Continuous function over the real plain

[William Elliot]

Assume f:R^2 -> R is continuous and
for all rectangles R with area 1, integral f dR
= integral f(x,y) dx dy over R = 0

Is f identically 0?

[quasi]

Are you posting a query from a poster in another forum?

If so, you should at least attribute the problem to that
poster.

For example, you could post:

Poster x in forum y asks the following:

... question ...

Moreover, if someone in sci.math other than you successfully
answers the question, then, if you bring the solution back to
the original forum, you should credit the solver so as not to
give the impression that it was your solution.

Ditto for the x/o problem -- I suspect that problem wasn't
yours either.

Bottom line -- give credit both to the source and the solver.

quasi

[quasi]

The "real plain"?

It's really plain to see that the English language is not
one of your strong points.

This might help:

The rane in Spane stays manely in the plane.

quasi

[David C. Ullrich]

On Thu, 20 Dec 2012 22:32:49 -0800, William Elliot <ma...@panix.com>
wrote:

>Assume f:R^2 -> R is continuous and
>for all rectangles R with area 1, integral f dR
> = integral f(x,y) dx dy over R = 0

Too many R's here. The set of real numbers is not a rectangle.

>Is f identically 0?

[David C. Ullrich]

On Thu, 20 Dec 2012 22:32:49 -0800, William Elliot <ma...@panix.com>
wrote:

>Assume f:R^2 -> R is continuous and
>for all rectangles R with area 1, integral f dR

Also that "dR" makes no sense - for every rectangle
R, the integral with respect to R? You meant integral_R f.

[William Elliot]

On Fri, 21 Dec 2012, quasi wrote:
> William Elliot wrote:
>
> >Assume f:R^2 -> R is continuous and
> >for all rectangles R with area 1, integral f dR
> > = integral f(x,y) dx dy over R = 0
> >
> >Is f identically 0?
>
> Are you posting a query from a poster in another forum?
>

It turns out that it's problem A6 of the 2012 Putnam exams.

[William Elliot]

On Fri, 21 Dec 2012, David C. Ullrich wrote:
>
> >Assume f:R^2 -> R is continuous and
> >for all rectangles R with area 1, integral f dR
>
> Also that "dR" makes no sense - for every rectangle
> R, the integral with respect to R? You meant integral_R f.
>
> > = integral f(x,y) dx dy over R = 0
> >
> >Is f identically 0?

Assume f:R^2 -> R is continuous and for all

rectangles S with area 1, integral_R f = 0.

Is f identically 0?

Does this approach work?
In every rectangle S, there's a point p in S, for which f(p) = 0.
Can I construct of set of such points that is a dense subset of R^2?

[quasi]

Right.

So then next time, if it's not your own question, you should
credit the source.

quasi

[1treePetrifiedForestLane]

... or, deal with doctor Putnam, and the Cmte.!

[quasi]

My solution to Problem A6 on the 2012 Putnam Exam ...

Problem:

If f: R^2 -> R is a continuous function such that for all
rectangular regions D in R^2 of area 1, int_D f = 0,
must f = 0?

Answer: Yes.

Proof:

Suppose f satisfies the hypothesis but f != 0.

Then f(P) != 0 for some point P.

Replacing f by -f if necessary, we can assume f(P) > 0.

Since f is continuous, f is positive in a neighborhood of P,
hence, since any rectangular region D of area 1 containing P
in its interior has int_D f = 0, it follows that the interior
of D must also contain a point Q such that f(Q) < 0.

Without loss of generality, we can translate the coordinate
system so that the origin is at P and rotate the axes so that
Q lies on the positive x-axis.

Let d be the distance from P to Q.

Then P = (0,0) and Q = (d,0).

Some terminology ...

By "adjacent rectangles" we mean two non-degenerate closed
rectangular regions whose intersection is a common edge.

It's immediate that if D1 and D2 are adjacent rectangles
then D1 U D2 is a rectangular region and

int_(D1 U D2) f = (int_D1 f) + (int_D2 f)

By "complementary rectangles" we mean two adjacent rectangles
whose union is a rectangular region of area 1.

By hypothesis, if D1 and D2 are complementary rectangles then

int_D2 f = -(int_D1 f)

Since f is positive in a neighborhood of P and negative in a
neighborhood of Q, there exists w > 0 such that f is positive
on the w by w square with vertices

(-w/2,w/2) (w/2,w/2)
(-w/2,-w/2) (w/2,-w/2)

and f is negative on the w by w square with vertices

(q - w/2,w/2) (q + w/2,w/2)
(q - w/2,-w/2) (q + w/2,-w/2)

Since the function

M(x) = (1/2) + (1/2)*x^2 +1/(x*d) - x*d

is such that

M(x) --> oo as x --> 0+,

it follows that we can choose u with 0 < u <= w such that
M(u) is a positive integer, m say, with m > 1.

Let S be the u by u square with vertices

(-u/2,u/2) (u/2,u/2)
(-u/2,-u/2) (u/2,-u/2)

and let T be the u by u square with vertices

(q - u/2,u/2) (q + u/2,u/2)
(q - u/2,-u/2) (q + u/2,-u/2)

Let a = int_S f and let b = int_T f.

Then a > 0 and b < 0.

For each positive integer n, let G_n be the rectangular region
with vertices

upper left: (-u/2+(n-1)*(u/(1-u^2)),-u/2+1/u)
upper right: (u/2+(n-1)*(u/(1-u^2)),-u/2+1/u)
lower left: (-u/2+(n-1)*(u/(1-u^2)),u/2)
lower right: (u/2+(n-1)*(u/(1-u^2)),u/2)

and let H_n be the rectangular region with vertices

upper left: (u/2+(n-1)*(u/(1-u^2)),-u/2+1/u)
upper right: (u/2+(u^3/(1-u^2))+(n-1)*(u/(1-u^2)),-u/2+1/u)
lower left: (u/2+(n-1)*(u/(1-u^2)),u/2)
lower right: (u/2+(u^3/(1-u^2))+(n-1)*(u/(1-u^2)),u/2)

It's easily verified that the following pairs of rectangular
regions are complementary:

S, G_1

G_n, H_n, for all positive integers n

H_n, G_(n+1), for all positive integers n

It follows that

int_(G_n) f = -a, for all positive integers n

int_(H_n) f = a, for all positive integers n

In particular, int_(G_m) f = -a.

By choice of m, the rectangular regions T and G_m are
complementary, so int_(G_m) f = -a implies int_T f = a.

On the other hand, we have int_T f = b, so a = b,
contradiction, since a > 0 and b < 0.

Therefore f = 0, as claimed.

quasi

[quasi]

Correction: Replace "q" by "d".

[quasi]

The same correction is needed above: Replace "q" by "d".

[quasi]

I see that my proof has still another error, so I'll post a
(hopefully) fully corrected version in a few minutes ...

quasi

[quasi]

Corrected version:

(d - w/2,w/2) (d + w/2,w/2)
(d - w/2,-w/2) (d + w/2,-w/2)

Since the function

M(x) = 1 - d*x + d/x

is such that

M(x) --> oo as x --> 0+,

it follows that we can choose u with 0 < u <= w such that
M(u) is a positive integer, m say, with m > 1.

Let S be the u by u square with vertices

(-u/2,u/2) (u/2,u/2)
(-u/2,-u/2) (u/2,-u/2)

and let T be the u by u square with vertices

(d - u/2,u/2) (d + u/2,u/2)
(d - u/2,-u/2) (d + u/2,-u/2)

[Ross A. Finlayson]

On Dec 23, 4:19 am, quasi <qu...@null.set> wrote:
> Corrected version:
>
> My solution to Problem A6 on the 2012 Putnam Exam ...
>
> Problem:
>
> If f: R^2 -> R is a continuous function such that for all
> rectangular regions D in R^2 of area 1, int_D f = 0,
> must f = 0?
>

What about a periodic wave with period 1, in two dimensions? In one
dimension, its integral is zero for any bounds of integration with
length 1. Then, given that the side length has zero under it, what
properties might hold that the unit square's function has area one and
integral zero, and this tiles the continuum and is continuous? Does
that not satisfy the hypothesis?

f_x(x) = periodic function with period one and unit area, ranging
from -1/2 to 1/2 in (-1/2, 1/2)
f(x,y) = f_x(x), periodic function in y as above translated to start
at f_x(-1/2)

Here the cross-section in x, of length 1, has area one, and,
integrating in y for each point in x has area one, what properties
hold that this 1x1=1?

A simple continuous function like 0-/\-1:

f(x) =
2x - 1/2 : 0 < x < 1/2
-(2x-1) + 1/2 : 1/2 < x < 1
f(x,y) =
f(y) - f(x): f(y) > 0, f(x) > 0
f(y) + f(x): f(y) < 0, f(x) > 0
f(y) + f(x): f(y) > 0, f(x) < 0
f(y) - f(x): f(y) < 0; f(x) < 0

Tiling the plane it is continuous, via symmetry the volume under f is
zero, for any square region of area one,

Is it not so that the integral of this function in (0,0), (1,1)
evaluates to zero? How would it be be reparable or repaired if it is
not?

In the extreme, and modeling non-real functions, what general
properties of functions would see meeting the hypothesis?

Regards,

Ross Finlayson

[quasi]

Ross A. Finlayson wrote:

quasi wrote:
>>
>> Problem:
>>
>> If f: R^2 -> R is a continuous function such that for all
>> rectangular regions D in R^2 of area 1, int_D f = 0,
>> must f = 0?
>
>What about a periodic wave with period 1, in two dimensions?
>In one dimension, its integral is zero for any bounds of
>integration with length 1. Then, given that the side length
>has zero under it, what properties might hold that the unit
>square's function has area one and integral zero, and this
>tiles the continuum and is continuous? Does that not satisfy
>the hypothesis?

If I understand your intent correctly, you are proposing a
doubly periodic continuous function f(x,y) -- periodic with
period 1 in the x-direction (for fixed y) and periodic with
period 1 in the y-direction (for fixed x), such that the
integral of f over the standard unit square is zero. Such a
function f doesn't automatically satisfy the hypothesis since
it's not clear that the integral of f is zero over all
rectangular regions of area 1.

Moreover, as I've shown in my posted proof, there are no
nonzero functions which satisfy the hypothesis.

quasi

[David C. Ullrich]

On Fri, 21 Dec 2012 20:06:02 -0800, William Elliot <ma...@panix.com>
wrote:

The question is whether you can _prove_ that _this_ set of
points is dense, not whether you can "construct" such a set.

It may well be that that set has to be dense (in fact it seems likely,
simply because it seems likely that the answer to the original
question is yes). But density certainly does not follow
from the fact that it intersects every rectangle of area 1.
This is staggeringly obvious: Let E be any set with
non-empty interior, such that E contains no rectangle
of area 1. Consider the complement of E.

[Ross A. Finlayson]

Seems clear. Basically that the tiling is regular, has that for any
given 1x1 window on the plane, it is a copy of the unit square's
window, under volume-preserving translation to integer coordinates
corners, as it is continuous. Then I was speaking to the regions of
integration being squares or rectangles with an integral side length,
not any with area one, misread the hypothesis.

Obviously this wouldn't be so for regions that are not squares with
integral multiples of the period as side length, or rectangles with
side length of integral multiples of the period.

Here the case is that the complementary rectangles you define are
disjoint and have as union area one, but, for them to be symmetrical
that int_D2 = -int_D1, the hypothesis doesn't establish that they are
complementary unless their union is a square.

Now the hypothesis has that it is for rectangular regions with area
one, is this reconstructable to square regions with area one? No,
tiling the rectangles with side length not-necessarily-one doesn't
leave the function continuous, for doubly-periodic functions with
period one and as above. If the hypothesis were:

"If f: R^2 -> R is a continuous function such that for all _square_

regions D in R^2 of area 1, int_D f = 0, must f = 0?"

then the answer would be no. As well, if the hypothesis were

"If f: R^2 -> R is a continuous function such that for _given
rectangular_ regions _of fixed dimensions_ D in R^2 of area 1, int_D f

= 0, must f = 0?"

then the answer would also be no. Also:

"If f: R^2 -> R is a continuous function such that for all

_rectangular_ regions _with an integral side length_ D in R^2 of area

1, int_D f = 0, must f = 0?"

then again the answer would be no.

I agree that for this continuous tiling of the plane, that there are
rectangular regions that as tiled, are not continuous. For simple
conditions on the regions and function, there are tilings of the plane
with are solutions that aren't necessarily trivial.

"In the extreme, and modeling non-real functions, what general
properties of functions would see meeting the hypothesis?"

Basically this is for that: modifying the region to be the unit circle
instead of the unit square, they aren't readily apparent properties of
symmetry in the square coordinates of the x,y plane that would see a
periodic function in x, and in y given x, define a continuous (square)
tiling of regions with the properties, and symmetry about any point in
the plane, for unit square regions, for unit circle regions. Yet,
there are, in the extreme.

So, in real functions, square regions, no, rectangular regions with
square integral multiples of side-lengths, no, rectangles, yes,
circles, yes. Then in non-real functions, basically periodic
functions with dense crests, no.

Regards,

Ross Finlayson

[Mike Terry]

"quasi" <qu...@null.set> wrote in message
news:k6tdd897t057g58mr...@4ax.com...

Hi quasi,

I like this proof very much, but I feel you have not presented it as well as
you could. The basic idea is very simple, but your presentation is full of
mysterious coordinates, and very little motivating text that would explain
how the proof actually works. This would surely put off a lot of people
from following your proof, so I'll have a go at explaining the simple idea
behind it.

The basic idea is that *if* we have a small square A of length u where
Integral f >0 on A, then we can construct another square E, offset from A by
a small amount, where also Integral f > 0 on E. (Then by repetition, a
whole sequence of further squares where Integral f > 0 on each square).

To construct E given A, first construct a long thin rectangle B above A so
that area (A+B) = 1.

Then we construct a thin rectangle C to the right of B, so that area (B+C) =
1, and then rectangle D to the right of C so that area (C+D) = 1.

Finally we construct a rectangle E beneath D so that area (D+E) = 1.

Here's a picture:

BBBCDDD
BBBCDDD
BBBCDDD
BBBCDDD
BBBCDDD
BBBCDDD
BBBCDDD
BBBCDDD
AAA EEE
AAA EEE
AAA EEE

A & B are examples of complementary rectangles, in your terminology.
Similarly B&C, C&D, and D&E are complementary.

Clearly B and D are similar, and so A and E are similar, i.e. E is another
square
like A, but shifted to the right a bit. Looking at the integrals of f on
each region, we have:

Int_A f = -Int_B f = Int_C f = -Int_D f = Int_E f

[using Int_X f to mean the integral of f over rectangle X]

So E, like A has a positive integral for f, as claimed.

The above process can be repeated to make further new squares with positive
integral for f, stepping to the right by the same amount with each
repetition of the process.

Once this basic "stepping" idea is understood, the proof is straight
forward:

1. If f is not identically zero, there is a point where it is >0
2. ..which in turn implies there is another point where it is <0
(as you explain perfectly well, so I won't repeat it)
3. Continuity of f means we can arrange things and choose coords
etc. so that there is a little square at the origin where f>0
and another square somewhere off to the right where f<0.
We take the offset distance to be d.
4. But now we can choose an even smaller square A at the origin
where f>0, and if we follow the stepping procedure above we
can show there are a sequence of squares offset from A where
f>0 on each square, and we can choose the size of A so that
one of these squares lands smack on our square where supposedly
f<0, contradiction.

The unclear bit is maybe in (4), but it's easy when we see what we're trying
to do! :)

If our square A has length u, what is the "offset" to our next square E (in
the diagram above)?

A has length u
B has width u, length (1/u)-u = (1 - u^2)/u
so B+C has width u/(1 - u^2), which is our offset value

If our square where supposedly f<0 is offset from A by distance d, we simply
have to choose u so that

m * u/(1 - u^2) = d

for a suitable positive integer m. m will be the number of times we need to
repeat the stepping process to reach the square which reveals the
contradiction. We can choose such an m because the offset is a continuous
function of u, converging to zero as u tends to zero...

(Hmmm, this doesn't look quite the same as your calculation for suitable m
????)


Anyway, I hope this saves people the trouble of trying to work out the logic
behind your proof from the coordinate expressions you give, as the logic is
simple, but the expressions in your proof do not really bring this out.

I hasten to add that all the above is there in your proof and I'm just
explaining it - I don't know whether I'd have worked this out without seeing
your proof, so all credit goes to you! :)

Regards,
Mike.

[Mike Terry]

"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message
news:-_ednfU2lOrMxUrN...@brightview.co.uk...

It occurs to me that since the stepping procedure can be applied both
horizontally and vertically, an adaptation of quasi's proof solves the
problem with slightly weaker hypotheses:

If f: R^2 -> R is a continuous function such that for all

*ALLIGNED* rectangular regions D in R^2 of area 1, int_D f = 0,
must f = 0?

i.e. the result remains true if we are just given int_D f = 0 for rectangles
D that are products of intervals [a,b] and [c,d] in the underlying
coordinates of R^2. (Quasi's proof uses a rotation at one point to bring
two positive and negative points of f onto the x-axis, but we can work
around this without using any rotation by "stepping" both horizontally and
vertically rather than just horizontally in the proof. With this change,
all the rectangles used in the proof are "alligned".)

Also, the same "stepping" argument works fine for higher dimensional R^n,
but luckily the stepping process breaks down in R^1. (Luckily, as the
analogous result is clearly false for R^1! :))


Mike.

[quasi]

Mike Terry wrote:

Yes, I left out much of the motivation (for lack of time), and
felt guilty about it, so I'm glad that you worked through the
proof and recovered the underlying ideas.

>The basic idea is that *if* we have a small square A of length
>u where Integral f > 0 on A, then we can construct another
>square E, offset from A by a small amount, where also Integral
>f > 0 on E. (Then by repetition, a whole sequence of further
>squares where Integral f > 0 on each square).
>
>To construct E given A, first construct a long thin rectangle
>B above A so that area (A+B) = 1.
>
>Then we construct a thin rectangle C to the right of B, so that
>area (B+C) = 1, and then rectangle D to the right of C so that
>area (C+D) = 1.
>
>Finally we construct a rectangle E beneath D so that area
>(D+E) = 1.
>
>Here's a picture:
>
> BBBCDDD
> BBBCDDD
> BBBCDDD
> BBBCDDD
> BBBCDDD
> BBBCDDD
> BBBCDDD
> BBBCDDD
> AAA EEE
> AAA EEE
> AAA EEE

Nice diagram.

The idea is right, but I differ on the details.

The number of B type rectangles is one more than the number of
C type rectangles. Also, d is the distance between the centers
of the initial and final u by u squares, so the total width of
the B and C type rectangles is u more than d. Thus, to solve
for m, the correct equation is

(m-1)*(u/(1 - u^2)) + u = d + u

or equivalently,

(m-1)*(u/(1 - u^2)) = d

>for a suitable positive integer m. m will be the number of
>times we need to repeat the stepping process to reach the
>square which reveals the contradiction. We can choose such
>an m because the offset is a continuous function of u,
>converging to zero as u tends to zero...

Or as I explained it, m is a continuous function of u which
approaches infinity as u approaches 0 from the right, hence
we can choose u sufficiently small to guarantee both

f > 0 on A

m is an integer greater than 1

>(Hmmm, this doesn't look quite the same as your calculation
>for suitable m ????)

I think the discrepancy is explained by the error I noted
in your equation for m.

>Anyway, I hope this saves people the trouble of trying to work
>out the logic behind your proof from the coordinate expressions
>you give, as the logic is simple, but the expressions in your
>proof do not really bring this out.
>
>I hasten to add that all the above is there in your proof and
>I'm just explaining it - I don't know whether I'd have worked
>this out without seeing your proof, so all credit goes to you!
>:)

Nice work.

You definitely provided a clear and understandable road map,
and motivated the key ideas very well.

Thank you.

quasi

[quasi]

Mike Terry wrote:
>
>It occurs to me that since the stepping procedure can be
>applied both horizontally and vertically, an adaptation of
>quasi's proof solves the problem with slightly weaker
>hypotheses:
>
> If f: R^2 -> R is a continuous function such that for all

> *ALIGNED* rectangular regions D in R^2 of area 1,

> int_D f = 0, must f = 0?
>
>i.e. the result remains true if we are just given int_D f = 0
>for rectangles D that are products of intervals [a,b] and [c,d]
>in the underlying coordinates of R^2. (Quasi's proof uses a
>rotation at one point to bring two positive and negative points
>of f onto the x-axis, but we can work around this without using
>any rotation by "stepping" both horizontally and vertically
>rather than just horizontally in the proof.

Yes, that avoids the need for rotation -- nice.

>With this change, all the rectangles used in the proof are

> aligned".)

>
>Also, the same "stepping" argument works fine for higher
>dimensional R^n, but luckily the stepping process breaks down
>in R^1. (Luckily, as the analogous result is clearly false
>for R^1! :))

Agreed, the analogous result is true in R^n for all n > 1.

A followup question:

If f: R^2 -> R is a continuous function such that int_D f = 0
for all square regions D of area 1 with edges parallel to the
axes, must f = 0?

I suspect the answer is yes.

quasi

[quasi]

quasi wrote:
>
>A followup question:
>
>If f: R^2 -> R is a continuous function such that int_D f = 0
>for all square regions D of area 1 with edges parallel to the
>axes, must f = 0?
>
>I suspect the answer is yes.

No, that's silly.

Ross Finlayson's doubly periodic idea yields a counterexample.

Ok, then how about this question instead ...

If f: R^2 -> R is a continuous function such that int_D f = 0
for all square regions D of area 1 with edges parallel to the

axes, must it be true that f(x+1,y) = f(x,y) = f(x,y+1)
for all x,y?

quasi

[William Elliot]

On Sun, 23 Dec 2012, David C. Ullrich wrote:

> >Assume f:R^2 -> R is continuous and for all

> >rectangles S with area 1, integral_R f = 0.
> >
> >Is f identically 0?
> >
> >Does this approach work?
> >In every rectangle S, there's a point p in S, for which f(p) = 0.
> >Can I construct of set of such points that is a dense subset of R^2?
>
> The question is whether you can _prove_ that _this_ set of
> points is dense, not whether you can "construct" such a set.
>
> It may well be that that set has to be dense (in fact it seems likely,
> simply because it seems likely that the answer to the original
> question is yes). But density certainly does not follow
> from the fact that it intersects every rectangle of area 1.

> This is staggeringly obvious: Let E be any set with
> non-empty interior, such that E contains no rectangle
> of area 1. Consider the complement of E.

E^c may or may not contain a rectangle of area one.
Even so, E may contain a zero point (a point p with f(p) = 0).
So what does that show?

[David C. Ullrich]

On Sun, 23 Dec 2012 18:55:29 -0800, William Elliot <ma...@panix.com>
wrote:

Jeez. The point is that E^c intersects every rectangle
of area 1, but is not dense.

[William Elliot]

On Mon, 24 Dec 2012, David C. Ullrich wrote:
> >
> >> >Assume f:R^2 -> R is continuous and for all
> >> >rectangles S with area 1, integral_R f = 0.
> >> >
> >> >Is f identically 0?
> >> >
> >> >Does this approach work?
> >> >In every rectangle S, there's a point p in S, for which f(p) = 0.

> >> >Can I prove the set of such points is a dense subset of R^2?

> >>
> >> It may well be that that set has to be dense (in fact it seems likely,
> >> simply because it seems likely that the answer to the original
> >> question is yes). But density certainly does not follow
> >> from the fact that it intersects every rectangle of area 1.
> >
> >> This is staggeringly obvious: Let E be any set with
> >> non-empty interior, such that E contains no rectangle
> >> of area 1. Consider the complement of E.
> >
> >E^c may or may not contain a rectangle of area one.
> >Even so, E may contain a zero point (a point p with f(p) = 0).
> >So what does that show?
>
> Jeez. The point is that E^c intersects every rectangle
> of area 1, but is not dense.

Even so, E^c doesn't have to contain all the zero points.
What are you trying to show?
That D = { p | f(p) = 0 } isn't dense?
I don't see you've done that.

[David C. Ullrich]

On Tue, 25 Dec 2012 00:50:42 -0800, William Elliot <ma...@panix.com>
wrote:

Of course I haven't done that. I didn't say I was doing that.
I said exactly what I was trying to show, in very simple
English: The fact that D intersects every rectangle of
area 1 does not imply that D is dense.

[William Elliot]

Of course not; that collection of sets isn't a base, it's
a subbase. To proof it's dense requires something more.