1
$\begingroup$

Let $f$ be a function from defined on subset of a closed inteveral $[a, b]$ to the real line $\mathbb{R}$, such that $f$ has finite or infinite derivative everywhere in $(a, b)$ and $f$ has finite or infinite one-side derivative in the border.

Let $\overline{\mathbb{R}}=\mathbb{R} \cup \{+\infty, -\infty\}$ be the extended real line, i.e. $\mathbb{R}$ with the usual topology basis extedend with the open rays $(+\infty, r)$ and $(r, -\infty)$ for every $r \in \mathbb{R}$.

In other words we're asking that for every $c \in [a, b]$ it must be possible to define a funcion $g_c(x)$ in a neighborhood $U(c)$ of $c$ in $[a, b]$ such that $g_c(x)$ is continuous in $c$, $g_c(x)$ defined in this way:

$g_c:U(c) \rightarrow \overline{\mathbb{R}} \\ g_c(x)=\frac{f(x) - f(c)}{x-c}$

My question is about the set $S=\{c \ \ |\ \ g_c(c) \in \{+\infty, -\infty \}\}$, what can we say about $S$? Is this set compact? Is totally disconnected? Every point of $S$ is isolated?

.

$\endgroup$
1
$\begingroup$

The set you mention can be dense.

Consider the function $$ f(x) = \sum_{n=0}^\infty 2^{-n} \text{sign}(x-r_n) |x - r_n|^{1/2} $$ where $\text{sign}(x) = 1$ if $x \ge 0$, or $-1$ otherwise, and $(r_n)_{n \in \mathbb{N}}$ is an ordering of $\mathbb{Q} \cap [0, 1]$. It can be checked that this series is uniformly convergent on every interval of $\mathbb{R}$, and that $f$ will have infinite slope at every $r_n$.

So in this example $S$ contains $\mathbb{Q} \cap [0, 1]$. I don't know whether it is larger.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Are you sure $g_c$ exists for all $c$? $\endgroup$ – Hagen von Eitzen Apr 23 at 10:39
  • 2
    $\begingroup$ On pp. 39-40 of the survey paper Derivatives by Bruckner/Leonard it is mentioned that the set where a function has an infinite (bilateral) derivative has Lebesgue measure zero (this was proved in the 1910s by Denjoy), and conversely, given any Lebesgue measure zero set $E$ there exists a function $f$ (even a strictly increasing continuous function) such that $f(x) = +\infty$ for each $x \in E$ (and possibly also at other points not in $E).$ (continued) $\endgroup$ – Dave L. Renfro Apr 23 at 11:09
  • 1
    $\begingroup$ Note that a measure zero set can have uncountable intersection with every open interval (this is much more "denser" than the usual notion of dense), for example by putting a scaled/translated copy of the Cantor set in every open interval with rational endpoints and then taking the union of these countably many copies of the Cantor set. Even more, the intersection with every open interval can have Hausdorff dimension $1$ (this is much stronger than being "uncountably dense") and also have a first (Baire) category complement (this last part isn't automatic from the Hausdorff dimension part). $\endgroup$ – Dave L. Renfro Apr 23 at 11:16
  • $\begingroup$ @Dave thanks, if you make an answer out of your comment i'll be glad to accept it. $\endgroup$ – Giovanni Barbarani Apr 24 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.