Question regarding weak convergence
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I'm doing my bachelor's thesis and I have stumbled upon a certain reoccurring thing in some of the proofs I have to make. The book that I use keeps on arguing that it is possible to assume weak convergence, but I can't quite see how they can argue that. I can give an example: In a proof, it has been shown that
$$||varphi^{delta}||leq ||varphi||$$
here $varphi^{delta}$ is a perturbed solution and $varphi$ is the true solution. In the book, they then proceed to say that because of this inequality, weak convergence $varphi_nrightharpoonupvarphi_0in X$ can be assumed ($X$ is a Hilbert space). Furthermore, in the proof ${varphi_{n}}$ is a sequence where $varphi_{n}:=varphi^{delta_n}$, where ${delta_n}$ is a null sequence.
If anyone can help me realize how weak convergence can be assumed that would be highly appreciated. Also, if more information about the statement of weak convergence is needed, then please let me know.
Thanks in advance!
functional-analysis regularization
asked Dec 17 '18 at 12:51
JamesJames
11910
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add a comment |
$begingroup$
I'm doing my bachelor's thesis and I have stumbled upon a certain reoccurring thing in some of the proofs I have to make. The book that I use keeps on arguing that it is possible to assume weak convergence, but I can't quite see how they can argue that. I can give an example: In a proof, it has been shown that
$$||varphi^{delta}||leq ||varphi||$$
here $varphi^{delta}$ is a perturbed solution and $varphi$ is the true solution. In the book, they then proceed to say that because of this inequality, weak convergence $varphi_nrightharpoonupvarphi_0in X$ can be assumed ($X$ is a Hilbert space). Furthermore, in the proof ${varphi_{n}}$ is a sequence where $varphi_{n}:=varphi^{delta_n}$, where ${delta_n}$ is a null sequence.
If anyone can help me realize how weak convergence can be assumed that would be highly appreciated. Also, if more information about the statement of weak convergence is needed, then please let me know.
Thanks in advance!
functional-analysis regularization
asked Dec 17 '18 at 12:51
JamesJames
11910
$endgroup$
add a comment |
$begingroup$
I'm doing my bachelor's thesis and I have stumbled upon a certain reoccurring thing in some of the proofs I have to make. The book that I use keeps on arguing that it is possible to assume weak convergence, but I can't quite see how they can argue that. I can give an example: In a proof, it has been shown that
$$||varphi^{delta}||leq ||varphi||$$
here $varphi^{delta}$ is a perturbed solution and $varphi$ is the true solution. In the book, they then proceed to say that because of this inequality, weak convergence $varphi_nrightharpoonupvarphi_0in X$ can be assumed ($X$ is a Hilbert space). Furthermore, in the proof ${varphi_{n}}$ is a sequence where $varphi_{n}:=varphi^{delta_n}$, where ${delta_n}$ is a null sequence.
If anyone can help me realize how weak convergence can be assumed that would be highly appreciated. Also, if more information about the statement of weak convergence is needed, then please let me know.
Thanks in advance!
functional-analysis regularization
asked Dec 17 '18 at 12:51
JamesJames
11910
$endgroup$
I'm doing my bachelor's thesis and I have stumbled upon a certain reoccurring thing in some of the proofs I have to make. The book that I use keeps on arguing that it is possible to assume weak convergence, but I can't quite see how they can argue that. I can give an example: In a proof, it has been shown that
$$||varphi^{delta}||leq ||varphi||$$
here $varphi^{delta}$ is a perturbed solution and $varphi$ is the true solution. In the book, they then proceed to say that because of this inequality, weak convergence $varphi_nrightharpoonupvarphi_0in X$ can be assumed ($X$ is a Hilbert space). Furthermore, in the proof ${varphi_{n}}$ is a sequence where $varphi_{n}:=varphi^{delta_n}$, where ${delta_n}$ is a null sequence.
If anyone can help me realize how weak convergence can be assumed that would be highly appreciated. Also, if more information about the statement of weak convergence is needed, then please let me know.
Thanks in advance!
functional-analysis regularization
functional-analysis regularization
asked Dec 17 '18 at 12:51
JamesJames
11910
asked Dec 17 '18 at 12:51
JamesJames
11910
asked Dec 17 '18 at 12:51
JamesJames
11910
asked Dec 17 '18 at 12:51
JamesJames
11910
asked Dec 17 '18 at 12:51
JamesJames
11910
11910
add a comment |
add a comment |
1 Answer
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Any Hilbert space is a reflexive Banach space. In a reflexive Banach space, any closed ball is weakly sequentially compact by Eberlein-Smulian theorem. Hence, if you pick $${varphi^{delta_n}}_{ngeq 1},$$ such that $delta_n to 0$, you can find a subsequence $n(k)$ such that ${varphi_{n(k)}}_{kgeq 1}$ is weakly convergent.
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
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1 Answer
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1 Answer
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$begingroup$
Any Hilbert space is a reflexive Banach space. In a reflexive Banach space, any closed ball is weakly sequentially compact by Eberlein-Smulian theorem. Hence, if you pick $${varphi^{delta_n}}_{ngeq 1},$$ such that $delta_n to 0$, you can find a subsequence $n(k)$ such that ${varphi_{n(k)}}_{kgeq 1}$ is weakly convergent.
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
$endgroup$
add a comment |
$begingroup$
Any Hilbert space is a reflexive Banach space. In a reflexive Banach space, any closed ball is weakly sequentially compact by Eberlein-Smulian theorem. Hence, if you pick $${varphi^{delta_n}}_{ngeq 1},$$ such that $delta_n to 0$, you can find a subsequence $n(k)$ such that ${varphi_{n(k)}}_{kgeq 1}$ is weakly convergent.
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
$endgroup$
add a comment |
$begingroup$
Any Hilbert space is a reflexive Banach space. In a reflexive Banach space, any closed ball is weakly sequentially compact by Eberlein-Smulian theorem. Hence, if you pick $${varphi^{delta_n}}_{ngeq 1},$$ such that $delta_n to 0$, you can find a subsequence $n(k)$ such that ${varphi_{n(k)}}_{kgeq 1}$ is weakly convergent.
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
$endgroup$
Any Hilbert space is a reflexive Banach space. In a reflexive Banach space, any closed ball is weakly sequentially compact by Eberlein-Smulian theorem. Hence, if you pick $${varphi^{delta_n}}_{ngeq 1},$$ such that $delta_n to 0$, you can find a subsequence $n(k)$ such that ${varphi_{n(k)}}_{kgeq 1}$ is weakly convergent.
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
answered Dec 17 '18 at 13:09
SongSong
18.3k21549
18.3k21549
add a comment |
add a comment |
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