Oscillation of function
Agapito Martinez
given x, does it make sense to define its oscillation at that point.
Is it infinite, or just undefined?
Dave L. Renfro
[sci.math 24 Apr 2002 12:34:39 -0700]
http://mathforum.org/epigone/sci.math/smixspenkhur
wrote
> If a function f is defined but unbounded in all intervals
> containing a given x, does it make sense to define its
> oscillation at that point. Is it infinite, or just undefined?
There might be situations where you'd say it was undefined,
I suppose, but I can't think of any nontrivial reason why
you'd want to do that. On the other hand, there are a lot of
reasons why you'd want to say the oscillation is infinite.
For example, f is continuous at x=b iff the oscillation of
f at x=b is zero. We'd like to be able to say that f is not
continuous at x=b iff the oscillation of f at x=b is nonzero.
However, if the oscillation can be undefined, then we'd have to
use something more awkward, such as "f is not continuous at x=b
iff the oscillation of f at x=b is nonzero or the oscillation
of f at x=b is undefined".
Let O(f,b) be the oscillation of f at x=b. Then O(f,b) is a
measure of the discontinuity of f at x=b. We can think of
O(f,b) = oo as a way of saying that f is maximally discontinuous
at x=b.
Here are some things sci.math pathologists might want to try
their hand at --->>>
1. Show that {x: O(f,x) = oo} is always a closed set.
2. Given any closed set C, does there exist a function f such
that {x: O(f,x) = oo} = C?
3. (a) If the answer to #2 is "YES", can you still do this if
f is required to be a measurable function? (b) What if f is
required to be a Baire two function? (c) What if f is required
to be a Baire one function?
[[ "f is Baire one" means that f is a pointwise limit of
continuous functions. "f is Baire two" means that f is
a pointwise limit of Baire one functions. ]]
Let f be a function and E be a nonempty set. Define O(f,E) to be
the oscillation of f on E. Thus, O(f,E) is the vertical width
of the graph of y = f(x) restricted to E, with the understanding
that this width might not be achieved (i.e. a least upper bound
is involved). Note that
O(f,b) = inf { O(f,E): E is an open interval containing b}.
This allows us to consider various refinements of oscillation
at a point, and hence more extreme discontinuity notions.
O+(f,b) = inf { O(f,E): E is a closed interval with left endpoint b}
O-(f,b) = inf { O(f,E): E is a closed interval with right endpoint b}
Op(f,b) = inf { O(f,E): E is a perfect set containing b}
Om(f,b) = inf { O(f,E): E is a positive measure set containing b}
As expected, we have O(f,b) = max{ O+(f,b), O-(f,b) }.
[[ This isn't completely trivial since I used open
intervals to define O(f,b) and closed intervals
to define the unilateral versions. ]]
4. Does there exist a function f such that
{x: O+(f,b) not equal to O-(f,b) } is uncountable?
5. Consider problems #1-3 when O(f,b) is replaced with O+(f,b).
6. Consider problems #1-3 when O(f,b) is replaced with Op(f,b).
7. Consider problems #1-3 when O(f,b) is replaced with Om(f,b).
Dave L. Renfro
David C. Ullrich
wrote:
>Agapito Martinez <agapi...@aol.com>
>[sci.math 24 Apr 2002 12:34:39 -0700]
>http://mathforum.org/epigone/sci.math/smixspenkhur
>
>wrote
>
>> If a function f is defined but unbounded in all intervals
>> containing a given x, does it make sense to define its
>> oscillation at that point. Is it infinite, or just undefined?
>
>There might be situations where you'd say it was undefined,
>I suppose, but I can't think of any nontrivial reason why
>you'd want to do that. On the other hand, there are a lot of
>reasons why you'd want to say the oscillation is infinite.
>For example, f is continuous at x=b iff the oscillation of
>f at x=b is zero. We'd like to be able to say that f is not
>continuous at x=b iff the oscillation of f at x=b is nonzero.
>However, if the oscillation can be undefined, then we'd have to
>use something more awkward, such as "f is not continuous at x=b
>iff the oscillation of f at x=b is nonzero or the oscillation
>of f at x=b is undefined".
>
>Let O(f,b) be the oscillation of f at x=b. Then O(f,b) is a
>measure of the discontinuity of f at x=b. We can think of
>O(f,b) = oo as a way of saying that f is maximally discontinuous
>at x=b.
>
>Here are some things sci.math pathologists might want to try
>their hand at --->>>
You called? (Although "pathologist" doesn't seem like quite
the right word - a pathologist diagnoses pathology, right?
Wonder what the right word is.)
>1. Show that {x: O(f,x) = oo} is always a closed set.
Ok, I've done that.
>2. Given any closed set C, does there exist a function f such
> that {x: O(f,x) = oo} = C?
Yes. (On the line, if the components of the complement of
C are the open intervals I_1, ... define f to be continuous
on each I_j behaving at the endpoints like sin(1/x)/x does
at the origin. If C has interior then let f be any bizarre
thing in the interior, like maybe f(x) = 0 if x is irrational
and q if x = p/q in lowest terms.)
>3. (a) If the answer to #2 is "YES", can you still do this if
> f is required to be a measurable function? (b) What if f is
> required to be a Baire two function? (c) What if f is required
> to be a Baire one function?
If C has empty interior it seems to me that the above gives
a Baire 1 function, and if C has nonempty interior it seems
like the above gives a Baire 2 function (because a function that
vanishes except at finitely many points is Baire 1). Probably
if C has interior a category argument shows you can't get a
Baire 1 function.
> [[ "f is Baire one" means that f is a pointwise limit of
> continuous functions. "f is Baire two" means that f is
> a pointwise limit of Baire one functions. ]]
>
>Let f be a function and E be a nonempty set. Define O(f,E) to be
>the oscillation of f on E. Thus, O(f,E) is the vertical width
>of the graph of y = f(x) restricted to E, with the understanding
>that this width might not be achieved (i.e. a least upper bound
>is involved). Note that
>
>O(f,b) = inf { O(f,E): E is an open interval containing b}.
>
>This allows us to consider various refinements of oscillation
>at a point, and hence more extreme discontinuity notions.
>
>O+(f,b) = inf { O(f,E): E is a closed interval with left endpoint b}
>
>O-(f,b) = inf { O(f,E): E is a closed interval with right endpoint b}
>
>Op(f,b) = inf { O(f,E): E is a perfect set containing b}
>
>Om(f,b) = inf { O(f,E): E is a positive measure set containing b}
>
>As expected, we have O(f,b) = max{ O+(f,b), O-(f,b) }.
>
> [[ This isn't completely trivial since I used open
> intervals to define O(f,b) and closed intervals
> to define the unilateral versions. ]]
>
>4. Does there exist a function f such that
> {x: O+(f,b) not equal to O-(f,b) } is uncountable?
Can't decide. Seems like maybe this can happen for all
but countably many points of the Cantor set, by a
simple modification of the above construction, but
I'm not sure.
>5. Consider problems #1-3 when O(f,b) is replaced with O+(f,b).
Surely it's still clear that the set where O+ is infinite is
closed, and the construction above shows that any closed set
is the set where some function has O+ = infinity?
(If x is an interior point of C we have Q+(f,x) = infinity,
and if x is in the complement of C f is continuous at x.
Say x is a boundary point of C. If there is a sequence of
points of the complement decreasing to x then either x is
an endpoint of a component of the complement or there is
a sequence of components of the complement decreasing
to x, and in either case O+(f,x) = infinity. If on the
other hand there is no sequence of points of the complement
decreasing to x then some [x, x+delta] is contained
in C, and the construction of f in the interior of C
shows that again O+(f,x) = infinity.)
>6. Consider problems #1-3 when O(f,b) is replaced with Op(f,b).
>
>7. Consider problems #1-3 when O(f,b) is replaced with Om(f,b).
It seems that the set where Om(f) = infinity is either empty
or all of R: If there exists A with m(A) > 0 such that O(f,A)
is finite then for every x the set A_x = {x} union A shows
that Om(f,x) is finite; otoh if there is no such A then
O(f,x) is infinite for every x.
Probably you get a better question if you define Om in
terms of measurable sets that have density 1 at x? Then
the set where Om = infinity need not be closed.
>Dave L. Renfro
David C. Ullrich
Dave L. Renfro
[sci.math 25 Apr 2002 07:42:43 -0700]
http://mathforum.org/epigone/sci.math/smixspenkhur
wrote
[stuff about the oscillation of a function snipped]
I thought it might be useful to put some stuff I wrote last
year about the oscillation of a function in this thread.
Dave L. Renfro
***************************************
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http://mathforum.org/epigone/sci.math/bydwehkheh/9p9ip8...@forum.mathforum.com
Subject: Re: Boundary of a Boundary?
Author: Dave L. Renfro <ren...@central.edu>
Date: 7 Apr 01 09:10:05 -0400 (EDT)
Bill Taylor <mat...@math.canterbury.ac.nz>
[sci.math 5 Apr 2001 04:49:21 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/bydwehkheh>
wrote
> In topology the boundary operator, D, is quite amusing.
[snip]
> I find this very amusing indeed. AFAIK, there is no other
> "natural" context in standard math, where we have an operator
> such that...
>
> O^3 = O^2 =/= O.
>
> i.e. - to get the "full effect" of the operator, you have to
> apply it twice! So it's not idempotent, but rather close.
> We could call it "sesqui-potent" !
Another example is the "saltus function" associated with a
real-valued function defined on an interval. Let I be an interval
and f: I --> R be a real-valued function on I. Define f^: I --> R
by f^(x) = the oscillation of f at x.
Then f^^ can differ from f^, but f^^^ = f^^ always holds. That is,
the process stablizes after two iterations, but not always after
one iteration.
Sierpinski proved this in 1910 [Bull. Acad. Sci. Cracovie, 633-634].
The result was later generalized by Henry Blumberg [Proc. National
Acad. Sci. 2 (1916), 646-649; Amer. J. Math. 41 (1919), 183-190;
Annals of Math. 18 (1916-1917), 147-160] to cases in which the
saltus is computed by allowing for the neglect of countable,
measure zero, and first category sets.
Dave L. Renfro
***************************************
***************************************
http://mathforum.org/epigone/sci.math/bydwehkheh/czomcw...@forum.mathforum.com
Subject: Re: Boundary of a Boundary?
Author: Dave L. Renfro <ren...@central.edu>
Date: 12 Apr 01 10:43:58 -0400 (EDT)
Bill Taylor <mat...@math.canterbury.ac.nz>
[sci.math 12 Apr 2001 05:01:43 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/bydwehkheh>
wrote, in reply to an earlier post of mine:
> ren...@central.edu (Dave L. Renfro) writes:
> |>
> |> Another example is the "saltus function"...
> |> ...by f^(x) = the oscillation of f at x.
>
> Sounds fun, but I'm not familiar with the term.
> What is the oscillation of f at x ?
>
> I'd guess the (limsup f - liminf f) with lim as -->x;
> but whatever it is, please tell us.
Yes, that's it. It's also equal to the limsup as delta --> 0
of |f(s) - f(t)| for s,t in the delta-neighborhood of x.
Most people encounter this in a proof that the discontinuities of
a Riemann integrable function forms a set of measure zero. [One
observes that f is continuous at x <==> the oscillation of f at x
is zero. Hence, the points of discontinuity belong to the union
as n=1 to infinity of the set of points at which the oscillation
is >= 1/n. Each of these sets can be shown to have measure zero
when f is Riemann integrable (in fact, each set is also nowhere
dense), and so their union has measure zero.]
Letting f^(x) be the oscillation of a function f at x, I mentioned
earlier that Sierpinski (1910) proved f^^ = f^^^. A reference for
this that's a bit easier to dig up than Sierpinski paper's is
Donald C. Benson, "A property of the saltus operator", Amer. Math.
Monthly 67 (1960), page 869.
There's something a bit ironic in Benson's paper, by the way.
Benson writes (using S(f) for f^):
"In this note we shall prove a conjecture of S. K. Stein,
namely that S satisfies the relationship S^2 = S^3; [...]
While this relationship is easily proved, and is possibly
well known, it does not seem to be stated explicitly in
the literature."
The irony comes in when you notice that Benson gives exactly one
reference, Goffman's 1953 text "Real Functions" (for a proof of
the fact that f^ is upper semicontinuous), and Goffman happened
to be one of Henry Blumberg's Ph.D. students. In fact, Goffman's
1953 text was dedicated to Henry Blumberg. As you may recall,
I mentioned that several of Blumberg's papers dealt with
generalizations of f^^ = f^^^. [Papers that appeared in journals
that are well known and readily available in the U.S., I might add.
(Benson was at Univ. of Calif. at Davis when his paper appeared.)]
EXTRA CREDIT: Write an essay explaining how f^^ = f^^^ can be
incorporated into logic based probability using the proximity
function.
Dave L. Renfro
***************************************
***************************************
If my comment about the "proximity function" has you curious,
then you may want to read through a few of the posts here --->>>
http://groups.google.com/groups?as_epq=proximity%20function&as_ugroup=sci.math&hl=en
Bill Taylor
|> > In topology the boundary operator, D, is quite amusing.
|> >
|> > a "natural" context in standard math, where we have an operator such that...
|> > O^3 = O^2 =/= O.
|> > i.e. - to get the "full effect" of the operator, you have to
|> > apply it twice! So it's not idempotent, but rather close.
|> > We could call it "sesqui-potent" !
|>
|> Another example is the "saltus function" associated with a
|> real-valued function defined on an interval.
|> f^(x) = the oscillation of f at x.
|> Then f^^ can differ from f^, but f^^^ = f^^ always holds.
I was almost going to ask for an example, but finally observed that
e.g. f = indicator_Q is a case in point. f^ is constant 1; f^^ is constant 0.
It's intriguing to note that this has a very similar smell to the "standard"
example for the boundary operator, namely a disk of doubly-rational points.
The two operators work very similarly - smoothing out the grosser pathologies
of the example in question first, then blanking over the remaining triviality.
I wonder if there is any toplogical/categorical connection between the two cases?
Now to finish up with a question. Either for Dave or anyone else.
The saltus operator takes f to f^; where f^(x) = limsup(|f(x+e1)-f(x+e2),
limsup as e1 & e2 --> 0.
I would like to know, is there any other neat characterization of the functions
which are the ^ of some other function? There is a trivial characterization
of being a boundary, in the other case:- any closed set is a boundary!
(At least for complete metric spaces - not sure in general.)
So is there a similar neat characterization for being a saltus function?
------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
------------------------------------------------------------------------------
French politics: vote for a crook to keep out a fascist
------------------------------------------------------------------------------
David C. Ullrich
The word "boundary" means many different things. If we mean "boundary"
as in point-set topology then the boundaries are exactly the closed
sets with empty interior.
>(At least for complete metric spaces - not sure in general.)
If X is any topological space and C is a closed subset with empty
interior it's very easy to show that C is the boundary of C:
The boundary of S is the intersection of the closure of S and the
closure of the complement of S. Say C is a closed set with empty
interior and let S be the complement of C in X. The fact that C
has empty interior shows that C is contained in the closure of
S; since C is certainly contained in the closure of C, C is
contained in the boundary of C. On the other hand the boundary
of C is a subset of the closure of C, which is C, since C is
closed.
>So is there a similar neat characterization for being a saltus function?
>
>------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
>------------------------------------------------------------------------------
> French politics: vote for a crook to keep out a fascist
>------------------------------------------------------------------------------
David C. Ullrich
Dave L. Renfro
[sci.math Mon, 29 Apr 2002 08:10:23 +0000 (UTC)]
http://mathforum.org/epigone/sci.math/smixspenkhur/aaiv5f$1l5$1...@cantuc.canterbury.ac.nz
wrote (in part):
> The saltus operator takes f to f^; where
> f^(x) = limsup(|f(x+e1)-f(x+e2), limsup as e1 & e2 --> 0.
>
> I would like to know, is there any other neat characterization
> of the functions which are the ^ of some other function? There
> is a trivial characterization of being a boundary, in the other
> case:- any closed set is a boundary! (At least for complete
> metric spaces - not sure in general.)
>
> So is there a similar neat characterization for being a
> saltus function?
I've been swamped with final exams and other end-of-the-semester
activities (why I haven't followed up on David C. Ullrich's
April 25'th post yet), but this question I can answer, although
not as completely as I'd like to. [Among other things, I wouldn't
recommend the references further on down if you're looking for a
proof of the characterization I give in the next paragraph.]
The oscillation is always upper semicontinuous (the *other*
kind of semicontinuity that arc length is), so the place to
start is by asking if every upper semicontinuous function is
the oscillation of some function. The answer is YES, but
unfortunately I wasn't able to track down who first proved this.
It's something I expected to find in Hobson's treatise ("The
Theory of Functions of a Real Variable"), Kuratowski's "Topology",
or in a few other places that I quickly looked at just now, but I
wasn't able to find it in any of these.
However, here are some references for more general results.
If f:X --> R is a real-valued function on a topological space X, we
say that F is an omega-primitive of f if the oscillation of F is f.
Every upper semicontinuous function f has an omega-primitive if
X is a Baire metric space (proved in [1]). Reference [2] generalized
this to something they call massive metric spaces, and reference [3]
gives further results for topological spaces defined by various
technical conditions ("various technical conditions" is short for
"you *really* don't want to know", or "don't ask and I won't tell").
[1] Pavel Kostyrko, "Some properties of oscillation", Math. Slovaca
30 (1980), 157-162.
[2] Zbigniew Duszynski, Zbigniew Grande, and Stanislaw P. Ponomarev,
"On the omega-primitive", Math. Slovaca 51 (2001), 469-476.
[3] Janina Ewert and Stanislaw P. Ponomarev, "Oscillation and
omega-primitives", Real Analysis Exchange 26 (2000-01), 687-702.
Dave L. Renfro
Dave L. Renfro
[sci.math Mon, 29 Apr 2002 08:10:23 +0000 (UTC)]
http://mathforum.org/epigone/sci.math/smixspenkhur
wrote (in part):
> It's intriguing to note that this has a very similar smell
> to the "standard" example for the boundary operator, namely
> a disk of doubly-rational points. The two operators work very
> similarly - smoothing out the grosser pathologies of the
> example in question first, then blanking over the remaining
> triviality.
I just read something in a math education discussion group
that explains a student error in a way I've never considered
before. It's funny enough that I thought I'd share it with
this group. I'm sticking it in this thread because it has some
connections with Bill's quasi-idempotent operator thoughts
(connections that occur on several different levels, in fact).
Dave L. Renfro
****************************************
Shelley Walsh <She...@shells.demon.co.uk>
[k12.ed.math Fri, 03 May 2002 15:26:44 GMT]
http://mathforum.org/epigone/k12.ed.math/zhaldkhaukrung
wrote (in part):
> I often think they think that it should be clear to me why they
> would think these things, and I am deliberately being hard by
> not being more understanding, but actually I am far too stupid
> to understand their way of looking at it.
>
> Here's one that I finally did get.
>
> 5-7 and 5-(-7) obviously mean the same thing.
>
> For the longest time I haven't had a clue about this. The problem
> was that to me, it seems natural, or at least it has been
> ingrained in me for as long as I can remember that in mathematical
> notation all symbols make a difference. What appears to be happening
> is that for many students this is not the case. For the student I
> was talking to about it yesterday, once it is negative it can't get
> any more negative, so the second minus sign has no effect.
****************************************
David C. Ullrich
wrote:
>Bill Taylor <mat...@math.canterbury.ac.nz>
And all these years I thought I was the only one who was that
stupid...
>> Here's one that I finally did get.
>>
>> 5-7 and 5-(-7) obviously mean the same thing.
>>
>> For the longest time I haven't had a clue about this. The problem
>> was that to me, it seems natural, or at least it has been
>> ingrained in me for as long as I can remember that in mathematical
>> notation all symbols make a difference. What appears to be happening
>> is that for many students this is not the case. For the student I
>> was talking to about it yesterday, once it is negative it can't get
>> any more negative, so the second minus sign has no effect.
Heh-heh. Your task now is to come up with a real number x such that
the numbers (-1)^n*x are all distinct.
Maybe when you finish grading those finals.
> ****************************************
David C. Ullrich