Re: 4 real periods

On Mon, 24 Sep 2007 18:59:21 EDT, tommy1729 <tommy1729@xxxxxxxxx>
wrote:

david wrote:

On Fri, 21 Sep 2007 18:01:43 EDT, tommy1729
<tommy1729@xxxxxxxxx>
wrote:

Assume A has the property
forall eps>0: exists P>0: forall x: |A(x)-A(x+P)|
P)| < eps
(aka. almost periodic)
and B is continuous and periodic with period Q.
Then C:=A+B is almost periodic.
Proof:
Assume eps>0 given.
Let eps_n be a sequence of positive numbers
converging to 0.
For each n, there is an eps_n-almost period P_n of
A.
The sequence P_n mod Q has an accumulation point
R.
Wlog. P_n mod Q converges to R.
Let k_m be a sequence of positive integers such
that
k_m*R mod Q converges to 0 (or Q if you like).
For each m, we have eps_n < eps/(2*k_m) for n big
enough.
By taking a subsequence of the P_n, we may assume
wlog. that
eps_n < eps/(2*k_n) for all n.
Again by taking a subsequence of the P_n (but not
the
k_n),
we may assume that k_n*P_n mod Q converges to 0
(i.e.
the
difference between k_n*P_n and the closest integer
multiple of Q
converges to 0).
Observe that k_n*P_n is an eps/2-almost period of
A
per telescope summing.

Next, the Q-periodic functions
B_n(x):=B(x)-B(x+k_n*P_n)
converge pointwise to 0.
The B_n are clearly equicontinuous and uniformly
bounded.
Hence, by Arzela-Ascoli, some subsequence
converges
uniformly
(of course also towards the pointwise limit)
Wlog. B_n -> 0 uniformly.
Thus, for n big enough, |B_n(x)| < eps/2 for all
x.
|C(x) - C(x+k_n*P_n)| <= |A(x)-A(x+k_n*P_n)| +
|B(x)-B(x+k_n*P_n)|
< eps/2 + eps/2 = eps for all x.
Thus C is almost periodic. QED
(Gee, I'm so glad eps/2 worked out. I hate it when
you
note that you have to go back and replace all
eps/2
by eps/3) ;)


Corollary: The sum for finitely many continuous
periodic functions
is almost periodic.
Proof: trivial induction.

lol

continuous :)

where did i state that restriction in the original
post :)

nice try hagman

better math than david

respect , but still not good enough to beat tommy :)

or to answer the questions from the OP.


I'm sorry that my proof is so lengthy ansd
unelegant,
but
I have read this thread and especially the
definition
of almost
periodic just 10 minutes ago and I'm rather
algebraist than analyst.

thats ok


Someone who has considered the original problem
thoroughly for some time and can tell immediately
when D. Ullrich
is wrong

i can tell you that :)

, could surely make that proof more
stringent.

continuous :)

i never restricted to that :)

even with the help of a real mathematician like
hagman , david's ideas cannot be defended :)

at least he doesnt have an additude of

i dont know and wont admit , and if i did know , i
refuse to help you and call it trivial or insult you.

You really have no idea how funny it is for _you_ to
complain about
people being insulting?

No, you never said the functions you were talking
about were
continuous.


AHA , you admit for once :-)


You just said

A = a_0 + a_1 sin(P_1 x) + a_2 sin(2 * P_1 x) + a_3
...
+ A_1 cos(P_1 x) + A_2 cos(2 * P_1 x) + A_3
... ,

which is more or less meaningless without some
comment

no it is more or less not restriction too continuous...

it is more "general".

not meaningless.

there you insulting me again ...

claiming i posted meaningless stuff.

yet you admitted you wrongly restricted to continuous and are beginning to understand the mistake in your reply.


on what sort of function you're talking about and/or
what sort of convergence you have in mind. I assumed
you

i was talking about the functions i defined of course ...

That doesn't answer the question.

in general a questions refers to the definitions given before it.

:-)


were talking about continuous functions, that being
sort
of the simplest case.

you mean : the simplest approach you had.

way to simple.


If you explain what you did
have
in mind there will be a version of almost-periodicity
that handles that case.

explained what i had in mind ????

the functions where DEFINED !!!

if you start having fantasies about things i did not say , thats your responsibility , not mine.

everything was CLEARLY defined by the series.



so regards
tommy1729


************************

David C. Ullrich

well at least your making progress by seeing your mistake

so

regards
tommy1729


************************

David C. Ullrich
.

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