Mixing
Help showing derivative is compact
$begingroup$
I'm struggling with proving the following theorem:
$textit{Let }A:Usubset Xrightarrow Ytextit{ be a completely continuous operator from an open subset U of a normed}$
$textit{space X into a Banach space Y and assume A to be Fréchet differentiable at }psiin U.textit{ Then the}$
$textit{derivative }A'_{psi}textit{ is compact.}$
I'm not sure how to go about this. I was thinking of trying to do it by contradiction, i.e., assuming that $A'_{psi}$ is not compact, but I'm not really sure how to proceed from there... I guess what I am asking for is a hint or a strategy for trying to prove $A'_{psi}$. Any help is highly appreciated.
Thanks in advance.
real-analysis functional-analysis frechet-derivative
asked Jan 9 at 8:12
JamesJames
909
$endgroup$
add a comment |
$begingroup$
I'm struggling with proving the following theorem:
$textit{Let }A:Usubset Xrightarrow Ytextit{ be a completely continuous operator from an open subset U of a normed}$
$textit{space X into a Banach space Y and assume A to be Fréchet differentiable at }psiin U.textit{ Then the}$
$textit{derivative }A'_{psi}textit{ is compact.}$
I'm not sure how to go about this. I was thinking of trying to do it by contradiction, i.e., assuming that $A'_{psi}$ is not compact, but I'm not really sure how to proceed from there... I guess what I am asking for is a hint or a strategy for trying to prove $A'_{psi}$. Any help is highly appreciated.
Thanks in advance.
real-analysis functional-analysis frechet-derivative
asked Jan 9 at 8:12
JamesJames
909
$endgroup$
add a comment |
$begingroup$
I'm struggling with proving the following theorem:
$textit{Let }A:Usubset Xrightarrow Ytextit{ be a completely continuous operator from an open subset U of a normed}$
$textit{space X into a Banach space Y and assume A to be Fréchet differentiable at }psiin U.textit{ Then the}$
$textit{derivative }A'_{psi}textit{ is compact.}$
I'm not sure how to go about this. I was thinking of trying to do it by contradiction, i.e., assuming that $A'_{psi}$ is not compact, but I'm not really sure how to proceed from there... I guess what I am asking for is a hint or a strategy for trying to prove $A'_{psi}$. Any help is highly appreciated.
Thanks in advance.
real-analysis functional-analysis frechet-derivative
asked Jan 9 at 8:12
JamesJames
909
$endgroup$
I'm struggling with proving the following theorem:
$textit{Let }A:Usubset Xrightarrow Ytextit{ be a completely continuous operator from an open subset U of a normed}$
$textit{space X into a Banach space Y and assume A to be Fréchet differentiable at }psiin U.textit{ Then the}$
$textit{derivative }A'_{psi}textit{ is compact.}$
I'm not sure how to go about this. I was thinking of trying to do it by contradiction, i.e., assuming that $A'_{psi}$ is not compact, but I'm not really sure how to proceed from there... I guess what I am asking for is a hint or a strategy for trying to prove $A'_{psi}$. Any help is highly appreciated.
Thanks in advance.
real-analysis functional-analysis frechet-derivative
real-analysis functional-analysis frechet-derivative
asked Jan 9 at 8:12
JamesJames
909
asked Jan 9 at 8:12
JamesJames
909
asked Jan 9 at 8:12
JamesJames
909
asked Jan 9 at 8:12
JamesJames
909
asked Jan 9 at 8:12
JamesJames
909
909
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $T=A'_psi$. By the assumption, we have for some $delta>0$,
$$
A(psi + h) = A(psi) +T h + epsilon(h),quad forall |h|<delta
$$ where $frac{|epsilon(h)|}{|h|}to 0$ as $|h|to 0$. Without loss of generality, we may assume that $psi =0$ and $A(psi)=0$. Let $(x_n)subset X$ be a sequence with $sup_n |x_n|le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $epsilon(cdot)$, for any given $eta >0$, there exists (sufficiently large) $Minmathbb{N}$ such that$$Big|epsilonleft(frac{1}{M}x_nright)Big|le etaBig|frac{1}{M}x_nBig|$$ for all $nge 1$. This impiles that$$
MBig|epsilonleft(frac{1}{M}x_nright)Big|le eta|x_n|leq eta.
$$
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,ldots$, the sequence
$$
Aleft(frac{1}{j}x_{n(k)}right)
$$ converges to some $y_jin Y$.
From the fact that$$
Aleft(frac{1}{M}x_nright) = frac{1}{M}Tx_n+epsilonleft(frac{1}{M}x_nright),
$$ it follows
$$begin{eqnarray}
|Tx_{n(k)}-My_M |&=& Big|MAleft(frac{1}{M}x_{n(k)}right)-My_M-Mepsilonleft(frac{1}{M}x_{n(k)}right)Big|\
&le&MBig|Aleft(frac{1}{M}x_{n(k)}right)-y_MBig|+eta.
end{eqnarray}$$ By taking $ktoinfty$, we get
$$
limsup_{ktoinfty}|Tx_{n(k)}-My_M |le eta
$$ and
$$
limsup_{k,ltoinfty}|Tx_{n(k)}-Tx_{n(l)} |le 2eta.
$$Since $eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
answered Jan 9 at 9:22
SongSong
11.3k628
$endgroup$
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
$endgroup$
– Song
Jan 9 at 10:32
$begingroup$
The previous version of the answer was also very good, I was asking out of my ignorance.
$endgroup$
– Giuseppe Negro
Jan 9 at 11:41
$begingroup$
@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
$endgroup$
– Song
Jan 9 at 12:43
|
show 7 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $T=A'_psi$. By the assumption, we have for some $delta>0$,
$$
A(psi + h) = A(psi) +T h + epsilon(h),quad forall |h|<delta
$$ where $frac{|epsilon(h)|}{|h|}to 0$ as $|h|to 0$. Without loss of generality, we may assume that $psi =0$ and $A(psi)=0$. Let $(x_n)subset X$ be a sequence with $sup_n |x_n|le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $epsilon(cdot)$, for any given $eta >0$, there exists (sufficiently large) $Minmathbb{N}$ such that$$Big|epsilonleft(frac{1}{M}x_nright)Big|le etaBig|frac{1}{M}x_nBig|$$ for all $nge 1$. This impiles that$$
MBig|epsilonleft(frac{1}{M}x_nright)Big|le eta|x_n|leq eta.
$$
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,ldots$, the sequence
$$
Aleft(frac{1}{j}x_{n(k)}right)
$$ converges to some $y_jin Y$.
From the fact that$$
Aleft(frac{1}{M}x_nright) = frac{1}{M}Tx_n+epsilonleft(frac{1}{M}x_nright),
$$ it follows
$$begin{eqnarray}
|Tx_{n(k)}-My_M |&=& Big|MAleft(frac{1}{M}x_{n(k)}right)-My_M-Mepsilonleft(frac{1}{M}x_{n(k)}right)Big|\
&le&MBig|Aleft(frac{1}{M}x_{n(k)}right)-y_MBig|+eta.
end{eqnarray}$$ By taking $ktoinfty$, we get
$$
limsup_{ktoinfty}|Tx_{n(k)}-My_M |le eta
$$ and
$$
limsup_{k,ltoinfty}|Tx_{n(k)}-Tx_{n(l)} |le 2eta.
$$Since $eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
answered Jan 9 at 9:22
SongSong
11.3k628
$endgroup$
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
$endgroup$
– Song
Jan 9 at 10:32
$begingroup$
The previous version of the answer was also very good, I was asking out of my ignorance.
$endgroup$
– Giuseppe Negro
Jan 9 at 11:41
$begingroup$
@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
$endgroup$
– Song
Jan 9 at 12:43
|
show 7 more comments
$begingroup$
Let $T=A'_psi$. By the assumption, we have for some $delta>0$,
$$
A(psi + h) = A(psi) +T h + epsilon(h),quad forall |h|<delta
$$ where $frac{|epsilon(h)|}{|h|}to 0$ as $|h|to 0$. Without loss of generality, we may assume that $psi =0$ and $A(psi)=0$. Let $(x_n)subset X$ be a sequence with $sup_n |x_n|le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $epsilon(cdot)$, for any given $eta >0$, there exists (sufficiently large) $Minmathbb{N}$ such that$$Big|epsilonleft(frac{1}{M}x_nright)Big|le etaBig|frac{1}{M}x_nBig|$$ for all $nge 1$. This impiles that$$
MBig|epsilonleft(frac{1}{M}x_nright)Big|le eta|x_n|leq eta.
$$
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,ldots$, the sequence
$$
Aleft(frac{1}{j}x_{n(k)}right)
$$ converges to some $y_jin Y$.
From the fact that$$
Aleft(frac{1}{M}x_nright) = frac{1}{M}Tx_n+epsilonleft(frac{1}{M}x_nright),
$$ it follows
$$begin{eqnarray}
|Tx_{n(k)}-My_M |&=& Big|MAleft(frac{1}{M}x_{n(k)}right)-My_M-Mepsilonleft(frac{1}{M}x_{n(k)}right)Big|\
&le&MBig|Aleft(frac{1}{M}x_{n(k)}right)-y_MBig|+eta.
end{eqnarray}$$ By taking $ktoinfty$, we get
$$
limsup_{ktoinfty}|Tx_{n(k)}-My_M |le eta
$$ and
$$
limsup_{k,ltoinfty}|Tx_{n(k)}-Tx_{n(l)} |le 2eta.
$$Since $eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
answered Jan 9 at 9:22
SongSong
11.3k628
$endgroup$
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
$endgroup$
– Song
Jan 9 at 10:32
$begingroup$
The previous version of the answer was also very good, I was asking out of my ignorance.
$endgroup$
– Giuseppe Negro
Jan 9 at 11:41
$begingroup$
@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
$endgroup$
– Song
Jan 9 at 12:43
|
show 7 more comments
$begingroup$
Let $T=A'_psi$. By the assumption, we have for some $delta>0$,
$$
A(psi + h) = A(psi) +T h + epsilon(h),quad forall |h|<delta
$$ where $frac{|epsilon(h)|}{|h|}to 0$ as $|h|to 0$. Without loss of generality, we may assume that $psi =0$ and $A(psi)=0$. Let $(x_n)subset X$ be a sequence with $sup_n |x_n|le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $epsilon(cdot)$, for any given $eta >0$, there exists (sufficiently large) $Minmathbb{N}$ such that$$Big|epsilonleft(frac{1}{M}x_nright)Big|le etaBig|frac{1}{M}x_nBig|$$ for all $nge 1$. This impiles that$$
MBig|epsilonleft(frac{1}{M}x_nright)Big|le eta|x_n|leq eta.
$$
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,ldots$, the sequence
$$
Aleft(frac{1}{j}x_{n(k)}right)
$$ converges to some $y_jin Y$.
From the fact that$$
Aleft(frac{1}{M}x_nright) = frac{1}{M}Tx_n+epsilonleft(frac{1}{M}x_nright),
$$ it follows
$$begin{eqnarray}
|Tx_{n(k)}-My_M |&=& Big|MAleft(frac{1}{M}x_{n(k)}right)-My_M-Mepsilonleft(frac{1}{M}x_{n(k)}right)Big|\
&le&MBig|Aleft(frac{1}{M}x_{n(k)}right)-y_MBig|+eta.
end{eqnarray}$$ By taking $ktoinfty$, we get
$$
limsup_{ktoinfty}|Tx_{n(k)}-My_M |le eta
$$ and
$$
limsup_{k,ltoinfty}|Tx_{n(k)}-Tx_{n(l)} |le 2eta.
$$Since $eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
answered Jan 9 at 9:22
SongSong
11.3k628
$endgroup$
Let $T=A'_psi$. By the assumption, we have for some $delta>0$,
$$
A(psi + h) = A(psi) +T h + epsilon(h),quad forall |h|<delta
$$ where $frac{|epsilon(h)|}{|h|}to 0$ as $|h|to 0$. Without loss of generality, we may assume that $psi =0$ and $A(psi)=0$. Let $(x_n)subset X$ be a sequence with $sup_n |x_n|le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $epsilon(cdot)$, for any given $eta >0$, there exists (sufficiently large) $Minmathbb{N}$ such that$$Big|epsilonleft(frac{1}{M}x_nright)Big|le etaBig|frac{1}{M}x_nBig|$$ for all $nge 1$. This impiles that$$
MBig|epsilonleft(frac{1}{M}x_nright)Big|le eta|x_n|leq eta.
$$
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,ldots$, the sequence
$$
Aleft(frac{1}{j}x_{n(k)}right)
$$ converges to some $y_jin Y$.
From the fact that$$
Aleft(frac{1}{M}x_nright) = frac{1}{M}Tx_n+epsilonleft(frac{1}{M}x_nright),
$$ it follows
$$begin{eqnarray}
|Tx_{n(k)}-My_M |&=& Big|MAleft(frac{1}{M}x_{n(k)}right)-My_M-Mepsilonleft(frac{1}{M}x_{n(k)}right)Big|\
&le&MBig|Aleft(frac{1}{M}x_{n(k)}right)-y_MBig|+eta.
end{eqnarray}$$ By taking $ktoinfty$, we get
$$
limsup_{ktoinfty}|Tx_{n(k)}-My_M |le eta
$$ and
$$
limsup_{k,ltoinfty}|Tx_{n(k)}-Tx_{n(l)} |le 2eta.
$$Since $eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
answered Jan 9 at 9:22
SongSong
11.3k628
edited Jan 9 at 12:02
answered Jan 9 at 9:22
SongSong
11.3k628
answered Jan 9 at 9:22
SongSong
11.3k628
answered Jan 9 at 9:22
SongSong
11.3k628
11.3k628
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
$endgroup$
– Song
Jan 9 at 10:32
$begingroup$
The previous version of the answer was also very good, I was asking out of my ignorance.
$endgroup$
– Giuseppe Negro
Jan 9 at 11:41
$begingroup$
@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
$endgroup$
– Song
Jan 9 at 12:43
|
show 7 more comments
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
$endgroup$
– Song
Jan 9 at 10:32
$begingroup$
The previous version of the answer was also very good, I was asking out of my ignorance.
$endgroup$
– Giuseppe Negro
Jan 9 at 11:41
$begingroup$
@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
$endgroup$
– Song
Jan 9 at 12:43
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
Very good answer, but can you please spend a couple more words on the fact that $A(x_n)to 0$? To me, a completely continuous operator is a continuous operator that maps bounded sets to compact sets. Is it the case that this operator is weak-norm continuous? Thank you.
$endgroup$
– Giuseppe Negro
Jan 9 at 9:55
$begingroup$
@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
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– Song
Jan 9 at 10:32
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@GiuseppeNegro I'm sorry for the confusion, sir ... I've looked up precise definition of completely continuous operators, and it turns out that I used the wrong definition that $A$ sends weakly converging sequence to norm convergent sequence. Now, I edited my answer according to the correct definition that you gave. I appreciate your comment!
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– Song
Jan 9 at 10:32
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The previous version of the answer was also very good, I was asking out of my ignorance.
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– Giuseppe Negro
Jan 9 at 11:41
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The previous version of the answer was also very good, I was asking out of my ignorance.
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– Giuseppe Negro
Jan 9 at 11:41
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@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
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– Song
Jan 9 at 12:43
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@James Note that $A$ is not assumed to be a linear map. We are dealing with a possibly non-linear, Frechet differentiable function $A:Uto Y$. But its derivative $A'$ is of course linear. About the definition: if $A$ linear (which is not suitable for our case), '$A$ sending weakly convergent sequence to norm-convergent sequence' is the definition of completely continuous function. And it is in general stronger than the compactness of $A$ which is saying 'every bounded set is mapped to a relatively compact via $A$'.
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– Song
Jan 9 at 12:43
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@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
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– Song
Jan 9 at 12:43
$begingroup$
@James However, for a non-linear map $A$, I found in link.springer.com/chapter/10.1007%2F978-94-015-9986-3_3 that the definition of complete continuity is more like that of compactness of a linear operator. It probably is more universally accepted definition, but I'm not sure. The proof I gave is based on this weaker assumption, and should also be true under the alternative definition.
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– Song
Jan 9 at 12:43
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