Calculate the limit of the limits of sequences
Suposse the sequence $x^{n}$ such that
$$x^{n}= (x^{n}_1, x^{n}_2,..., x^{n}_m,...),quad 0 leq x^{n}_m leq 1, forall n,m in mathbb{N}$$
$$lim_{mto infty} x^{n}_m = 1, quad forall n in mathbb{N}$$
Also, suposse that $lim_{nto infty} x^{n}_m = x_m$. I want to prove that $lim_{mto infty} x_m = 1$. The following array is more explanatory:
begin{array}{ccccccc}
x^{1}_1 & x^{1}_2 & x^{1}_3 & ... & x^{1}_m & cdots to & 1 \
x^{2}_1 & x^{2}_2 & x^{2}_3 & ... & x^{2}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
x^{n}_1 & x^{n}_2 & x^{n}_3 & ... & x^{n}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
downarrow & downarrow & downarrow & ... & downarrow & cdots & vdots \
x_1 & x_2 & x_3 & ... & x_m & cdots to & ? \
end{array}
I tried to start with the classical approach: if $epsilon > 0$, I want find some $m_0in mathbb{N}$ such that
$$|1 -x_m |< epsilon, quad forall m> m_0$$. But I can not engage the triangular inequality because of the two indices. I would just like an initial hint. Some ideia or other approach?
real-analysis sequences-and-series limits
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
asked Dec 9 at 9:06
Fam
32
add a comment |
Suposse the sequence $x^{n}$ such that
$$x^{n}= (x^{n}_1, x^{n}_2,..., x^{n}_m,...),quad 0 leq x^{n}_m leq 1, forall n,m in mathbb{N}$$
$$lim_{mto infty} x^{n}_m = 1, quad forall n in mathbb{N}$$
Also, suposse that $lim_{nto infty} x^{n}_m = x_m$. I want to prove that $lim_{mto infty} x_m = 1$. The following array is more explanatory:
begin{array}{ccccccc}
x^{1}_1 & x^{1}_2 & x^{1}_3 & ... & x^{1}_m & cdots to & 1 \
x^{2}_1 & x^{2}_2 & x^{2}_3 & ... & x^{2}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
x^{n}_1 & x^{n}_2 & x^{n}_3 & ... & x^{n}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
downarrow & downarrow & downarrow & ... & downarrow & cdots & vdots \
x_1 & x_2 & x_3 & ... & x_m & cdots to & ? \
end{array}
I tried to start with the classical approach: if $epsilon > 0$, I want find some $m_0in mathbb{N}$ such that
$$|1 -x_m |< epsilon, quad forall m> m_0$$. But I can not engage the triangular inequality because of the two indices. I would just like an initial hint. Some ideia or other approach?
real-analysis sequences-and-series limits
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
asked Dec 9 at 9:06
Fam
32
Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03
add a comment |
Suposse the sequence $x^{n}$ such that
$$x^{n}= (x^{n}_1, x^{n}_2,..., x^{n}_m,...),quad 0 leq x^{n}_m leq 1, forall n,m in mathbb{N}$$
$$lim_{mto infty} x^{n}_m = 1, quad forall n in mathbb{N}$$
Also, suposse that $lim_{nto infty} x^{n}_m = x_m$. I want to prove that $lim_{mto infty} x_m = 1$. The following array is more explanatory:
begin{array}{ccccccc}
x^{1}_1 & x^{1}_2 & x^{1}_3 & ... & x^{1}_m & cdots to & 1 \
x^{2}_1 & x^{2}_2 & x^{2}_3 & ... & x^{2}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
x^{n}_1 & x^{n}_2 & x^{n}_3 & ... & x^{n}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
downarrow & downarrow & downarrow & ... & downarrow & cdots & vdots \
x_1 & x_2 & x_3 & ... & x_m & cdots to & ? \
end{array}
I tried to start with the classical approach: if $epsilon > 0$, I want find some $m_0in mathbb{N}$ such that
$$|1 -x_m |< epsilon, quad forall m> m_0$$. But I can not engage the triangular inequality because of the two indices. I would just like an initial hint. Some ideia or other approach?
real-analysis sequences-and-series limits
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
asked Dec 9 at 9:06
Fam
32
Suposse the sequence $x^{n}$ such that
$$x^{n}= (x^{n}_1, x^{n}_2,..., x^{n}_m,...),quad 0 leq x^{n}_m leq 1, forall n,m in mathbb{N}$$
$$lim_{mto infty} x^{n}_m = 1, quad forall n in mathbb{N}$$
Also, suposse that $lim_{nto infty} x^{n}_m = x_m$. I want to prove that $lim_{mto infty} x_m = 1$. The following array is more explanatory:
begin{array}{ccccccc}
x^{1}_1 & x^{1}_2 & x^{1}_3 & ... & x^{1}_m & cdots to & 1 \
x^{2}_1 & x^{2}_2 & x^{2}_3 & ... & x^{2}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
x^{n}_1 & x^{n}_2 & x^{n}_3 & ... & x^{n}_m & cdots to & 1 \
vdots & vdots & vdots & & vdots & vdots &vdots & \
downarrow & downarrow & downarrow & ... & downarrow & cdots & vdots \
x_1 & x_2 & x_3 & ... & x_m & cdots to & ? \
end{array}
I tried to start with the classical approach: if $epsilon > 0$, I want find some $m_0in mathbb{N}$ such that
$$|1 -x_m |< epsilon, quad forall m> m_0$$. But I can not engage the triangular inequality because of the two indices. I would just like an initial hint. Some ideia or other approach?
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
asked Dec 9 at 9:06
Fam
32
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
asked Dec 9 at 9:06
Fam
32
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
edited Dec 9 at 9:37
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Dec 9 at 9:06
Fam
32
asked Dec 9 at 9:06
Fam
32
asked Dec 9 at 9:06
Fam
32
32
Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03
add a comment |
Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03
Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03
Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03
add a comment |
2 Answers
2
active
oldest
votes
Unfortunately, what you are trying to prove
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{mtoinfty}lim_{ntoinfty}x^n_mquadcdots(*)
$$, which is changing the order of the limits, is not true in general. We can see this
by the example $x^n_m = 1_{m>n}$ giving us the result
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}1=1,
$$while
$$
lim_{mtoinfty}lim_{ntoinfty}x^n_m = lim_{mtoinfty}0=0.
$$ However, uniform convergence of $x^n_m$ in one of the variables $n$ or $m$ can provide a sufficient condition for $(*)$ to be true. It is a much stronger condition than one of the (or both) limits in $(*)$ does exist. You can see some definitions and further explanation in https://en.wikipedia.org/wiki/Uniform_convergence, for example.
answered Dec 9 at 9:33
Song
4,200316
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
|
show 2 more comments
As already stated, it is not true what you want to proof:
begin{array}{cccccccr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 1 & 0 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 1 & 0 & ldots & to & 0 \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 1setminus 0 \
end{array}
Here anothe example where the columns do not converge at all:
begin{array}{rrrrrrrr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
-2& -2 & 0 & 0 & 0 & 0 & ldots & to & 0 \
3& 3 & 3 & 0 & 0 & 0 & ldots & to &0 \
-4& -4 & -4 & -4 & 0 & 0 & ldots & to &0 \
5& 5 & 5 & 5 & 5 & 0 & ldots & to & 0 \
-6& -6 & -6 & -6 & -6 & -6 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
?& ? & ? & ? & ? & ? & ldots & to & ?setminus 0 \
end{array}
answered Dec 9 at 10:28
miracle173
7,32222247
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Unfortunately, what you are trying to prove
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{mtoinfty}lim_{ntoinfty}x^n_mquadcdots(*)
$$, which is changing the order of the limits, is not true in general. We can see this
by the example $x^n_m = 1_{m>n}$ giving us the result
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}1=1,
$$while
$$
lim_{mtoinfty}lim_{ntoinfty}x^n_m = lim_{mtoinfty}0=0.
$$ However, uniform convergence of $x^n_m$ in one of the variables $n$ or $m$ can provide a sufficient condition for $(*)$ to be true. It is a much stronger condition than one of the (or both) limits in $(*)$ does exist. You can see some definitions and further explanation in https://en.wikipedia.org/wiki/Uniform_convergence, for example.
answered Dec 9 at 9:33
Song
4,200316
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
|
show 2 more comments
Unfortunately, what you are trying to prove
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{mtoinfty}lim_{ntoinfty}x^n_mquadcdots(*)
$$, which is changing the order of the limits, is not true in general. We can see this
by the example $x^n_m = 1_{m>n}$ giving us the result
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}1=1,
$$while
$$
lim_{mtoinfty}lim_{ntoinfty}x^n_m = lim_{mtoinfty}0=0.
$$ However, uniform convergence of $x^n_m$ in one of the variables $n$ or $m$ can provide a sufficient condition for $(*)$ to be true. It is a much stronger condition than one of the (or both) limits in $(*)$ does exist. You can see some definitions and further explanation in https://en.wikipedia.org/wiki/Uniform_convergence, for example.
answered Dec 9 at 9:33
Song
4,200316
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
|
show 2 more comments
Unfortunately, what you are trying to prove
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{mtoinfty}lim_{ntoinfty}x^n_mquadcdots(*)
$$, which is changing the order of the limits, is not true in general. We can see this
by the example $x^n_m = 1_{m>n}$ giving us the result
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}1=1,
$$while
$$
lim_{mtoinfty}lim_{ntoinfty}x^n_m = lim_{mtoinfty}0=0.
$$ However, uniform convergence of $x^n_m$ in one of the variables $n$ or $m$ can provide a sufficient condition for $(*)$ to be true. It is a much stronger condition than one of the (or both) limits in $(*)$ does exist. You can see some definitions and further explanation in https://en.wikipedia.org/wiki/Uniform_convergence, for example.
answered Dec 9 at 9:33
Song
4,200316
Unfortunately, what you are trying to prove
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{mtoinfty}lim_{ntoinfty}x^n_mquadcdots(*)
$$, which is changing the order of the limits, is not true in general. We can see this
by the example $x^n_m = 1_{m>n}$ giving us the result
$$
lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}1=1,
$$while
$$
lim_{mtoinfty}lim_{ntoinfty}x^n_m = lim_{mtoinfty}0=0.
$$ However, uniform convergence of $x^n_m$ in one of the variables $n$ or $m$ can provide a sufficient condition for $(*)$ to be true. It is a much stronger condition than one of the (or both) limits in $(*)$ does exist. You can see some definitions and further explanation in https://en.wikipedia.org/wiki/Uniform_convergence, for example.
answered Dec 9 at 9:33
Song
4,200316
edited Dec 9 at 10:25
answered Dec 9 at 9:33
Song
4,200316
answered Dec 9 at 9:33
Song
4,200316
answered Dec 9 at 9:33
Song
4,200316
4,200316
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
|
show 2 more comments
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
you are wrong: $lim_{ntoinfty}lim_{mtoinfty}x^n_m = lim_{ntoinfty}frac{1}{n}=0$
– miracle173
Dec 9 at 10:15
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
Thank you for your comment. I should have been more careful choosing an example. Maybe I was confused with something like $x^n_m = frac{m}{n}$.
– Song
Dec 9 at 11:08
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:49
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
Well, actually your set is weakly closed because any convex subset of a Banach space is weakly closed if and only if it is norm closed. Where the problem occurs is that $lim_{ntoinfty}x^n_m = x^infty_m$ for all $mgeq 1$ does not mean $x^n to x^infty$ (coordinatewise convergence is not sufficient for weak convergence.) If we identify $l^infty(mathbb{N}) sim C(betamathbb{N})$ where $betamathbb{N}$ is the Stone-Cech compactification, this is almost straightforward.
– Song
Dec 9 at 23:30
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
You're right. I forgot the basic classes of real analysis. Thank you
– Fam
Dec 10 at 0:12
|
show 2 more comments
As already stated, it is not true what you want to proof:
begin{array}{cccccccr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 1 & 0 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 1 & 0 & ldots & to & 0 \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 1setminus 0 \
end{array}
Here anothe example where the columns do not converge at all:
begin{array}{rrrrrrrr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
-2& -2 & 0 & 0 & 0 & 0 & ldots & to & 0 \
3& 3 & 3 & 0 & 0 & 0 & ldots & to &0 \
-4& -4 & -4 & -4 & 0 & 0 & ldots & to &0 \
5& 5 & 5 & 5 & 5 & 0 & ldots & to & 0 \
-6& -6 & -6 & -6 & -6 & -6 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
?& ? & ? & ? & ? & ? & ldots & to & ?setminus 0 \
end{array}
answered Dec 9 at 10:28
miracle173
7,32222247
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
add a comment |
As already stated, it is not true what you want to proof:
begin{array}{cccccccr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 1 & 0 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 1 & 0 & ldots & to & 0 \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 1setminus 0 \
end{array}
Here anothe example where the columns do not converge at all:
begin{array}{rrrrrrrr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
-2& -2 & 0 & 0 & 0 & 0 & ldots & to & 0 \
3& 3 & 3 & 0 & 0 & 0 & ldots & to &0 \
-4& -4 & -4 & -4 & 0 & 0 & ldots & to &0 \
5& 5 & 5 & 5 & 5 & 0 & ldots & to & 0 \
-6& -6 & -6 & -6 & -6 & -6 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
?& ? & ? & ? & ? & ? & ldots & to & ?setminus 0 \
end{array}
answered Dec 9 at 10:28
miracle173
7,32222247
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
add a comment |
As already stated, it is not true what you want to proof:
begin{array}{cccccccr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 1 & 0 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 1 & 0 & ldots & to & 0 \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 1setminus 0 \
end{array}
Here anothe example where the columns do not converge at all:
begin{array}{rrrrrrrr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
-2& -2 & 0 & 0 & 0 & 0 & ldots & to & 0 \
3& 3 & 3 & 0 & 0 & 0 & ldots & to &0 \
-4& -4 & -4 & -4 & 0 & 0 & ldots & to &0 \
5& 5 & 5 & 5 & 5 & 0 & ldots & to & 0 \
-6& -6 & -6 & -6 & -6 & -6 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
?& ? & ? & ? & ? & ? & ldots & to & ?setminus 0 \
end{array}
answered Dec 9 at 10:28
miracle173
7,32222247
As already stated, it is not true what you want to proof:
begin{array}{cccccccr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 0 & 0 & 0 & 0 & ldots & to & 0 \
1& 1 & 1 & 0 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 0 & 0 & ldots & to &0 \
1& 1 & 1 & 1 & 1 & 0 & ldots & to & 0 \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
1& 1 & 1 & 1 & 1 & 1 & ldots & to & 1setminus 0 \
end{array}
Here anothe example where the columns do not converge at all:
begin{array}{rrrrrrrr}
1& 0 & 0 & 0 & 0 & 0 & ldots & to & 0 \
-2& -2 & 0 & 0 & 0 & 0 & ldots & to & 0 \
3& 3 & 3 & 0 & 0 & 0 & ldots & to &0 \
-4& -4 & -4 & -4 & 0 & 0 & ldots & to &0 \
5& 5 & 5 & 5 & 5 & 0 & ldots & to & 0 \
-6& -6 & -6 & -6 & -6 & -6 & ldots & to & 0 \
vdots& vdots & vdots & vdots & vdots & vdots & ddots &&vdots\
downarrow & downarrow &downarrow & downarrow & downarrow &downarrow& &&downarrow \
?& ? & ? & ? & ? & ? & ldots & to & ?setminus 0 \
end{array}
answered Dec 9 at 10:28
miracle173
7,32222247
edited Dec 9 at 22:36
answered Dec 9 at 10:28
miracle173
7,32222247
answered Dec 9 at 10:28
miracle173
7,32222247
answered Dec 9 at 10:28
miracle173
7,32222247
7,32222247
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
add a comment |
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
In fact, I was trying to prove that the set ${x in l^{infty} : x_m in [0,1], lim_{m to infty} x_m = 1 }$ is weakly closed. $x^m to x$ in the weak topology if $f(x^m) to f(x)$, for all $f$ in the dual of $l^{infty}$. Taking $f_m ((x_1, x_2,...x_m,...))= x_m$, I believed that my problem rested on the proposed question.
– Fam
Dec 9 at 20:55
add a comment |
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Try $x_m^n=1$ if $m>n$, $x_m^n=0$ if $mleqslant n$ (and reconsider the result you wish to prove...).
– Did
Dec 9 at 10:03