Set of discontinuities of a derivative
Amanda
Suppose f:I -> R is differentiable on an open interval I of R. Let D be
the set of points where f is discontinuous. We know D is meager and,
therefore, has an empty interior. I have 2 questions, about which I
couldn't come to a conclusion:
a) Is it possible that D is dense in I? Since f has the intermediate
value property, it seems to me it can't, but I couldn't come to a
conclusion.
b) D can have positive Lebesgue measure, right?
Thank you.
Amanda
Dave L. Renfro
I think you meant "... where f' is discontinuous" in the
first paragraph and "Since f' has the ..." in the second
paragraph.
The continuity set of a derivative on an open interval J
is dense in J. In fact, the continuity set has cardinality c
in every subinterval of J. On the other hand, the discontinuity
set of a derivative can have the following properties --->
1. dense in the reals
2. cardinality c in every interval
3. positive measure (hence, not Riemann integrable)
4. positive measure in every interval
5. full measure in every interval (i.e. measure zero complement)
6. have a Hausdorff dimension zero complement
Briefly, a subset D of the reals is a discontinuity set for some
bounded derivative if and only if D is an F_sigma first category
set. Moreover, derivatives for which D is large are plentiful in
the sense of Baire category (use the sup norm on the collection
of bounded derivatives) -- For most bounded derivatives, the set D
is the complement of a measure zero set. Note that this is much
stronger than simply saying that D has positive measure (i.e. that
the derivative isn't Riemann integrable). Note also that D and
the complement of D, for any of these Baire-typical bounded
derivatives, gives a partition of the reals into a measure zero set
and a first category set. In 1993 Bernd Kirchheim strengthened
this by proving that, given any Hausdorff measure function h,
for most bounded derivatives the set D is the complement of
a set that has Hausdorff h-measure zero. (I seem to have
left out Kirchheim's result in the first post below.)
HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES
http://groups.google.com/group/sci.math/msg/814be41b1ea8c024
Proof that derivatives have the intermediate value property
http://mathforum.org/kb/thread.jspa?messageID=5259756
A couple of years ago I was hard at work on a revision of
the "historical essay" post above, but then I had to stop and
work on other things (such as my Fall classes, which began
before I had finished). I haven't gotten back to it since
then, but it's something I intend on returning to eventually.
The revision would be several times longer and more detailed,
and what I've done so far is over twice as long already.
Dave L. Renfro
The World Wide Wade
<1162838322.9...@f16g2000cwb.googlegroups.com>,
"Amanda" <sc...@hotmail.com> wrote:
> Hello
>
> Suppose f:I -> R is differentiable on an open interval I of R. Let D be
> the set of points where f is discontinuous.
You mean f' in that last line.
> We know D is meager and,
> therefore, has an empty interior. I have 2 questions, about which I
> couldn't come to a conclusion:
>
> a) Is it possible that D is dense in I?
Yes. A basic theorem here is: Suppose f_n is a sequence of
differentiable functions on [a,b]. If f_n, f_n' converge
uniformly to f, g resp. on [a,b], then f' = g on [a,b].
So given a countable set D = {d_1, ...} in R, we can choose
functions f_n, differentiable everywhere, such that f_n' is
continuous everywhere except at d_n, and such that |f_n|, |f_n'|
< 1/2^n on R. f = f_1 + f_2 + ... will be differentiable on R,
and f' will be discontinuous precisely on D. D of course could be
dense in R.
> Since f has the intermediate
> value property, it seems to me it can't, but I couldn't come to a
> conclusion.
>
> b) D can have positive Lebesgue measure, right?
Yes, it can have full measure. Meaning, we can have R = D U E,
where m(E) = 0. The idea would be to take D = U K_n, where the
K_n's are pairwise disjoint compact sets with no interior (aka
fat Cantor sets). If you set f_n(x) =
d(x,K_n)^2*sin(1/d(x,K_n))*2^(-n), and add them up, you'll get an
example. Almost. Maybe not quite. You'd have to play around with
it a little.
> Thank you.
> Amanda
Amanda
Amanda