Lipschitz continous?
Anders Gorst-Rasmussen
differentiable - but does the converse hold? If so, references would be
appreciated.
best,
Anders G-R.
Niclas Carlsson
>It is true, that a Lipschitz continous map f : I->R is almost everywhere
>differentiable - but does the converse hold? If so, references would be
>appreciated.
The devil's staircase (also known as the Cantor function) has f'(x) = 0 a.e.
I'd say it isn't Lipschitz.
Cheers,
Nicke
--
"A witty saying proves nothing."
- Voltaire (1694-1778)
Zdislav V. Kovarik
No, big time no.
They didn't tell you more of the story.
A Lipschitz continuous function (on an interval) is
(1) absolutely continuous,
(2) with an a.e. bounded a.e. derivative.
Violate any of (1), (2) by a function differentiable
a.e. for a counterexample. (sign(x) on [-1,1] etc.)
There are collections of counterexamples in analysis,
such as Gelbaum and Olmsted.
Cheers, ZVK(Slavek).
The World Wide Wade
"Anders Gorst-Rasmussen" <ago...@mail1.stofanet.dk> wrote:
> It is true, that a Lipschitz continous map f : I->R is almost everywhere
> differentiable - but does the converse hold? If so, references would be
> appreciated.
No, in fact f can be differentiable everywhere on I and fail to be
Lipschitz continous on I. For example, let f(x) = (x^2)sin(1/x^2). It's
easy to check that f'(x) exists for all x, but f'(x) is unbounded in every
neighborhood of 0, hence cannot be Lipshcitz in any such neighborhood.
David C. Ullrich
Exercise: Modify this example to give f which is differentiable
everywhere, such that f' is bounded but not Riemann integrable
(so in particular the RI-Fundamental Theorem of Calculus doesn't
hold.)
David C. Ullrich
The World Wide Wade
David C. Ullrich <ull...@math.okstate.edu> wrote:
> Exercise: Modify this example to give f which is differentiable
> everywhere, such that f' is bounded but not Riemann integrable
> (so in particular the RI-Fundamental Theorem of Calculus doesn't
> hold.)
I modified it and it didn't do that. How about a hint?
David C. Ullrich
Two hints:
(i) a bounded function is RI if and only if it is ae continuous
(ii) fat Cantor set.
David C. Ullrich
The World Wide Wade
David C. Ullrich <ull...@math.okstate.edu> wrote:
> Two hints:
>
> (i) a bounded function is RI if and only if it is ae continuous
> (ii) fat Cantor set.
But of course. Something like [d(x)^2]*sin(d(x)), where d(x) is the
distance from x to a fat Cantor set, should work (mmm... smooth it at the
midpoints of the complementary intervals). Nice example.
Is it possible for f to be differentialbe everwhere, with bounded
derivative that is discontinuous everwhere?
Daniel Grubb
>Is it possible for f to be differentialbe everwhere, with bounded
>derivative that is discontinuous everwhere?
Yes. Check out Stromberg's 'An Introduction to Classical Real Analysis'
in the excercises after the section entitled 'Termwise Differentiation
of Sequences'. An explicit (although complicated) example is given.
--Dan Grubb
Zdislav V. Kovarik
Am I missing something?
The "everywhere existing derivative" is a pointwise limit of a
sequence of continuous functions (f(x+1/n)-f(x))/(1/n), with
slight adjustments at the endpoints.
Such functions (of "first Baire class") are known to have
plenty of points of continuity (a set of second Baire category).
Cheers, ZVK(Slavek).
David C. Ullrich
<wadera...@attbi.remove13.com> wrote:
>In article <6pdbruk6trd8197u9...@4ax.com>,
> David C. Ullrich <ull...@math.okstate.edu> wrote:
>
>> Two hints:
>>
>> (i) a bounded function is RI if and only if it is ae continuous
>> (ii) fat Cantor set.
>
>But of course. Something like [d(x)^2]*sin(d(x)), where d(x) is the
>distance from x to a fat Cantor set, should work
Something like that - I suspect you meant something more like
[d(x)^2]*sin(1/d(x)). Something which is O(d(x)^2), hence has
derivative = 0 on the Cantor set, which is smooth in each
complementary interval, and which has a bounded derivative
with no limit at any endpoint of any such interval.
>(mmm... smooth it at the
>midpoints of the complementary intervals). Nice example.
>
>Is it possible for f to be differentialbe everwhere, with bounded
>derivative that is discontinuous everwhere?
(As has already been pointed out) no, because a derivative is the
pointwise limit of a sequence of continuous functions. (Either
Grubb was confused or thinking of a different question or ZK and
I are both very confused.)
David C. Ullrich
The World Wide Wade
David C. Ullrich <ull...@math.okstate.edu> wrote:
> >But of course. Something like [d(x)^2]*sin(d(x)), where d(x) is the
> >distance from x to a fat Cantor set, should work
>
> Something like that - I suspect you meant something more like
> [d(x)^2]*sin(1/d(x))
But of course. Thanks to Zdislav for the other answer.
Daniel Grubb
Sorry, you are right. The example in the book is not continuous
on any interval.
--Dan Grubb
Anders Gorst-Rasmussen
"Zdislav V. Kovarik" <kov...@mcmail.cis.mcmaster.ca> skrev i en meddelelse
news:Pine.SOL.4.33.02102...@mcmail.cis.mcmaster.ca...
> A Lipschitz continuous function (on an interval) is
>
> (1) absolutely continuous,
> (2) with an a.e. bounded a.e. derivative.
>
> Violate any of (1), (2) by a function differentiable
> a.e. for a counterexample. (sign(x) on [-1,1] etc.)
>
> There are collections of counterexamples in analysis,
> such as Gelbaum and Olmsted.
OK, I see this. My particular problem is, however, not that complicated. Let
I be an interval and suppose the function g(x) is
piecewise differentiable (with piecewise continous derivative) on I. If I
remember correctly, g(x) is in fact Lipschitz continous and we can
take the Lipschitz constant L to be
L=max(|g'(x)|: x \in I).
How does one prove this?
best,
Anders G-R
Dave L. Renfro
[sci.math Oct 23 2002 3:20:30:000PM]
http://mathforum.org/discuss/sci.math/m/451663/452331
wrote
> Am I missing something?
> The "everywhere existing derivative" is a pointwise limit of a
> sequence of continuous functions (f(x+1/n)-f(x))/(1/n), with
> slight adjustments at the endpoints.
> Such functions (of "first Baire class") are known to have
> plenty of points of continuity (a set of second Baire category).
Although this matter seems to have been taken care of already
in this thread (I've been away from sci.math a bit lately, working
on other things), those interested in a more complete story of
the continuity set of a derivative will want to see the following
sci.math post of mine --->>>
HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES [Jan. 23, 2000]
http://mathforum.org/discuss/sci.math/t/251588
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,
the set D is the complement of a set that has Hausdorff h-measure
zero for most bounded derivatives. (I seem to have left out
Kirchheim's result in the above essay.)
For those that are interested, note that I've included some
additional information in the references section of this post.
While I'm here, I suppose I may as well plug some of my other
more extensive posts concerning real analysis pathology --->>>
HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS [May 1, 2000]
http://mathforum.org/discuss/sci.math/t/267778
ESSAY ON NON-DIFFERENTIABILITY POINTS OF MONOTONE FUNCTIONS
[Nov. 4, 2000]
http://mathforum.org/discuss/sci.math/t/301615
ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS [May 9 and 20, 2002]
http://mathforum.org/discuss/sci.math/t/410034
Dave L. Renfro