Re: sequence of functionals in l^2

On 29-04-2006 16:06, eugene wrote:

Let f_n be the linear functionals on l^2

f_n(x) = ( x_1 + x_2 + ... + x_n ) / n^(1/2).

It is easy to see that || f_n || = 1.

Does (f_n) converge poinwise to 0?
Is the problem too easy?
Yes, for both questions [I should have think more
before posting].
(it is sufficient to approximate x with a finite
supported sequence).

Why is it sufficient to approximate _x_ with
finite
supported sequences?
Yes, it's easy to see that, if _x_ is a finite
supported sequence, then
lim_n f_n(x) = 0, but how do you get the general
case
from that?

But if we denote f=lim f_n, then as well as f_n are
all
norm-bounded then by Boundedness Principle(l^2 is
Banach
space ), we have that f is also continious as all
f_n.

What do mean with f = lim_n f_n? The sequence (f_n)_n
does not
converge in (l^2)*.

Yes, i was wrong here, f(x)=lim f_n(x) wasn't supposed to exist for all x from l^2.


Another thing: why are you using the uniform
boundness principle?

i was going to deduce from it that since all f_n are continious linear functionals, the their pointwise limit f(x)=lim_n f_n(x) would also be continious, and after it since f(S)=0, where S is a set of finitely-supported sequences in l^2, we would be able to say that f(x)=0 for all x from l^2, since S is dense in l^2.

But it surely goes nowhere because of the indefinite definion of f(x), unless there were given some reasons for convergence of f_n(x) for all x from l^2.

The only thing that you can deduce from this
principle here is that
either the set { ||f_n|| | n natural } is bounded or
there's a dense
set of elements _x_ in l^2 such that the set {
|f_n(x)| | n natural }
is not bounded. Since you have already been told that
||f_n|| = 1
for each natural _n_, I really can't see which
information you expect
to extract from the principle.

Best regards,

Jose Carlos Santos
.

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