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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Sat Jul 15, 2006 12:10 am Post subject:
Re: analysis with uniform convergence..



"José Carlos Santos" <jcsantos@fc.up.pt> wrote in message
news:4hq0evFot49U1@individual.net...
Quote:  mina_world@hanmail.net wrote:
for each x,
there exists some M >0 such that
f_n(x) <= M , (n=1,2...)
for each e >0,
there exists d >0 such that f_n(x)  f_n(y)<e
whenever xy < d for all n.
show that lim{n>00} sup f_n(x) is continouos.

this is a exercise of uniform convergence unit.
is this a possible problem ?
what's lim{n>00} sup f_n(x) ?
i can't understand why use the sup.
What you're talking about here is the superior limit of a sequence,
which is denoted by lim sup. It is defined by
lim sup_n (a_n)_n = inf_n sup_{p >= n} a_p
It can also be seen as the greatest limit of a subsequence of (a_n)_n.
Now that you know what lim sup is, can you solve the problem?
i know the fact that
there exists a uniformly convergent subsequence of f_n
by AscoliArzela theorem.
Forget ArzelaAscoli. It's much more simple than that.
but from this,
i can't deduce that lim sup f_n(x) is continuous.
maybe, i must show that f_n(x) converges uniformly.
if f_n(x) converges pointwise, i can show it.
but this problem does not have this condition.
so, i need your advice.
Define g = sup_n f_n. Then _g_ is a real function; in fact, g(x) <= M,
for each real _x_. I'll prove that _g_ is continuous. Let _x_ be a real
number and take e > 0. There's some natural number _p_ such that
g(x)  e/2 < f_p(x) < g(x). There's also some d > 0 such that, for every
real _y_ and for every natural _n_
y  x < d => f_n(y)  f_n(x) < e/2.
Then, for each _y_ such that y  x < d,
f_p(y)  g(x) <= f_p(y)  f_p(x) + f_p(y)  g(x) < 2e/3.
In other words,
g(x)  e < f_p(y) < g(x) + e
which, together with the fact that f_p(y) <= g(y), implies that
g(x)  e < g(y). (*)
Since the number _d_ does not depend upon _x_, the same argument proves
that
g(y)  e < g(x) (**)
and (*) and (**) together are equivalent to g(y)  g(x) < e.
So, I proved that sup_n f_n is continuous and the same argument shows
that inf_n f_n is also continuous.
Now, since lim sup_n f_n = inf_p sup_{n >= p} f_n...

oh...my god. fantastic...
thank you very much. 

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José Carlos Santos science forum Guru
Joined: 25 Mar 2005
Posts: 1111

Posted: Fri Jul 14, 2006 4:52 pm Post subject:
Re: analysis with uniform convergence..



mina_world@hanmail.net wrote:
Quote:  for each x,
there exists some M >0 such that
f_n(x) <= M , (n=1,2...)
for each e >0,
there exists d >0 such that f_n(x)  f_n(y)<e
whenever xy < d for all n.
show that lim{n>00} sup f_n(x) is continouos.

this is a exercise of uniform convergence unit.
is this a possible problem ?
what's lim{n>00} sup f_n(x) ?
i can't understand why use the sup.
What you're talking about here is the superior limit of a sequence,
which is denoted by lim sup. It is defined by
lim sup_n (a_n)_n = inf_n sup_{p >= n} a_p
It can also be seen as the greatest limit of a subsequence of (a_n)_n.
Now that you know what lim sup is, can you solve the problem?
i know the fact that
there exists a uniformly convergent subsequence of f_n
by AscoliArzela theorem.

Forget ArzelaAscoli. It's much more simple than that.
Quote:  but from this,
i can't deduce that lim sup f_n(x) is continuous.
maybe, i must show that f_n(x) converges uniformly.
if f_n(x) converges pointwise, i can show it.
but this problem does not have this condition.
so, i need your advice.

Define g = sup_n f_n. Then _g_ is a real function; in fact, g(x) <= M,
for each real _x_. I'll prove that _g_ is continuous. Let _x_ be a real
number and take e > 0. There's some natural number _p_ such that
g(x)  e/2 < f_p(x) < g(x). There's also some d > 0 such that, for every
real _y_ and for every natural _n_
y  x < d => f_n(y)  f_n(x) < e/2.
Then, for each _y_ such that y  x < d,
f_p(y)  g(x) <= f_p(y)  f_p(x) + f_p(y)  g(x) < 2e/3.
In other words,
g(x)  e < f_p(y) < g(x) + e
which, together with the fact that f_p(y) <= g(y), implies that
g(x)  e < g(y). (*)
Since the number _d_ does not depend upon _x_, the same argument proves
that
g(y)  e < g(x) (**)
and (*) and (**) together are equivalent to g(y)  g(x) < e.
So, I proved that sup_n f_n is continuous and the same argument shows
that inf_n f_n is also continuous.
Now, since lim sup_n f_n = inf_p sup_{n >= p} f_n...
Best regards,
Jose Carlos Santos 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Fri Jul 14, 2006 1:19 pm Post subject:
Re: analysis with uniform convergence..



JosÃ© Carlos Santos ìž‘ì„±:
Quote:  On 14072006 9:26, mina_world wrote:
for each x,
there exists some M >0 such that
f_n(x) <= M , (n=1,2...)
for each e >0,
there exists d >0 such that f_n(x)  f_n(y)<e
whenever xy < d for all n.
show that lim{n>00} sup f_n(x) is continouos.

this is a exercise of uniform convergence unit.
is this a possible problem ?
what's lim{n>00} sup f_n(x) ?
i can't understand why use the sup.
What you're talking about here is the superior limit of a sequence,
which is denoted by lim sup. It is defined by
lim sup_n (a_n)_n = inf_n sup_{p >= n} a_p
It can also be seen as the greatest limit of a subsequence of (a_n)_n.
Now that you know what lim sup is, can you solve the problem?

i know the fact that
there exists a uniformly convergent subsequence of f_n
by AscoliArzela theorem.
but from this,
i can't deduce that lim sup f_n(x) is continuous.
maybe, i must show that f_n(x) converges uniformly.
if f_n(x) converges pointwise, i can show it.
but this problem does not have this condition.
so, i need your advice. 

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José Carlos Santos science forum Guru
Joined: 25 Mar 2005
Posts: 1111

Posted: Fri Jul 14, 2006 10:15 am Post subject:
Re: analysis with uniform convergence..



On 14072006 9:26, mina_world wrote:
Quote:  for each x,
there exists some M >0 such that
f_n(x) <= M , (n=1,2...)
for each e >0,
there exists d >0 such that f_n(x)  f_n(y)<e
whenever xy < d for all n.
show that lim{n>00} sup f_n(x) is continouos.

this is a exercise of uniform convergence unit.
is this a possible problem ?
what's lim{n>00} sup f_n(x) ?
i can't understand why use the sup.

What you're talking about here is the superior limit of a sequence,
which is denoted by lim sup. It is defined by
lim sup_n (a_n)_n = inf_n sup_{p >= n} a_p
It can also be seen as the greatest limit of a subsequence of (a_n)_n.
Now that you know what lim sup is, can you solve the problem?
Best regards,
Jose Carlos Santos 

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mina_world science forum Guru Wannabe
Joined: 20 Jul 2005
Posts: 186

Posted: Fri Jul 14, 2006 8:26 am Post subject:
analysis with uniform convergence..



hello sir~
for each x,
there exists some M >0 such that
f_n(x) <= M , (n=1,2...)
for each e >0,
there exists d >0 such that f_n(x)  f_n(y)<e
whenever xy < d for all n.
show that lim{n>00} sup f_n(x) is continouos.

this is a exercise of uniform convergence unit.
is this a possible problem ?
what's lim{n>00} sup f_n(x) ?
i can't understand why use the sup.
i know the fact that
Let {f_n} be an equicontinous sequence of fucntions.
if f_n(x) > f(x) for every x, the f_n(x) uniformly converge to f(x).
so, f(x) is continuous.
but my problem does not have the condition that
f_n(x) is pointwise convergence.
so, i think...that my problem is stange.
how do you think about it ? 

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