Weak* Convergence exercise
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I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:
Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that
begin{equation}
x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
end{equation}
for fixed $k in mathbb{N}$.
functional-analysis weak-convergence
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up vote
0
down vote
favorite
I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:
Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that
begin{equation}
x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
end{equation}
for fixed $k in mathbb{N}$.
functional-analysis weak-convergence
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:
Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that
begin{equation}
x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
end{equation}
for fixed $k in mathbb{N}$.
functional-analysis weak-convergence
I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:
Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that
begin{equation}
x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
end{equation}
for fixed $k in mathbb{N}$.
functional-analysis weak-convergence
functional-analysis weak-convergence
asked Nov 15 at 11:19
James Arten
579
579
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2 Answers
2
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oldest
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up vote
2
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What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?
[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].
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up vote
1
down vote
To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:
Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.
The proof is straightforward if you know the definition of equicontinuity.
Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?
[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].
add a comment |
up vote
2
down vote
What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?
[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].
add a comment |
up vote
2
down vote
up vote
2
down vote
What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?
[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].
What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?
[I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].
answered Nov 15 at 12:04
Kavi Rama Murthy
43k31751
43k31751
add a comment |
add a comment |
up vote
1
down vote
To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:
Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.
The proof is straightforward if you know the definition of equicontinuity.
Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).
add a comment |
up vote
1
down vote
To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:
Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.
The proof is straightforward if you know the definition of equicontinuity.
Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).
add a comment |
up vote
1
down vote
up vote
1
down vote
To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:
Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.
The proof is straightforward if you know the definition of equicontinuity.
Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).
To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:
Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.
The proof is straightforward if you know the definition of equicontinuity.
Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).
answered Nov 15 at 14:08
MaoWao
2,268416
2,268416
add a comment |
add a comment |
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