# second partial derivatives commute (Clairaut's Thm.)

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### G Patel

Oct 29, 2006, 9:34:28 PM10/29/06
to

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

Oct 30, 2006, 12:49:58 AM10/30/06
to

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

### smn

Oct 30, 2006, 5:57:16 AM10/30/06
to

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

### G Patel

Oct 30, 2006, 7:31:44 AM10/30/06
to

smn wrote:
>
> 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).
>

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

### smn

Oct 30, 2006, 5:35:06 PM10/30/06
to

G Patel wrote:
> smn wrote:
> >
> > 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).
> >
>
> If second partials exist in a neighbourhood of (a,b) and are continuous

> 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.

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

> > 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

### Dave L. Renfro

Oct 30, 2006, 7:23:18 PM10/30/06
to
G Patel wrote:

>> 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

### G Patel

Oct 30, 2006, 9:43:59 PM10/30/06
to

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

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}.

### smn

Oct 30, 2006, 11:42:42 PM10/30/06
to

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

### Dave L. Renfro

Nov 25, 2006, 2:46:07 PM11/25/06
to
Dave L. Renfro wrote (in part):

> 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

### Dave L. Renfro

Dec 21, 2006, 12:29:27 PM12/21/06
to
Dave L. Renfro wrote:

[snip G. P. Tolstov references on partial derivative commutativity]

I came across something that I thought would be useful

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

Feb 16, 2015, 10:47:15 AM2/16/15
to
A question: Suppose f : R^2 --> R is a function such that
D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2
and
D_1 D_2 f is continuous at (0,0).
Does it follow that (D_2 D_1 f)(0,0) exists and is equal
to (D_1 D_2 f)(0,0) ?

Thanks!

### Justin Thyme

Feb 16, 2015, 11:14:50 AM2/16/15
to
> A question: Suppose f : R^2 --> R is a function such that
> D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2
> and
> D_1 D_2 f is continuous at (0,0).

Don't you need D_2 D_1 f is continuous at (0,0) as well? With that
extra assumption D_1 and D_2 do commute. Consider

double_difference f = (f(x+h,y+h)-f(x+h,y))-(f(x,y+h)-f(x,y))

and compute lim (double_difference f/h^2) as h->0 in two ways to show
that it equals both D_1 D_2 f(x,y) and D_2 D_1 f(x,y).

> Does it follow that (D_2 D_1 f)(0,0) exists and is equal
> to (D_1 D_2 f)(0,0) ?
>
> Thanks!
>

--
Sorrow in all lands, and grievous omens.
Great anger in the dragon of the hills,
And silent now the earth's green oracles
That will not speak again of innocence.
David Sutton -- Geomancies

### Justin Thyme

Feb 16, 2015, 11:35:13 AM2/16/15
to
Justin Thyme wrote:
>> A question: Suppose f : R^2 --> R is a function such that
>> D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2
>> and
>> D_1 D_2 f is continuous at (0,0).
>
> Don't you need D_2 D_1 f is continuous at (0,0) as well? With that
> extra assumption D_1 and D_2 do commute. Consider
>
> double_difference f = (f(x+h,y+h)-f(x+h,y))-(f(x,y+h)-f(x,y))
>
> and compute lim (double_difference f/h^2) as h->0 in two ways to show
> that it equals both D_1 D_2 f(x,y) and D_2 D_1 f(x,y).
>
>> Does it follow that (D_2 D_1 f)(0,0) exists and is equal
>> to (D_1 D_2 f)(0,0) ?
>>
>> Thanks!

Note that "D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2"
is unnecessarily restrictive, being defined at (0,0) is enough.

Feb 16, 2015, 1:13:56 PM2/16/15
to
On Monday, February 16, 2015 at 10:14:50 AM UTC-6, Justin Thyme wrote:
> Don't you need D_2 D_1 f is continuous at (0,0) as well? With that
> extra assumption D_1 and D_2 do commute. Consider

I simply don't know whether continuity of D_2 D_1 f at (0,0) is needed. That's exactly the question. Can we drop that hypothesis?

Thanks!

Feb 16, 2015, 1:15:47 PM2/16/15
to
On Monday, February 16, 2015 at 10:35:13 AM UTC-6, Justin Thyme wrote:
> Note that "D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2"
> is unnecessarily restrictive, being defined at (0,0) is enough.

Are you assuming that D_2 D_1 f is continuous at (0,0)? I want to drop that, if possible. I do want to assume that D_1 D_2 f is continuous at (0,0), though.

- Scot

### Justin Thyme

Feb 16, 2015, 2:05:31 PM2/16/15
to
> On Monday, February 16, 2015 at 10:35:13 AM UTC-6, Justin Thyme wrote:
>> Note that "D_1 f, D_2 f, D_1 D_2 f are all three defined on all of R^2"
>> is unnecessarily restrictive, being defined at (0,0) is enough.
>
> Are you assuming that D_2 D_1 f is continuous at (0,0)?

Yes.

> I want to drop that, if possible. I do want to assume that D_1 D_2 f is continuous at (0,0), though.

That is an interesting question and I don't know the answer. The result
I know is that if D_1 D_2 f and D_2 D_1 f are both continuous at the
point in question, then D_1 and D_2 commute.

### Waldek Hebisch

Feb 19, 2015, 10:05:52 AM2/19/15
to
Yes. It is enough to assume that D_1 f, D_2 f, D_1 D_2 f are
defined in a neighbourhood of (0,0) and D_1 D_2 f is continuous at (0,0).

We need the following:

Fact1: if F' is bounded on [a, b] then F(b) - F(a) = \int_a^b F'(s) ds

Fact2: if g(x) = \int_0^x h(s) ds and g is differentiable at 0,
then lim inf_{s -> 0+} h(s) \leq g'(0) \leq lim sup_{s -> 0+} h(s)

Now:

1) D_1 D_2 f is continuous at (0,0) => D_1 D_2 f is bounded in
a a neighbourhood of (0,0)
2) D_1 D_2 f bounded in a rectangle [-a, a]x[-a, a] =>

(D_2f)(x, y) - (D_2f)(0, y) = \int_0^x (D_1 D_2 f)(s, y)ds

for x, y in [-a, a] =>

(f(x, y) - f(0, y)) - (f(x, 0) - f(0, 0)) =
\int_0^y\int_0^x (D_1 D_2 f)(s, t)dsdt

for x, y in [a, -a] =>

f(x, y) = \int_0^y\int_0^x (D_1 D_2 f)(s, t)dsdt +
f(x, 0) + f(0, y) - f(0, 0)

(Use twice the Fact1).

3) Rearranging 2) we get:

(f(x, y) - f(x, 0)) = \int_0^x\int_0^y(D_1 D_2 f)(s, t)dsdt +
(f(0, y) - f(0, 0))

4)
lim inf_{x -> 0+} \int_0^y (D_1 D_2 f)(x, t)dsdt \leq
(D_1f)(0, y) - (D_1f)(0, 0) \leq
lim sup_{x -> 0+} \int_0^y (D_1 D_2 f)(x, t)dsdt

5) Since D_1 D_2 f is continouos at 0 we have:

lim_{y -> 0} y^{-1}lim inf_{x -> 0+} \int_0^y (D_1 D_2 f)(x, t)dsdt =
lim_{y -> 0} y^{-1}lim sup_{x -> 0+} \int_0^y (D_1 D_2 f)(x, t)dsdt
= (D_1 D_2 f)(0, 0)

6) By pinching theorem

(D_2D_1f)(0, 0) = lim_{y -> 0} y^{-1}((D_1f)(0, y) - (D_1f)(0, 0)) =
(D_1 D_2 f)(0, 0)

--
Waldek Hebisch
heb...@math.uni.wroc.pl

Feb 19, 2015, 4:54:47 PM2/19/15
to
Very nice argument! Thanks.

Just a few remarks. Let me know if I make any mistakes.

Fact 1 is a Lebesgue integration version of the Fundamental
Theorem of Calculus. It isn't obvious, but follows from
Th 8.21, p. 179, of Rudin's "Real and Complex Analysis".
I'm sure there are many other references.

Fact 2 looks easy to prove directly from definitions.

A small typo: In Parts 4) and 5), "ds dt" should be "dt".

By Fact 2, I think you can go directly from
(f(x, y) - f(0, y)) - (f(x, 0) - f(0, 0)) =
\int_0^y \int_0^x (D_1 D_2 f)(s,t) ds dt,
for all x, y in [a, -a]
to Part 4). This skips the last half of Part 2) and all of Part 3).

Part 5) looks easy to prove directly from definitions.

I think all the one-sided limsups at 0+
could be replaced by two-sided limsups at 0.
I think all the one-sided liminfs at 0+
could be replaced by two-sided liminfs at 0.

------

Let's now *only* assume that D_1 D_2 f is continuous at (0,0)
and that (D_1 f)(0,0) exists.

By boundedness of D_1 D_2 f near (0,0), by the Mean Value Theorem,
and by the Dominated Convergence Theorem, we see:
For all y sufficiently near 0, the limit, as h --> 0, of
\int_0^y [1/h] [ ((D_2 f)(h,t)) - ((D_2 f)(0,t)) ] dt
exists, and is equal to \int_0^y (D_1 D_2 f)(0,t) dt.

By Fact 1, we see, for all (h,y) sufficiently near (0,0), that
\int_0^y [1/h] [ ((D_2 f)(h,t)) - ((D_2 f)(0,t)) ] dt
is equal to
[1/h] [ f(h,y) - f(0,y) ]
- [1/h] [ f(h,0) - f(0,0) ].
Consequently, for all y sufficiently near 0,
the limit, as h --> 0, of
[1/h] [ f(h,y) - f(0,y) ]
- [1/h] [ f(h,0) - f(0,0) ]
exists, and is equal to \int_0^y (D_1 D_2 f)(0,t) dt.

By definition of D_1, the limit, as h --> 0, of
[1/h] [ f(h,0) - f(0,0) ]
exists, and is equal to (D_1 f)(0,0).

Adding, we find, for all y sufficiently near 0, that
the limit, as h --> 0, of
[1/h] [ f(h,y) - f(0,y) ]
exists and is equal to
[ \int_0^y (D_1 D_2 f)(0,t) dt ] + [ (D_1 f)(0,0) ].
Thus, for all y sufficiently near 0, (D_1 f)(0,y) exists.

Once we know this, we can apply the argument from the previous post
(thanks, again!) to show that
(D_2 D_1 f)(0,0) exists, and is equal to
(D_1 D_2 f)(0,0).

-------

This yields, the following, which is, I think,
from a certain point of view, sharp:

Thm: Let f be a function defined on a subset of R^2.
Suppose (D_1 f)(0,0) exists.
Suppose D_1 D_2 f is continuous at (0,0).
Then (D_2 D_1 f)(0,0) exists, and is equal to
(D_1 D_2 f)(0,0).

Anyone see any mistake?

Is there a reference for this? A simpler proof?

### Waldek Hebisch

Feb 19, 2015, 11:32:25 PM2/19/15
to
Yes.

> By Fact 2, I think you can go directly from
> (f(x, y) - f(0, y)) - (f(x, 0) - f(0, 0)) =
> \int_0^y \int_0^x (D_1 D_2 f)(s,t) ds dt,
> for all x, y in [a, -a]
> to Part 4). This skips the last half of Part 2) and all of Part 3).

Well, we need to change order of integrals -- easy enough to
not write it explicitely but logically necessary...
Actually, to finish this argument we need only to know that

lim_{h->0} [1/h]\int_0^h g(t) dt = g(0)

when g is continuous at 0 and apply this to

[1/h]((D_1 f)(0, h) - (D_1 f)(0, 0)) = [1/h]\int_0^h (D_1 D_2 f)(0,t) dt

> -------
>
> This yields, the following, which is, I think,
> from a certain point of view, sharp:
>
> Thm: Let f be a function defined on a subset of R^2.
> Suppose (D_1 f)(0,0) exists.
> Suppose D_1 D_2 f is continuous at (0,0).
> Then (D_2 D_1 f)(0,0) exists, and is equal to
> (D_1 D_2 f)(0,0).
>
> Anyone see any mistake?
>
> Is there a reference for this? A simpler proof?

--
Waldek Hebisch
heb...@math.uni.wroc.pl

Feb 20, 2015, 5:08:43 PM2/20/15
to
Yes, true. I forgot to say that.

> Actually, to finish this argument we need only to know that
> lim_{h->0} [1/h]\int_0^h g(t) dt = g(0)
> when g is continuous at 0 and apply this to
> [1/h]((D_1 f)(0, h) - (D_1 f)(0, 0)) = [1/h]\int_0^h (D_1 D_2 f)(0,t) dt

But is it easy to see that
[1/h]((D_1 f)(0, h) - (D_1 f)(0, 0)) = [1/h]\int_0^h (D_1 D_2 f)(0,t) dt
?

I feel as if your original argument is needed to finish the proof, but it'd be great if there were something simpler. (Perhaps I'm misunderstanding what you're saying -- if so, sorry.)

- Scot

Feb 21, 2015, 11:03:35 AM2/21/15
to
Now I understand what you're saying. I was thinking that my argument only showed the existence of
(D_1 f)(0,h)
for h sufficiently near 0, but now I see: The same argument also gives, for small y, a formula for
(D_1 f)(0,y)
and, if, in that formula, you replace y by h, subtract (D_1 f)(0,0) and divide by h, you get
[1/h][((D_1 f)(0,h)) - ((D_1 f)(0,0))] = [1/h]\int_0^h (D_1 D_2 f)(0,t) dt
for h sufficiently near 0. Now just let h-->0.

So this uses some graduate level mathematics, but it does give the nice result:
Existence of (D_1 f)(0,0) and
continuity at (0,0) of (D_1 D_2 f)(0,0)
implies:
(D_2 D_1 f)(0,0) exists and is equal to (D_2 D_1 f)(0,0).

Still wondering about a reference and/or an undergraduate level argument for the same result.

Apr 3, 2015, 3:39:55 PM4/3/15
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> Still wondering about a reference and/or an undergraduate level argument for the same result.

Commutativity of derivatives (under the assumption that the function is twice differentiable, appropriately defined) appears as Cor (partComm) at the very end of http://www.math.umn.edu/~adams005/MATH4604/differentiability.pdf . This writeup is appropriate for an undergraduate "Advanced Calculus" course.

May 10, 2016, 3:32:03 PM5/10/16
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Essentially the same result appears in Charles Pugh's book "Real Mathematical Analysis". See Theorem 16 in Chapter 5 on p. 292 of the 2nd edition.
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Apr 3, 2018, 12:47:08 PM4/3/18
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On Saturday, April 1, 2017 at 4:14:53 PM UTC-5, adam...@umn.edu wrote:
> > 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?
>
> I have put my thoughts about this online. I would appreciate any comments or corrections (even typographical errors). See "Discussion about infinite directional differentiability" which begins near the bottom of p. 1 of
>
>
> The proof of "Fact 0.1" therein seems clear to me, although mistakes can always happen. I need to add more detail to some of the comments following that proof, but I think the statements are all correct. Again, comments/corrections are very much appreciated. I'll try to add more detail during the summer.
>
> Thanks.
>