written as a countable union of closed sets$^1$, thus so can $f^{-1}(V_n)\setminus \mathrm{Int}(f^{-1}(V_n))$, say it is equal to $\cup_k F_k$, where all $F_k$ are closed. Then each $F_k$ has no interior since we subtracted all interior from $f^{-1}(V_n)$. So, then $D$ is a countable union of nowhere dense sets. $\square$
$^1$: This is apparently denoted $f^{-1}(V_n)\in \Sigma_2^0$. Now, Baire Class 1 functions are actually defined in Kechris to just be functions such that $f^{-1}(U)\in \Sigma_2^0$ for open sets $U$, and it is shown that for real functions, these are the ones that are pointwise limits of continuous functions. This is the definition I learned from Wikipedia.
QUESTIONS:
Question 1: Am I even right in drawing the conclusion 3 in Update 1 above? And if so, is there a more straightforward way of showing this, using less involved arguments, i.e. an argument you would suspect someone in ca. day 3 of an introductory functional analysis course, who have just gone over Baire's Category theorem could make?
Question 2: In the proof attempt above, can I use that I have shown that open sets (and trivially closed ones) in a metric space can be written as a countable union of closed sets, and somehow see that $f^{-1}(V_n)$ must be either open, closed or clopen, and not neither, and thus conclude the same about $f^{-1}(V_n)\setminus \mathrm{Int}(f^{-1}(V_n))$. Basically, can I use the Baire 1:ness of $f$ in a less involved way here, and make this proof work for my purposes?
Question 3: Where does $Y$ being separable come into play in the proof of the theorem above?
