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moubinool.omarjee

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Sep 8, 2003, 2:26:36 PM9/8/03
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Let A >1 real

Find all f such that

f : R+---> R continous such that for any x in R+
f(x) = int_[0;Ax]f(t)dt

thank you m.o.

Robert Israel

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Sep 8, 2003, 6:05:49 PM9/8/03
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In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>,

moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:
>Let A >1 real

>Find all f such that

>f : R+---> R continous such that for any x in R+
>f(x) = int_[0;Ax]f(t)dt

f(0) = 0 and f'(x) = A f(Ax).
Then f''(x) = A d/dx f(Ax) = A^2 f'(Ax) = A^3 f(A^2 x)
and f^(n)(x) = A^(n(n+1)/2) f(A^n x).
In particular, f is C^infinity and all its derivatives at 0 are 0.
So the only possible analytic solution is f(x) = 0.
I don't know if there are non-analytic solutions.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2

dmn

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Sep 8, 2003, 6:19:37 PM9/8/03
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"moubinool.omarjee" <moubinoo...@wanadoo.fr> wrote in message news:<bjig6s$3nl$1...@news-reader3.wanadoo.fr>...

You could try differentiating both sides wrt x.

This gives f'(x) = A f(Ax), maybe this is easier to deal with.

The World Wide Wade

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Sep 8, 2003, 9:19:59 PM9/8/03
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In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>,
"moubinool.omarjee" <moubinoo...@wanadoo.fr> wrote:

> Let A >1 real
>
> Find all f such that
>
> f : R+---> R continous such that for any x in R+
> f(x) = int_[0;Ax]f(t)dt

Actually A > 1 is not needed.

Theorem: Let f : [0,oo) -> R be continuous. Let g be nonnegative,
increasing, and continuous on [0,oo), with g(0) = 0, and g(x) -> oo as x ->
oo. If

|f(x)| <= |int_[0, g(x)] f(t) dt| for all x in [0,oo),

then f is identically 0.

proof/ Let x* = sup {x >= 0 : f is 0 on [0,g(x)]}. (Note f(0) = 0, so the
sup is taken over a non-empty set.) Suppose x* < oo. Then f = 0 on [0,
g(x*)]. Choose x** > x* such that g(x**) - g(x*) < 1/2. Let M be the
maximum value of |f| on [g(x*),g(x**)]. Then M = |f(x)| for some x in
[g(x*),g(x**)]. We have

M = |f(x)| <= |int_[0, g(x)] f(t) dt|

= |int_[g(x*), g(x)] f(t) dt |

<= (g(x**) - g(x*))M <= M/2.

It follows that M = 0, a contradiction by the definition of x*. Therefore
x* = oo, finishing the proof.

Robert Israel

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Sep 9, 2003, 3:30:50 AM9/9/03
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In article <waderameyxiii-554...@news.supernews.com>,

The World Wide Wade <wadera...@comcast.remove13.net> wrote:

>Actually A > 1 is not needed.

>Theorem: Let f : [0,oo) -> R be continuous. Let g be nonnegative,
>increasing, and continuous on [0,oo), with g(0) = 0, and g(x) -> oo as x ->
>oo. If

> |f(x)| <= |int_[0, g(x)] f(t) dt| for all x in [0,oo),

>then f is identically 0.

Counterexample: f(x) = x, g(x) = sqrt(2 x).

moubinool.omarjee

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Sep 9, 2003, 5:02:32 AM9/9/03
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> Let a >1 real


>
> Find all f such that
>
> f : R+---> R continous such that for any x in R+
> f(x) = int_[0;Ax]f(t)dt
>

----------------------------------------------------

f(0)=0 , f '(x) = a.f(ax)

Lf '(x ) = laplace transform = x.Lf(x) with

Lf(x) = int_R+ f(t).exp(-xt)dt

x.Lf(x) = a.Lf(ax)
ax.Lf(ax) = aLf(a^2x)
................................
a^nxLf(a^nx) = aLf(a^(n+1)x)

multiply these equality

Lf(x ) = (1/x^(n+1))(1/a^(n(n+1)/2-n-1)Lf(a^(n+1)x)

suppose x>= 1 since a>1 we tend n--->+oo

Lf(x) = 0 for x>= 1
my question is can we deduce f is 0 on [1;+oo[ ?

Thnak for your anwsers
m.o.


Herman Rubin

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Sep 9, 2003, 9:40:33 AM9/9/03
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In article <bjiuft$s74$1...@nntp.itservices.ubc.ca>,

Robert Israel <isr...@math.ubc.ca> wrote:
>In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>,
>moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:
>>Let A >1 real

>>Find all f such that

>>f : R+---> R continous such that for any x in R+
>>f(x) = int_[0;Ax]f(t)dt

>f(0) = 0 and f'(x) = A f(Ax).
>Then f''(x) = A d/dx f(Ax) = A^2 f'(Ax) = A^3 f(A^2 x)
>and f^(n)(x) = A^(n(n+1)/2) f(A^n x).
>In particular, f is C^infinity and all its derivatives at 0 are 0.
>So the only possible analytic solution is f(x) = 0.
>I don't know if there are non-analytic solutions.

No. One can bound the derivatives, and the power series
must converge. See the theory of Volterra integral equations.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

Robert Israel

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Sep 9, 2003, 2:18:45 PM9/9/03
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In article <bjk3h6$3vj$1...@news-reader5.wanadoo.fr>,
moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:


>> Let a >1 real

>> Find all f such that

>> f : R+---> R continous such that for any x in R+
>> f(x) = int_[0;Ax]f(t)dt

>f(0)=0 , f '(x) = a.f(ax)

>Lf '(x ) = laplace transform = x.Lf(x) with

>Lf(x) = int_R+ f(t).exp(-xt)dt

Note that this integral won't necessarily converge, unless you
make an assumption such as |f(t)| <= exp(ct) where x > c.

...

>Lf(x) = 0 for x>= 1
>my question is can we deduce f is 0 on [1;+oo[ ?

We could deduce f is 0 on all of R+. But only on the assumption
of an exponential bound |f(t)| <= exp(ct) for some constant c.
Otherwise the Laplace transform of f might not be defined anywhere.

G. A. Edgar

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Sep 9, 2003, 2:20:38 PM9/9/03
to
In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>, moubinool.omarjee
<moubinoo...@wanadoo.fr> wrote:


It was interesting to read the amusing replies!

The Fabius function satisfies this with A=2. This function Fb(x) is
infinitely differentiable, but nowhere analytic. It is increasing on
[0,1], Fb(0)=0, Fb(1) = 1, Fb'(x) = 2 Fb(2x). Fabius [1] initially
constructs it for x in [0,1], but it can be extended easily to all real
x, using the defining equation for x>1, and by reflection for x<0 [2].

See a picture at <http://www.math.ohio-state.edu/~edgar/selfdiff/>

references:
[1] J. Fabius, "A probabilistic example of a nowhere analytic
C^\infty-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174.

[2] K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman 7 Hall, 1994), pp.
117--120.

I have a student with some surprising results for f'(x) = a f(2x) ...
for most a, this is quite different than the case a=2. [But note this
is not the original problem of this thread.]

Here is the construction of Fabius: Let (X_k) be an iid sequence of
random variables, each uniformly distributed on [0,1]. Let X = sum(k=1
to infinity) 2^{-k} X_k. Let Fb(x) be the cumulative distribution
function of X. Check Fb(x) = integral(0 to 2x) Fb(t) dt for all x in
[0,1/2].

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

Robert Israel

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Sep 9, 2003, 2:30:34 PM9/9/03
to
In article <bjkl8h$34...@odds.stat.purdue.edu>,

Herman Rubin <hru...@odds.stat.purdue.edu> wrote:
>In article <bjiuft$s74$1...@nntp.itservices.ubc.ca>,
>Robert Israel <isr...@math.ubc.ca> wrote:
>>In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>,
>>moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:
>>>Let A >1 real

>>>Find all f such that

>>>f : R+---> R continous such that for any x in R+
>>>f(x) = int_[0;Ax]f(t)dt

>>f(0) = 0 and f'(x) = A f(Ax).
>>Then f''(x) = A d/dx f(Ax) = A^2 f'(Ax) = A^3 f(A^2 x)
>>and f^(n)(x) = A^(n(n+1)/2) f(A^n x).
>>In particular, f is C^infinity and all its derivatives at 0 are 0.
>>So the only possible analytic solution is f(x) = 0.
>>I don't know if there are non-analytic solutions.

>No. One can bound the derivatives, and the power series
>must converge. See the theory of Volterra integral equations.

How do you get a bound on the derivatives, without an a priori
bound on the function f? E.g. in another part of this thread,
an exponential bound |f(x)| <= exp(c x) allows Laplace transform
methods to be used to show f = 0. But without any bound of that
sort, I don't see how you can proceed.

Robert Israel

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Sep 9, 2003, 2:47:42 PM9/9/03
to
In article <bjk3h6$3vj$1...@news-reader5.wanadoo.fr>,
moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:


>> Let a >1 real

>> Find all f such that

>> f : R+---> R continous such that for any x in R+
>> f(x) = int_[0;Ax]f(t)dt

>f(0)=0 , f '(x) = a.f(ax)

>Lf '(x ) = laplace transform = x.Lf(x) with

>Lf(x) = int_R+ f(t).exp(-xt)dt

>x.Lf(x) = a.Lf(ax)

Oops, no:
L(f')(x) = int_0^infinity exp(-xt) a f(at) dt
= int_0^infinity exp(-xu/a) f(u) du (taking u = at)
= L(f)(x/a)

And there's no reason to think L(f) = 0 (as in fact it need not be,
if you look at the Fabius function that Gerald Edgar mentioned).

Robert Israel

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Sep 9, 2003, 2:51:47 PM9/9/03
to
In article <bjl68a$ch$1...@nntp.itservices.ubc.ca>,

Robert Israel <isr...@math.ubc.ca> wrote:
>In article <bjkl8h$34...@odds.stat.purdue.edu>,
>Herman Rubin <hru...@odds.stat.purdue.edu> wrote:
>>In article <bjiuft$s74$1...@nntp.itservices.ubc.ca>,
>>Robert Israel <isr...@math.ubc.ca> wrote:
>>>In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>,
>>>moubinool.omarjee <moubinoo...@wanadoo.fr> wrote:
>>>>Let A >1 real
>
>>>>Find all f such that
>
>>>>f : R+---> R continous such that for any x in R+
>>>>f(x) = int_[0;Ax]f(t)dt
>
>>>f(0) = 0 and f'(x) = A f(Ax).
>>>Then f''(x) = A d/dx f(Ax) = A^2 f'(Ax) = A^3 f(A^2 x)
>>>and f^(n)(x) = A^(n(n+1)/2) f(A^n x).
>>>In particular, f is C^infinity and all its derivatives at 0 are 0.
>>>So the only possible analytic solution is f(x) = 0.
>>>I don't know if there are non-analytic solutions.
>
>>No. One can bound the derivatives, and the power series
>>must converge. See the theory of Volterra integral equations.
>
>How do you get a bound on the derivatives, without an a priori
>bound on the function f? E.g. in another part of this thread,
>an exponential bound |f(x)| <= exp(c x) allows Laplace transform
>methods to be used to show f = 0. But without any bound of that
>sort, I don't see how you can proceed.

I take that back, it doesn't show anything of the sort. There are,
as Gerald Edgar mentioned, bounded nonzero solutions.

The World Wide Wade

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Sep 9, 2003, 4:20:31 PM9/9/03
to
In article <bjjvja$ea5$1...@nntp.itservices.ubc.ca>,
isr...@math.ubc.ca (Robert Israel) wrote:

> >Actually A > 1 is not needed.
>
> >Theorem: Let f : [0,oo) -> R be continuous. Let g be nonnegative,
> >increasing, and continuous on [0,oo), with g(0) = 0, and g(x) -> oo as x ->
> >oo. If
>
> > |f(x)| <= |int_[0, g(x)] f(t) dt| for all x in [0,oo),
>
> >then f is identically 0.
>
> Counterexample: f(x) = x, g(x) = sqrt(2 x).

Thanks Robert. Interesting thread. Let me at least get something right
(although way less interesting):

Theorem: Let f : [0,oo) -> R be continuous. Let g: [0,oo) -> [0,oo) satisfy
g(x) <= x. If

|f(x)| <= |int_[0, g(x)] f(t) dt| for all x in [0,oo),

then f is identically 0.

proof/ It's not hard to show f = 0 on [0,1/2] (you get M <= M*(1/2), where
M is the maximum value of |f| on [0,1/2]). Now bootstrap this up to all of
[0,oo) using intervals of length 1/2 (or give a fancier all-in-one proof).

moubinool.omarjee

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Sep 10, 2003, 7:07:41 AM9/10/03
to
Thanks Robert Israel and G.A.Edgar for your answers
m.o.


Robert Israel

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Sep 10, 2003, 7:20:49 PM9/10/03
to
In article <090920031420385801%ydp4f...@sneakemail.com>,

G. A. Edgar <ydp4f...@sneakemail.com> wrote:
>In article <bjig6s$3nl$1...@news-reader3.wanadoo.fr>, moubinool.omarjee
><moubinoo...@wanadoo.fr> wrote:

>> Let A >1 real

>> Find all f such that

>> f : R+---> R continous such that for any x in R+
>> f(x) = int_[0;Ax]f(t)dt

>> thank you m.o.

>The Fabius function satisfies this with A=2. This function Fb(x) is
>infinitely differentiable, but nowhere analytic. It is increasing on
>[0,1], Fb(0)=0, Fb(1) = 1, Fb'(x) = 2 Fb(2x). Fabius [1] initially
>constructs it for x in [0,1], but it can be extended easily to all real
>x, using the defining equation for x>1, and by reflection for x<0 [2].

>references:
>[1] J. Fabius, "A probabilistic example of a nowhere analytic
>C^\infty-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174.

>[2] K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman 7 Hall, 1994), pp.
>117--120.

>Here is the construction of Fabius: Let (X_k) be an iid sequence of


>random variables, each uniformly distributed on [0,1]. Let X = sum(k=1
>to infinity) 2^{-k} X_k. Let Fb(x) be the cumulative distribution
>function of X. Check Fb(x) = integral(0 to 2x) Fb(t) dt for all x in
>[0,1/2].

Beautiful! And more generally, for any A > 1 we can define
X = sum_{k=1}^infinity A^(-k) X_k; its cumulative distribution function
F(x) = Prob(X <= x) satisfies
F(x) = int_{Ax-1}^{Ax} F(t) dt
(which we can get by conditioning on X_1)
and thus F'(x) = A (F(Ax) - F(Ax-1)).
Let R be the set of polynomials p(t) with coefficients in {0,1}.
For x >= 0, let f(x) = sum_{p in R} (-1)^p(1) F(x-p(A))
(note that on any bounded interval, there are only finitely many nonzero
terms in the sum).
Then f'(x) = A sum_{p in R} (-1)^p(1) (F(Ax-Ap(A)) - F(Ax-Ap(A)-1))
Noting that every p in R can be uniquely written in the form
p(t) = t q(t) or t q(t)+1 for q in R, we see that
f'(x) = A sum_{q in R} (-1)^q(1) F(Ax-q(A)) = A f(Ax)

So f is a nonzero solution of moubinool's problem.

Next question: are there more solutions, or is f unique (up to constant
multiples)?

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