Mixing
Weak convergence of sequence in Sobolev space implies uniform convergence
$begingroup$
Does weak convergence in a Sobolev space implies uniform convergence?
In the above question, a proof by contradiction is given from which it follows that if a sequence in $W^{1,p}$ over a nice, open and bounded domain in $mathbb{R}^n$ ($p>n$) is weakly convergent to $f$ in $W^{1,p}$, this sequence is uniformly convergent to $f$.
However, I don't really get the proof. Does the fact that $f_{n_k} rightarrow g$ in $C$ follow from the compact embedding? If yes, why is this necessarily the same subsequence as the subsequence that doesn't converge?
real-analysis functional-analysis sobolev-spaces uniform-convergence weak-convergence
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
$endgroup$
add a comment |
$begingroup$
Does weak convergence in a Sobolev space implies uniform convergence?
In the above question, a proof by contradiction is given from which it follows that if a sequence in $W^{1,p}$ over a nice, open and bounded domain in $mathbb{R}^n$ ($p>n$) is weakly convergent to $f$ in $W^{1,p}$, this sequence is uniformly convergent to $f$.
However, I don't really get the proof. Does the fact that $f_{n_k} rightarrow g$ in $C$ follow from the compact embedding? If yes, why is this necessarily the same subsequence as the subsequence that doesn't converge?
real-analysis functional-analysis sobolev-spaces uniform-convergence weak-convergence
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
$endgroup$
$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55
add a comment |
$begingroup$
Does weak convergence in a Sobolev space implies uniform convergence?
In the above question, a proof by contradiction is given from which it follows that if a sequence in $W^{1,p}$ over a nice, open and bounded domain in $mathbb{R}^n$ ($p>n$) is weakly convergent to $f$ in $W^{1,p}$, this sequence is uniformly convergent to $f$.
However, I don't really get the proof. Does the fact that $f_{n_k} rightarrow g$ in $C$ follow from the compact embedding? If yes, why is this necessarily the same subsequence as the subsequence that doesn't converge?
real-analysis functional-analysis sobolev-spaces uniform-convergence weak-convergence
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
$endgroup$
Does weak convergence in a Sobolev space implies uniform convergence?
In the above question, a proof by contradiction is given from which it follows that if a sequence in $W^{1,p}$ over a nice, open and bounded domain in $mathbb{R}^n$ ($p>n$) is weakly convergent to $f$ in $W^{1,p}$, this sequence is uniformly convergent to $f$.
However, I don't really get the proof. Does the fact that $f_{n_k} rightarrow g$ in $C$ follow from the compact embedding? If yes, why is this necessarily the same subsequence as the subsequence that doesn't converge?
real-analysis functional-analysis sobolev-spaces uniform-convergence weak-convergence
real-analysis functional-analysis sobolev-spaces uniform-convergence weak-convergence
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
asked Jan 24 at 10:44
CatemathicsCatemathics
1013
1013
$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55
add a comment |
$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55
$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can use the following, very useful little lemma:
A sequence $(x_n)$ converges to $x$ if and only if every subsequence
has a subsequence converging to $x$.
It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).
The application to the problem as hand works as follows. By point 2 from the linked question, $(f_n)$ has a subsequence converging uniformly to $f$. Of course, instead of starting with $(f_n)$, you can start with an arbitrary subsequence $(f_{n_k})$ and extract a subsequence converging uniformly to $f$. The lemma above then tells you that $f_nto f$ uniformly.
Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use the following, very useful little lemma:
A sequence $(x_n)$ converges to $x$ if and only if every subsequence
has a subsequence converging to $x$.
It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).
The application to the problem as hand works as follows. By point 2 from the linked question, $(f_n)$ has a subsequence converging uniformly to $f$. Of course, instead of starting with $(f_n)$, you can start with an arbitrary subsequence $(f_{n_k})$ and extract a subsequence converging uniformly to $f$. The lemma above then tells you that $f_nto f$ uniformly.
Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
$endgroup$
add a comment |
$begingroup$
You can use the following, very useful little lemma:
A sequence $(x_n)$ converges to $x$ if and only if every subsequence
has a subsequence converging to $x$.
It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).
The application to the problem as hand works as follows. By point 2 from the linked question, $(f_n)$ has a subsequence converging uniformly to $f$. Of course, instead of starting with $(f_n)$, you can start with an arbitrary subsequence $(f_{n_k})$ and extract a subsequence converging uniformly to $f$. The lemma above then tells you that $f_nto f$ uniformly.
Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
$endgroup$
add a comment |
$begingroup$
You can use the following, very useful little lemma:
A sequence $(x_n)$ converges to $x$ if and only if every subsequence
has a subsequence converging to $x$.
It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).
The application to the problem as hand works as follows. By point 2 from the linked question, $(f_n)$ has a subsequence converging uniformly to $f$. Of course, instead of starting with $(f_n)$, you can start with an arbitrary subsequence $(f_{n_k})$ and extract a subsequence converging uniformly to $f$. The lemma above then tells you that $f_nto f$ uniformly.
Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
$endgroup$
You can use the following, very useful little lemma:
A sequence $(x_n)$ converges to $x$ if and only if every subsequence
has a subsequence converging to $x$.
It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).
The application to the problem as hand works as follows. By point 2 from the linked question, $(f_n)$ has a subsequence converging uniformly to $f$. Of course, instead of starting with $(f_n)$, you can start with an arbitrary subsequence $(f_{n_k})$ and extract a subsequence converging uniformly to $f$. The lemma above then tells you that $f_nto f$ uniformly.
Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
answered Jan 24 at 12:34
MaoWaoMaoWao
3,838618
3,838618
add a comment |
add a comment |
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$begingroup$
Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear?
$endgroup$
– MaoWao
Jan 24 at 10:51
$begingroup$
Only the answer is unclear to me.
$endgroup$
– Catemathics
Jan 24 at 10:55