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From: Maury Barbato on 6 Aug 2010 04:12
achille wrote:

> On Aug 6, 12:33 am, Maury Barbato
> <mauriziobarb...(a)aruba.it> wrote:
> >
> > The problem is that it seems me quite hard to
> generalize
> > this argument to deal with a generic compact convex
> set.
> >
> > Best Regards,
> > Maury Barbato
>
>
> First, I don't know these stuff but how about this:
>
> Define a function \rho(x) on R^n by
>
> \rho(x) = inf { | x - z | : z \in R^n \ H }
>
> ie. \rho(x) measures how 'deep' a point x lies inside
> H.
> It is easy to check \rho(x) is continuous over R^n
> with
> support \subset H.
>
> If we can find a way to bound the 'volume' of the
> sets
>
> V_\epsilon := { x \in H : \rho(x) < \epsilon }.
>
> we can potentially let the function
>
> min( \rho(x)/\epsilon, 1 )
>
> takes the role of g(.) in Rudin's example.
>
> For simplicity, let us assume H contains B(0,r),
> a ball of radius r centered at origin 0, in its
> interior.
> For any 0 < \lambda < 1, it is easy to show
>
> \rho(x) >= (1-\lambda) r
>
> on \lambda H := { \lambda z : z \in H }.
> This means when \epsilon < r, we can 'sandwich'
> V_\epsilon between H and (1 - \epsilon/r) H, ie.
>
> V_\epsilon \subset H \ ( 1 - \epsilon/r) H
>
> and hence obtain a bound of its 'volume'.

I has a similar idea: considering homothetic imgages
of H (your \lambda H) is the most natural thing,
I believe. The point is that, given the fact that you
have to integrate with respect to each separate variable
you have to know something about "linear distances" and
not "volumes". To be more clear, of you look well at
Rudin example, the crucial point there is that you can
say "given y, the set of all x_k such that
F(y,x_k) =/= f(y,x_k) is contained in a segment whose
length is at most d". Knowing nothing about integration
in more dimensions condemn you to involved geometric
considerations ...
I wonder how Rudin coould consider this exercise a
trivial generalization of its example ...

Thank you very much, achille, for your attention.
Friendly Regards,
Maury Barbato
From: Maury Barbato on 6 Aug 2010 04:21
chip eastham wrote:

> On Aug 5, 12:33 pm, Maury Barbato
> <mauriziobarb...(a)aruba.it> wrote:
> > David Ullrich wrote:
> > > On Thu, 05 Aug 2010 05:41:30 EDT, Maury Barbato
> > > <mauriziobarb...(a)aruba.it> wrote:
> >
> > > >Hello,
> > > >could someone help me prove the following
> theorem?
> >
> > > >Let H be a compact convex set in R^k, with non
> empty
> > > >interior, and f a real function in C(H). Put
> f(x) =
> > > 0
> > > >in the complement of H, and define
> >
> > > >int_H f dx = int_{I^k} f dx,
> >
> > > >where I^k is a k-cell containing H. Then the
> > > integral
> > > >on the right side is well defined and it does
> not
> > > >depend on the order the k integrations are
> carried
> > > out.
> >
> > > ??? This is immediate from Fubini's Theorem. Are
> > > you stuck on some detail regarding the hypotheses
> of
> > > FT or something?
> >
> > > Or maybe you want a proof involving only the
> Riemann
> > > integral? If so, (i) why? The Lebesgue integral
> works
> > > better. (ii) It seems a simply modification of
> the
> > > standard
> > > proof for f continuous on I^k should suffice:
> >
> > > Say k = 2 and H is a compact subset of (0,1)^2.
> > > For each x let H_x be the set of y such that
> (x,y) is
> > > in H, and let f_x(y) = f(x,y). Then there exists
> > > [a,b] contained in (0,1) such that H_x = [a,b]
> > > and so that f_x is continuous on [a,b] and
> > > vanishes off [a,b]. Hence
> >
> > >   F(x) = int_0,1 f_x(y) dy
> >
> > > exists. It shouldn't be hard to show that F is
> also
> > > continuous on some compact subinterval of (0,1)
> > > and vanishes off this subinterval, so the
> iterated
> > > intergral exists.
> >
> > > To show the two iterated integrals are equal,
> > > again modify the proof for functions continuous
> > > on [0,1]^2, showing that the iterated integral
> > > is equal to the 2-dimensional Riemann integral.
> > > The proof should be the same, using the uniform
> > > continuity of f on H, except for a little fudge
> > > factor:
> > > Divide (0,1)^2 into n^s squares of equal size.
> > > It shouldn't be too hard to show(???) that at
> > > most O(n) of these squares intersect the
> > > boundarty of H. So those squares contribute
> > > at most O(1/n) to the 2-d integral and at
> > > most O(1/n) to the iterated integral. Otoh
> > > uniform continuity shows that the rest of the
> > > 2-d Riemann sum is withing epsilon of the
> > > corresponding part of the iterated integral...
> >
> > > >Thank you very much for your attention.
> > > >My Best Regards,
> > > >Maury Barbato
> >
> > > >PS This is an exercise in Rudin, Principles of
> > > >Mathematical Analysis (Exercise 1 in Chapter
> 10),
> > > and
> > > >it requires some explanations.
> > > >You are not allowed to use integration theory.
> All
> > > that
> > > >you know is the following.
> >
> > > >Let I^k be the k-cell in R^k defined be the
> > > inequalities
> >
> > > >a_i <= x_i <= b_i    (i=1,...,k).
> >
> > > >For every real function in C(I^k) define f =
> f_k,
> > > and
> >
> > > >f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f
> dx_k.
> >
> > > >Since f is uniformly continuous in I^k, f_{k-1}
> is
> > > >continuous. So you can repeat the process. After
> k
> > > steps
> > > >you arrive to a number f_0, and you set
> >
> > > >int_{I^k} f dx = f_0.
> >
> > > >You also know (Theorem 10.2 at page 246 of
> Rudin)
> > > that
> > > >for every f in C(I^k), int_{I^k} f dx does not
> > > depend
> > > >on the order the k integrations are carried out.
> >
> > > >Now, note that in our problem, f is in C(H), but
> of
> > > >course it can be discountuous in I^k. So neither
> the
> >
> > > >existence of the integral nor its independence
> of
> > > the
> > > >order of integration are trivial. Rudin gives
> the
> > > >following hint: approximate H by continuous
> > > functions
> > > >on R^k whose support is contained in H. Actually
> he
> > > uses
> > > >a similar technique in Example 10.4 at page 247,
> but
> > > >I can't see how it can works here.
> >
> > > >Though the exercise is thought to be trivial, I
> > > don't
> > > >know anyone who could solve it without using
> > > integration
> > > >theory!
> >
> > Thank you, prof. Ullrich for the reply, but if you
> > had read the PS of my post ... I state precisely
> > that no integration theory is allowed. All that you
> > know is Riemann integration in one dimension and
> the
> > few facts I recollected in my PS: this is the
> conundrum
> > of Rudin exercise!
> >
> > Rudin suggests to proceed as in his specific
> example,
> > which I reproduce here. Let H be the k-simplex in
> R^k
> > defined by the inequalities
> >
> > x_1 + ... + x_k <= 1,
> >
> > x_i >= 0 (i=1,...,k),
> >
> > and f a real function in C(H). Define f(x) = 0 in
> the
> > complement of H.Now let I^k be the unit cube
> defined by
> >
> > 0 <= x_i <= 1 (i=1,...,k).
> >
> > Suppose 0 < d < 1, and put
> >
> > g(t) = 1     if t <= 1 - d
> >
> > g(t) = (1-t)/d    if 1-d < t <= 1
> >
> > g(t) = 0     if 1 < t
> >
> > and define
> >
> > F(x) = g(x_1 + ... + x_k)*f(x)  
> >
> > for every x in I^k. So F is in C(I^k).
> > Put y=(x_1,...,x_{k-1}), and x=(y,x_k). For each y
> in
> > I^(k-1), the set of all x_k such that
> >
> > F(y,x_k) =/= f(y,x_k)
> >
> > is contained in a segment whose length does not
> exceed d.
> > Since 0 <= g <= 1, it follows that
> >
> > (I)  |F_{k-1}(y) - f_{k-1}(y)| < d*||f||,
> >
> > where ||f|| is the maximum of f over I^k, and
> >
> > F_{k-1}(y) = int_{0 to 1} F dx_k
> >
> > f_{k-1}(y) = int_{0 to 1} f dx_k.
> >
> > As d -> 0, (I) exhibits f_{k-1} as a uniform limit
> > of continuous functions. Thus f_{k-1} is in
> C(I^(k-1)),
> > and further integration present no problem. This
> proves
> > the existence of the integral
> >
> > int_{I^k} f dx.
> >
> > Moreover (I) shows that
> >
> > (II)  |int_{I^k} F dx - int_{I^k} f dx| <= d ||f||.
> >
> > Note that this inequaity is true, regardless of the
> order
> > in which the k single integrations are carried out
> > (repeat the preceding argument replacing x_k with
> x_i,
> > and then making the other (k-1) integrations to
> show that
> > (II) holds). Since F is in C(I^k), then for the
> Theorem
> > 10.2 (see my PS), int F is unaffected by any change
> in
> > this order. Hence (II) shows that the same is true
> of
> > int f.
> > QED
> >
> > The problem is that it seems me quite hard to
> generalize
> > this argument to deal with a generic compact convex
> set.
> >
> > Best Regards,
> > Maury Barbato
>
> It seems to me that the "missing ingredient" you
> need is the ability to decompose an arbitrary
> convex subset of R^k with nonempty interior into
> one or more k-simplices, which you've already
> shown (taking Rudin's hint) can be integrated.
>
> This is more or less an induction on dimension k.
>
> Given a decomposition of the boundary into {k-1}-
> simplices, pick a point in the interior of the
> convex subset of R^k and use that against each
> boundary {k-1}-simplex to get the induction step.
>
> regards, chip

Dear chip, unfortunately, your idea doesn't work.
A compact convex set C in R^k can be trinagulated
if and only if it is a convex polytope. Since, if
it's a convex polytope, then your inductive rasoning
shows that it admitas a trinaultaion, and conversely,
if it admits a triangulation, it is obviously a
polytope.
I think combinatorial arguments are quite unuseful in
this case, and I begin believing Rudin didn't
check this exercise before putting it into the text.

Best Regards,
Maury Barbato
From: David C. Ullrich on 6 Aug 2010 10:40
On Thu, 05 Aug 2010 12:33:58 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>David Ullrich wrote:
>
>> On Thu, 05 Aug 2010 05:41:30 EDT, Maury Barbato
>> <mauriziobarbato(a)aruba.it> wrote:
>>
>> >Hello,
>> >could someone help me prove the following theorem?
>> >
>> >Let H be a compact convex set in R^k, with non empty
>> >interior, and f a real function in C(H). Put f(x) =
>> 0
>> >in the complement of H, and define
>> >
>> >int_H f dx = int_{I^k} f dx,
>> >
>> >where I^k is a k-cell containing H. Then the
>> integral
>> >on the right side is well defined and it does not
>> >depend on the order the k integrations are carried
>> out.
>>
>> ??? This is immediate from Fubini's Theorem. Are
>> you stuck on some detail regarding the hypotheses of
>> FT or something?
>>
>> Or maybe you want a proof involving only the Riemann
>> integral? If so, (i) why? The Lebesgue integral works
>> better. (ii) It seems a simply modification of the
>> standard
>> proof for f continuous on I^k should suffice:
>>
>> Say k = 2 and H is a compact subset of (0,1)^2.
>> For each x let H_x be the set of y such that (x,y) is
>> in H, and let f_x(y) = f(x,y). Then there exists
>> [a,b] contained in (0,1) such that H_x = [a,b]
>> and so that f_x is continuous on [a,b] and
>> vanishes off [a,b]. Hence
>>
>> F(x) = int_0,1 f_x(y) dy
>>
>> exists. It shouldn't be hard to show that F is also
>> continuous on some compact subinterval of (0,1)
>> and vanishes off this subinterval, so the iterated
>> intergral exists.
>>
>> To show the two iterated integrals are equal,
>> again modify the proof for functions continuous
>> on [0,1]^2, showing that the iterated integral
>> is equal to the 2-dimensional Riemann integral.
>> The proof should be the same, using the uniform
>> continuity of f on H, except for a little fudge
>> factor:
>> Divide (0,1)^2 into n^s squares of equal size.
>> It shouldn't be too hard to show(???) that at
>> most O(n) of these squares intersect the
>> boundarty of H. So those squares contribute
>> at most O(1/n) to the 2-d integral and at
>> most O(1/n) to the iterated integral. Otoh
>> uniform continuity shows that the rest of the
>> 2-d Riemann sum is withing epsilon of the
>> corresponding part of the iterated integral...
>>
>>
>>
>> >Thank you very much for your attention.
>> >My Best Regards,
>> >Maury Barbato
>> >
>> >
>> >
>> >
>> >PS This is an exercise in Rudin, Principles of
>> >Mathematical Analysis (Exercise 1 in Chapter 10),
>> and
>> >it requires some explanations.
>> >You are not allowed to use integration theory. All
>> that
>> >you know is the following.
>> >
>> >Let I^k be the k-cell in R^k defined be the
>> inequalities
>> >
>> >a_i <= x_i <= b_i (i=1,...,k).
>> >
>> >For every real function in C(I^k) define f = f_k,
>> and
>> >
>> >f_{k-1} (x_1,..,x_{k-1}) = int_{a_k to b_k} f dx_k.
>> >
>> >Since f is uniformly continuous in I^k, f_{k-1} is
>> >continuous. So you can repeat the process. After k
>> steps
>> >you arrive to a number f_0, and you set
>> >
>> >int_{I^k} f dx = f_0.
>> >
>> >You also know (Theorem 10.2 at page 246 of Rudin)
>> that
>> >for every f in C(I^k), int_{I^k} f dx does not
>> depend
>> >on the order the k integrations are carried out.
>> >
>> >Now, note that in our problem, f is in C(H), but of
>> >course it can be discountuous in I^k. So neither the
>>
>> >existence of the integral nor its independence of
>> the
>> >order of integration are trivial. Rudin gives the
>> >following hint: approximate H by continuous
>> functions
>> >on R^k whose support is contained in H. Actually he
>> uses
>> >a similar technique in Example 10.4 at page 247, but
>> >I can't see how it can works here.
>> >
>> >Though the exercise is thought to be trivial, I
>> don't
>> >know anyone who could solve it without using
>> integration
>> >theory!
>>
>
>Thank you, prof. Ullrich for the reply, but if you
>had read the PS of my post ...

Sorry.

But in spite of not reading the PS I _did_ provide a
substantial hint. Did you read _it_? Which steps
are unclear or seem difficult?

First think about the proof in the case where f
is continuous in the square [0,1]^2. Then f
is uniformly continuous; let eps > 0, and choose
d > 0 such that etc. Now say S is a small square
S = [a, a+d/2] x [b, b+d/2]. If p is a point of S
then it's easy to see that

(*) |f(p) d^2/4 - int_a^{a+d/2} int_b^{b+d/2} f|

< eps d^2/4,

and similarly for the iterated intergral in the other
direction. Now divide [0,1]^2 into n^2 "small"
squares and apply the triangle inequality, and
you see that

|int_0^1 int_0^1 f - sum f(p) d^2/4| < eps,

and hence the difference between the two
iterated integrals is < 2 eps.

In your case you can apply (*) to the small
squares contained in H and also to the small
squares contained in the complement of H.
The extra "error" term comes from the small
squares that intersect the boundary of H.
Show that the sum of the areas of those
squares tends to 0 as n -> infinity and
you get the result (since f is bounded).

I could rephrase this a third time if you want,
but it would be better if you tried to work it out
first.

>I state precisely
>that no integration theory is allowed. All that you
>know is Riemann integration in one dimension and the
>few facts I recollected in my PS: this is the conundrum
>of Rudin exercise!
>
>Rudin suggests to proceed as in his specific example,
>which I reproduce here. Let H be the k-simplex in R^k
>defined by the inequalities
>
>x_1 + ... + x_k <= 1,
>
>x_i >= 0 (i=1,...,k),
>
>and f a real function in C(H). Define f(x) = 0 in the
>complement of H.Now let I^k be the unit cube defined by
>
>0 <= x_i <= 1 (i=1,...,k).
>
>Suppose 0 < d < 1, and put
>
>g(t) = 1 if t <= 1 - d
>
>g(t) = (1-t)/d if 1-d < t <= 1
>
>g(t) = 0 if 1 < t
>
>and define
>
>F(x) = g(x_1 + ... + x_k)*f(x)
>
>for every x in I^k. So F is in C(I^k).
>Put y=(x_1,...,x_{k-1}), and x=(y,x_k). For each y in
>I^(k-1), the set of all x_k such that
>
>F(y,x_k) =/= f(y,x_k)
>
>is contained in a segment whose length does not exceed d.
>Since 0 <= g <= 1, it follows that
>
>(I) |F_{k-1}(y) - f_{k-1}(y)| < d*||f||,
>
>where ||f|| is the maximum of f over I^k, and
>
>F_{k-1}(y) = int_{0 to 1} F dx_k
>
>f_{k-1}(y) = int_{0 to 1} f dx_k.
>
>As d -> 0, (I) exhibits f_{k-1} as a uniform limit
>of continuous functions. Thus f_{k-1} is in C(I^(k-1)),
>and further integration present no problem. This proves
>the existence of the integral
>
>int_{I^k} f dx.
>
>Moreover (I) shows that
>
>(II) |int_{I^k} F dx - int_{I^k} f dx| <= d ||f||.
>
>Note that this inequaity is true, regardless of the order
>in which the k single integrations are carried out
>(repeat the preceding argument replacing x_k with x_i,
>and then making the other (k-1) integrations to show that
>(II) holds). Since F is in C(I^k), then for the Theorem
>10.2 (see my PS), int F is unaffected by any change in
>this order. Hence (II) shows that the same is true of
>int f.
>QED
>
>The problem is that it seems me quite hard to generalize
>this argument to deal with a generic compact convex set.
>
>Best Regards,
>Maury Barbato

From: David C. Ullrich on 6 Aug 2010 10:43
On Fri, 06 Aug 2010 08:21:22 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>[...]I begin believing Rudin didn't
>check this exercise before putting it into the text.

He's proved the result already for globally continuous
functions. The solution to the exercise is straightforward;
the same idea applied to most terms plus a small
argument showing the terms where the same argument
doesn't work are not too large.

Concluding he didn't check the exercise just because you
don't see how to do it right away is a little arrogant.

>Best Regards,
>Maury Barbato

From: David C. Ullrich on 6 Aug 2010 10:44
On Fri, 06 Aug 2010 08:12:18 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>[...]
>I wonder how Rudin coould consider this exercise a
>trivial generalization of its example ...

Does he _say_ it's a trivia generalization? It is in
fact an easy generalization.

>Thank you very much, achille, for your attention.
>Friendly Regards,
>Maury Barbato

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