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analysis with uniform convergence..
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mina_world
science forum Guru Wannabe


Joined: 20 Jul 2005
Posts: 186

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

"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 |x-y| < 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 Ascoli-Arzela theorem.

Forget Arzela-Ascoli. 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

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

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 |x-y| < 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 Ascoli-Arzela theorem.

Forget Arzela-Ascoli. 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

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

José Carlos Santos 작성:

Quote:
On 14-07-2006 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 |x-y| < 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 Ascoli-Arzela 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.
Back to top
José Carlos Santos
science forum Guru


Joined: 25 Mar 2005
Posts: 1111

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

On 14-07-2006 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 |x-y| < 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

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

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 |x-y| < 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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