In Adams, "Calculus, A Complete Course" (third edition), sec.
13.4, the requirement is that f_{xy} and f_{yx} are
continuous at (a,b), while f, f_x and f_y are continuous in
a neighbourhood of (a,b).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
There is a proof in Advanced Calculus by Angus Taylor (later editions
added an author ) ,p220 which require the first partials only to exist
in a neighborhood of (a,b) (no continuity assumed) and the second
partials exist in a neighborhood and continuous at (a,b) (as you said).
There is a second theorem stated which require instead that the first
partials exist in a neighborhood of (a,b) and both be differentiable (
functions of (x,y)) at the point (a,b) .In this theorem no hypothesis
is required for 2nd Partials.Regards,Stuart M Newberger
If second partials exist in a neighbourhood of (a,b) and are continous
at (a,b), this automatically implies that the first partials are
continuous in same neighbourhood of (a,b) - right? So I don't see why
this book states the theorem this way.
>
> There is a second theorem stated which require instead that the first
> partials exist in a neighborhood of (a,b) and both be differentiable (
> functions of (x,y)) at the point (a,b) .In this theorem no hypothesis
> is required for 2nd Partials.Regards,Stuart M Newberger
But first partials being differentiable functions at (a,b) does say a
lot about the 2nd partials.
NO! Existence of partial derivatives of a function says nothing about
continuity of the function since partial derivatives only involve
function values on horizontal and vertical lines.
Also in the case at hand the existence only of the mixed 2nd partials
in a neighborhood is hypothesized in the first theorem,nothing needed
about f_xx and f_yy.
> > There is a second theorem stated which require instead that the first
> > partials exist in a neighborhood of (a,b) and both be differentiable (
> > functions of (x,y)) at the point (a,b) .In this theorem no hypothesis
> > is required for 2nd Partials.Regards,Stuart M Newberger
>
> But first partials being differentiable functions at (a,b) does say a
> lot about the 2nd partials.
It implies existence of all the second partials at (a,b) but says
nothing about the existence of any 2nd partials away from (a,b) .The 2
theorems are different,neither is stronger.
Another reference for all this (perhaps better is Apostol ,Tom
-Mathematical Analysis 2nd edition.This is an excellant reference for
alot of things.Regards,smn
>> To guarantee f_xy (a,b) = f_yx (a,b), do we require that
>> f_xy, f_yx be continuous at (a,b) or a stronger requirement
>> ( continuity in some neighbourhood of (a,b) )?
Robert Israel wrote:
> In Adams, "Calculus, A Complete Course" (third edition), sec.
> 13.4, the requirement is that f_{xy} and f_{yx} are
> continuous at (a,b), while f, f_x and f_y are continuous in
> a neighbourhood of (a,b).
In the 1940's the Russian mathematician Georgii Pavlovich
Tolstov [Tolstoff] published several papers on this topic.
One of his examples was a function f(x,y) with continuous
first order partial derivatives everywhere and such that
both mixed second order partial derivatives exist everywhere,
but the two mixed second order partial derivatives differ
on a set of positive measure. Another example was a function
g(x,y) with continuous first order partial derivatives
such that the mixed second order partial derivatives
exist and differ almost everywhere (i.e. on the complement
of a set of measure zero). The difference between the first
and second examples is that in his second example, at least
one of the mixed partial derivatives fails to exist at
some of the points that belong to the exceptional set of
measure zero.
In the positive direction, Tolstov proved that if all the
partial derivatives up to and including order n exist
throughout an open connected set, then: (a) each mixed partial
derivative of order less than n is independent of the order
of differentiation at each point in the domain; (b) each mixed
partial derivative of order n is independent of the order
of differentiation at almost every point in the domain;
(c) each partial derivative of order up to and including n
is a Baire one function. These results hold for functions
having an arbitrary finite number of variables.
I'm sure these results must have been improved in several
different directions by now (pun intended), but I don't
know any of the literature on this topic. The results I
gave for Tolstov are given on p. 209 of the following
paper.
A. N. Kolmogorov, V. Ya. Kozlov, D. E. Men'shov, and I. Ya
Verchenko, "Georgii Pavlovich Tolstov (on his sixtieth
birthday)", Russian Mathematical Surveys 27 #1 (1972), 207-217.
Dave L. Renfro
I had some confusions and your post forced me to clear up most of them.
I was kind of reading the theorem as all 2nd partials (not just mixed)
being continuous at (a,b).
So let me try and sum up the first theorem:
"If first partials of f are defined in some neighbourhood of (a,b) and
mixed second partials of f are continuous at (a,b), then these mixed
second partials are equal at (a,b)"
**Is there a requirement for f to be defined on some neighbourhood of
(a,b)? The version of this theorem I've encountered in school (last
sem.) also inluded this as a hypothesis.
Since f_x and f_y (only one of these needeed I believed) are defined on
some neighbourhood of (a,b), this implies that f is defined on some
neighbourhood of (a,b). Correct? Therefore I can cross the
redundant hypothesis out?
I originally had a gut feeling that this hypothesis was unecessary,
then I think I came up with a proof (I think):
f_x exists in neighbourhood of (a,b) implies that
f( p + h, q ) - f( p , q)
f_x( p,q ) = lim{h->0} ------------------------------------
h
exists for all (p,q) in some nighbourhood of (a,b)
And this limit only exists if the numerator approaches 0, which
requires the subtrahend of numerator, namely f(p,q) , exists [this can
be made more precise].
Thus f exists in some neighbourhood of (a,b).
ACTUALLY, by the same reasoning, can't we even cross out the hypothesis
dealing with f_x, f_y?
Because, f_xy being continuous at (a,b) implies:
lim{(x,y) -> (a,b)} f_xy (x,y) = f_xy (a,b)
which requires that f_xy be defined in a neighbourhood of (a,b). Which
then implies that f_x exists in some neighbourhood of (a,b) by similar
proof as above.
Similarly, continuity of f_yx at (a,b) => f_y exists in some
neighbourhood of (a,b)
So can't the theorem by stripped down to:
"If f has mixed second partials that are continuous at (a,b), then
these partials at (a,b) are equal"
Good?
And, thank you to everyone who has helped me so far {Robert Israel,
smn, Dave Renfro}.
Yes with the classical assumption that lim means limit over a full
neighborhood in R^2 .One can also define limits thru the domain of a
function even if the domain is not a full neighborhood (just that the
point where the limit is taken must be a limit point of this domain) so
authors add the redundent hypothesis for clarity.Regards,smn
> I'm sure these results must have been improved in several
> different directions by now (pun intended), but I don't
> know any of the literature on this topic. The results I
> gave for Tolstov are given on p. 209 of the following
> paper.
This was in reference to results I posted about some work
the Russian mathematician G. P. Tolstov did with mixed
partial derivatives in the 1940's.
It turns out that I do have some literature references on
this topic -- a few papers, and copies of Mathematical Reviews
for many more papers. However, essentially all of the papers
and reviews I have seem to deal with another result that
Tolstov proved in his 1949 paper, namely if f(x,y) is separately
continuous in each variable, then f(x,y) is determined by
the values it takes on a dense subset of its domain.
Nonetheless, my records show that an English translation
of Tolstov's paper exists (which I apparently don't have
a copy of).
Tolstov, "On partial derivatives" (Russian), Izvestiya Akad.
Nauk SSSR. Ser. Mat. 13 (1949), 425-446.
[MR 11,167b; Zbl 38.04003]
Tolstov, "On partial derivatives", American Mathematical
Society Translation 1952 (1952), no. 69, 30 pages.
[MR 13,926a]
Tolstov, "On partial derivatives", in "Translations, Series 1,
Volume 10: Functional Analysis and Measure Theory", American
Mathematical Society, 1962. [MR 38 #1985]
Dave L. Renfro
[snip G. P. Tolstov references on partial derivative commutativity]
I came across something that I thought would be useful
to archive in this thread.
What follows is from The American Mathematical Monthly
67 #8 (October 1960), 813-814.
----------------------------------------------------------
Discontinuous Function with Partial Derivatives Everywhere
4876 [1959, 921]. Proposed by Naoki Kimura, University of Washington
If a real valued function f(x,y) of two real variables
possesses all of its partial derivatives
( del^(m+n) )f(x,y) / (del x^m)(del y^n)
at every point, is it necessarily continuous?
Solution by John Burr, University of New England, Australia.
The following example shows that the function need not be
continuous. The function f(x,y) = exp(-x^2/y^2 - y^2/x^2)
(xy /= 0), f(0,y) = f(x,0) = 0, is discontinuous at (0,0),
since when t --> 0, f(t,t) --> exp(-2) /= f(0,0). Suppose
that it has been proved that a particular partial derivative
phi(x,y) has the properties (i) phi(0,y) = phi(x,0) = 0,
(ii) if xy /= 0 then phi(x,y) = R(x,y)*f(x,y) where R(x,y)
is a rational [function] with denominator of the form
(x^p)(y^q). Then, by (i), (phi_y)(0,y) = (phi_x)(x,0) = 0;
by (ii) (phi_x)(0,y) = 0 when y /= 0, since when x --> 0,
f(x,y) --> 0 more rapidly than any power of x; similarly
(phi_y)(x,0) = 0. Hence (phi_x)(x,y) and (phi_y)(x,y) both
have the property (i), and it is clear from (ii) that
they also have the property (ii). Since f(x,y) has these
properties, it follows by induction that all the partial
derivatives have them, and that these derivatives exist
at every point.
It may be noted that this function satisfies conditions
more stringent than those prescribed in the problem;
namely the additional conditions that the mixed partial
derivatives all exist, and are independent of the order
in which the differentiations are performed.
Also solved by George Piranian, and the proposer.
Editorial Note. Piranian asks: What are the point
sets E in the xy-plane for which there exists a function
(f(x,y); E) that is discontinuous everywhere on E and
possesses all partial derivatives everywhere in the plane?
For example, can E be everywhere dense? Can it be the
entire plane?
----------------------------------------------------------
Dave L. Renfro