a lemma of Zahorski (1946)
David Bernier
the first derivative of real-valued functions.
There was this thread from 2005:
http://groups.google.com/group/sci.math/browse_thread/thread/d0e1c36090f1071e/25121c9fcec86eda?#25121c9fcec86eda
Dave Renfro mentioned Zahorski here:
< http://groups.google.com/group/sci.math/msg/2e3098dc63bce380>
The Wikipedia page on Zahorski's Theorem,
< http://en.wikipedia.org/wiki/Zahorski_theorem >
mentions a 1946 paper of Zahorski in the Bulletin de la Soci�t� Math�matique de
France, where he proves the Theorem associated to his name.
In 1937, Zahorski had published an example of what's asked for in the 2005 post.
There's a footnote in the 1946 paper mentioning the 1937 paper and that
"When constructing the function psi_1(x) I followed in part Zah 1937 and ..."
The footnote is in the section on Lemma 5, which uses Lemma 4:
Lemma 4 from [Zahorski 1946]
Suppose a, b real numbers such that a< b. Suppose C is a non-empty
G_delta set of reals contained in [a, b] and that C has measure
zero. Then for every epsilon>0 there exists a function g(x)
which is differentiable, increasing and such that:
(1) g'(x) < 1 for all x in [a, b] .
(2) for x in [a, b], g'(x) = 0 iff x is an element of C.
(3) g(b) - g(a) = b - a - epsilon.
The proof mentions the Lusin-Menschoff Theorem (unknown to me) and
"premiere classe de Baire et Young" among other things.
As a concrete case, we could have a = 0, b = 1 and C be the
Cantor set. I haven't found a g(x) with the properties
in Lemma 4 so far.
David Bernier
David C. Ullrich
This is easy since the Cantor set is compact. It would be easier to
convince you of this if I could draw a picture...
Cover C by finitely many intervals I_j such that the total length of
the I_j is small. Say the complement of
the union of the I_j is the union of finitely many intervals J_k.
Start with a continuous funtion f such that f is differentiable at
every interior point of each I_j and at every interior point of each
J_k, with f' = 1 in the interior of each J_k and f' = 0 in the
interior of each I_j. That f does almosts everything required
except that it's not differentiable at the endpoints of the
I_j. So smooth it out a little near the endpoints to make it
differentiable on all of [0,1]. Now f(1) - f(0) > 1 - epsilon/2
(if the "small" above is small enough). Finally multiply f by a
suitable lambda < 1 so that f' < 1 everywhere and also
f(1) - f(0) = 1 - epsilon.
>
>David Bernier
David Bernier
It seems to me that with your construction, there would be points
x in [0, 1] \ C such that f'(x) = 0, because the set where
f'(x) = 0 would have non-empty interior .
I had another look at the Zahorski article and I'm quite sure that
his condition (2) amounts to:
{ x in [a, b] such that g'(x) = 0 } = C .
David Bernier
karl
He outlined the starting point of the construction.
David C. Ullrich
<davi...@videotron.ca> wrote:
Yes, I read the "iff" above as "if".
With the "iff" it's still trivial for any compact set of measure zero,
only perhaps slightly less triivial than with the "if".
Say C is compact and has measure zero. The complement of C
is a countable union of disjoint open intervals I_j.
(Open intervals except that at most two of them, the ones
containing 0 and 1 if 0 and 1 are not in C, may be only
half-open. I'll let you worry about the trivial modifications
of what I say below if it happens that 0 or 1 are in C.)
For each j choose a continuous function f_j such that
0 <= f_j < 1 and such that f_j(x) > 0 if and only if
x is in I_j. Take them such that
lim_j sup_x f_j(x) = 0.
Then f = sum f_j converges uniformly to a continuous
function which is zero precisely on C. Let g be the
indefinite integral of f. Then g is continuously differentiable,
0 <=gf' < 1, and g' = 0 precisely on C.
So we're done except for g(1) = 1 - epsilon. Note as before
that it's enough to get g(1) > 1 - epsilon.
Since C has measure zero there exists N such that the sum
of the length of I_j for 1 <= j <= N is larger than 1 - epsilon/2.
Now when you choose the f_j, make certain that for every
j <= N f_j is very close to 1 everywhere on I_j except
very near the endpoints. This makes the integral of f larger
than 1 - epsilon. (And since we're only imposing the extra
condition "f_j close to 1 except near the endpoints of I_j"
for finitely many values of j this extra condition has no
effect on the uniform convergence of sum f_j.)
(Note that at first I didn't recall it was "g" in the statement
of the lemma, so I typed the above with f' = g instead
of g'=f. I went back and swapped the g's and f's - if something
above looks wrong consider the possibility that it's just a g that
should be an f or vice versa.)
David Bernier
> On Wed, 13 Jan 2010 23:39:30 -0500, David Bernier
> <davi...@videotron.ca> wrote:
>
>> David C. Ullrich wrote:
>>> On Wed, 13 Jan 2010 02:15:23 -0500, David Bernier
>>> <davi...@videotron.ca> wrote:
Thanks. I think I understand your construction now, although
graphing g'(x) over [0, 1] for some epsilon < 0.001 would be
rather hard (or maybe interesting).
If I'm not mistaken, if Psi is a bump function such as
Psi(x) = exp(-1/(1- x^2)), if |x| < 1 and
Psi(x) = 0 if |x| >= 1 as in the first example here:
< http://en.wikipedia.org/wiki/Bump_function> ,
then taking an anti-derivative of Psi(x), scaling in x and y, plus
translating along the x and y-axes as needed, can yield a C^{oo}
function which is zero for x <= - epsilon and 1 for x >= epsilon,
given some arbitrarily small epsilon> 0.
The idea of integrating bump functions arrived when thinking about
your requirement:
[ David Ullrich: ]
"Now when you choose the f_j, make certain that for every
j <= N f_j is very close to 1 everywhere on I_j except
very near the endpoints."
David Bernier