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:

  1. $f'(x)$ is Baire Class 1 (simple, but unknown to me before)
  2. The set of points of continuity of Baire Class 1 functions is comeagre, by e.g. Thm 24.14 in Kechris
  3. 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.


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?

  • 1
    $\begingroup$ Ok, so loosely thinking about this on the subway, I realize I wouldn't even get a point p s.t. $|f'(p)|=\infty$, just another point in $A^c$ I guess, and I already had an infinite amount of those in $B_\delta(r)$. So, I'll think about this more later, I guess my mind has logged out from this one at the moment. $\endgroup$ Oct 31 '19 at 13:51
  • $\begingroup$ This is an interesting problem. In which context did you encounter it? $\endgroup$
    – Math1000
    Nov 1 '19 at 18:57
  • 1
    $\begingroup$ So, this is a (non-mandatory, bonus) homework assignment from the functional analysis part of a master course in real analysis. We use Avner Friedman, Foundations of Modern Analysis as a textbook. The reason I suspect it's simpler (or of course perhaps ill formed for our level), is that the other exercises accompanying (but not related to) this was much more straight forward. My guess is we should use Baire's Category theorem in some clever way. $\endgroup$ Nov 1 '19 at 19:13
  • 1
    $\begingroup$ For a more direct argument, you can look here. Also, all points of continuity of $f'$ belong to $A$, whence $\mathbb{R}\setminus A$ is meagre, and in a Baire space meagre sets have empty interior. $A$ being dense is the same as its complement having empty interior. $\endgroup$ Nov 3 '19 at 14:20
  • $\begingroup$ @DanielFischer Unless I'm misunderstanding: I would prove $\mathbb{R}$ could be written as a union of closed sets on which $f'$ bounded, as here, then Baire gives us that we have a non degenerate interval on which $f'$ is bounded. But if $A$ not dense, then I found that in any such interval, I could find an interval on which it is not bounded yielding a contradiction? Or did you mean something even more direct? I didn't really see where $A$ dense iff int of $A^c$ empty came into play $\endgroup$ Nov 3 '19 at 16:37

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