Mixing
Convergence of $A_nT$ to $AT$ in operator norm for compact $T$
$begingroup$
$A_n:Yrightarrow Z$ are operators that strongly converge to $A$. Also, $|A_n|_text {op}le c$ for $c>0$. Given a compact operator $T:Xrightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).
I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas?
functional-analysis convergence operator-theory compact-operators
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
$endgroup$
add a comment |
$begingroup$
$A_n:Yrightarrow Z$ are operators that strongly converge to $A$. Also, $|A_n|_text {op}le c$ for $c>0$. Given a compact operator $T:Xrightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).
I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas?
functional-analysis convergence operator-theory compact-operators
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
$endgroup$
$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31
add a comment |
$begingroup$
$A_n:Yrightarrow Z$ are operators that strongly converge to $A$. Also, $|A_n|_text {op}le c$ for $c>0$. Given a compact operator $T:Xrightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).
I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas?
functional-analysis convergence operator-theory compact-operators
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
$endgroup$
$A_n:Yrightarrow Z$ are operators that strongly converge to $A$. Also, $|A_n|_text {op}le c$ for $c>0$. Given a compact operator $T:Xrightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).
I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas?
functional-analysis convergence operator-theory compact-operators
functional-analysis convergence operator-theory compact-operators
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
edited Oct 18 '17 at 13:10
Davide Giraudo
127k16151264
127k16151264
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
asked Oct 18 '17 at 12:26
MasterJMasterJ
4921414
4921414
$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31
add a comment |
$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31
$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31
$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You need $T$ to be compact because otherwise, by taking $T=Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $|AT-A_nT|notto0$, this means that there exists a sequence of unit vectors $x_n$ so that $|(AT-A_nT)x_n|>epsilon$ for some $epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $|(A-A_n)x|to0$. Now lets combine our information:
$$|(AT-A_nT)x_n|=|(A-A_n)(Tx_n-x+x)|≤|(A-A_n)x|+|A_n-A|,|T x_n-x|.$$
Now $|A_n-A|≤(|A_n|+|A|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $|(AT-A_nT)x_n|>epsilon$ for all $n$.
edited Jan 16 at 15:29
el_tenedor
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
$endgroup$
add a comment |
$begingroup$
In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that
$$lim_{ntoinfty}sup_{x:lVert xrVertleqslant 1}leftlVert S_nx-SxrightrVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that
$$lim_{ntoinfty}sup_{yin T(B)}leftlVert A_ny-AyrightrVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $varepsilon$, there exists a finite set $Fsubset Y$ such that for any $yin Y$, there exists $y'in F$ such that $lVert y-y'rVertlt varepsilon$.
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
add a comment |
$begingroup$
Take a sequence ${x_n}$ with $|x_n|=1$ such that $|A_nTx_n-ATx_n|ge|A_nT-AT|-frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=|A_ny_n-Ay_n|to0$. Let ${s_{n_k}}$ be any subsequence. Since $T$ is compact, the sequence ${y_{n_k}}$ is relatively compact, so there is a further subsequence ${y_{n_{k_ell}}}$ such that $y_{n_{k_ell}}to y$. Then
$$s_{n_{k_ell}}le|A_{n_{k_ell}}(y_{n_{k_ell}}-y)|+|(A_{n_{k_ell}}-A)y|+|A(y_{n_{k_ell}}-y)|to0$$
where the first term can be seen to converge to zero since $|A_{n_{k_ell}}|le c$. Thus for any subsequence ${s_{n_k}}$, there is a further subsequence converging to zero, so in particular ${s_n}$ cannot have a subsequence bounded away from zero. This implies $s_nto0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
answered Oct 18 '17 at 13:49
JasonJason
12k11232
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
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$begingroup$
You need $T$ to be compact because otherwise, by taking $T=Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $|AT-A_nT|notto0$, this means that there exists a sequence of unit vectors $x_n$ so that $|(AT-A_nT)x_n|>epsilon$ for some $epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $|(A-A_n)x|to0$. Now lets combine our information:
$$|(AT-A_nT)x_n|=|(A-A_n)(Tx_n-x+x)|≤|(A-A_n)x|+|A_n-A|,|T x_n-x|.$$
Now $|A_n-A|≤(|A_n|+|A|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $|(AT-A_nT)x_n|>epsilon$ for all $n$.
edited Jan 16 at 15:29
el_tenedor
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
$endgroup$
add a comment |
$begingroup$
You need $T$ to be compact because otherwise, by taking $T=Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $|AT-A_nT|notto0$, this means that there exists a sequence of unit vectors $x_n$ so that $|(AT-A_nT)x_n|>epsilon$ for some $epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $|(A-A_n)x|to0$. Now lets combine our information:
$$|(AT-A_nT)x_n|=|(A-A_n)(Tx_n-x+x)|≤|(A-A_n)x|+|A_n-A|,|T x_n-x|.$$
Now $|A_n-A|≤(|A_n|+|A|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $|(AT-A_nT)x_n|>epsilon$ for all $n$.
edited Jan 16 at 15:29
el_tenedor
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
$endgroup$
add a comment |
$begingroup$
You need $T$ to be compact because otherwise, by taking $T=Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $|AT-A_nT|notto0$, this means that there exists a sequence of unit vectors $x_n$ so that $|(AT-A_nT)x_n|>epsilon$ for some $epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $|(A-A_n)x|to0$. Now lets combine our information:
$$|(AT-A_nT)x_n|=|(A-A_n)(Tx_n-x+x)|≤|(A-A_n)x|+|A_n-A|,|T x_n-x|.$$
Now $|A_n-A|≤(|A_n|+|A|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $|(AT-A_nT)x_n|>epsilon$ for all $n$.
edited Jan 16 at 15:29
el_tenedor
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
$endgroup$
You need $T$ to be compact because otherwise, by taking $T=Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $|AT-A_nT|notto0$, this means that there exists a sequence of unit vectors $x_n$ so that $|(AT-A_nT)x_n|>epsilon$ for some $epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $|(A-A_n)x|to0$. Now lets combine our information:
$$|(AT-A_nT)x_n|=|(A-A_n)(Tx_n-x+x)|≤|(A-A_n)x|+|A_n-A|,|T x_n-x|.$$
Now $|A_n-A|≤(|A_n|+|A|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $|(AT-A_nT)x_n|>epsilon$ for all $n$.
edited Jan 16 at 15:29
el_tenedor
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
edited Jan 16 at 15:29
el_tenedor
2,292921
edited Jan 16 at 15:29
el_tenedor
2,292921
edited Jan 16 at 15:29
el_tenedor
2,292921
2,292921
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
answered Oct 18 '17 at 12:49
s.harps.harp
8,45312249
8,45312249
add a comment |
add a comment |
$begingroup$
In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that
$$lim_{ntoinfty}sup_{x:lVert xrVertleqslant 1}leftlVert S_nx-SxrightrVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that
$$lim_{ntoinfty}sup_{yin T(B)}leftlVert A_ny-AyrightrVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $varepsilon$, there exists a finite set $Fsubset Y$ such that for any $yin Y$, there exists $y'in F$ such that $lVert y-y'rVertlt varepsilon$.
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
add a comment |
$begingroup$
In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that
$$lim_{ntoinfty}sup_{x:lVert xrVertleqslant 1}leftlVert S_nx-SxrightrVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that
$$lim_{ntoinfty}sup_{yin T(B)}leftlVert A_ny-AyrightrVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $varepsilon$, there exists a finite set $Fsubset Y$ such that for any $yin Y$, there exists $y'in F$ such that $lVert y-y'rVertlt varepsilon$.
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
add a comment |
$begingroup$
In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that
$$lim_{ntoinfty}sup_{x:lVert xrVertleqslant 1}leftlVert S_nx-SxrightrVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that
$$lim_{ntoinfty}sup_{yin T(B)}leftlVert A_ny-AyrightrVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $varepsilon$, there exists a finite set $Fsubset Y$ such that for any $yin Y$, there exists $y'in F$ such that $lVert y-y'rVertlt varepsilon$.
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
$endgroup$
In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that
$$lim_{ntoinfty}sup_{x:lVert xrVertleqslant 1}leftlVert S_nx-SxrightrVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that
$$lim_{ntoinfty}sup_{yin T(B)}leftlVert A_ny-AyrightrVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $varepsilon$, there exists a finite set $Fsubset Y$ such that for any $yin Y$, there exists $y'in F$ such that $lVert y-y'rVertlt varepsilon$.
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
answered Oct 18 '17 at 12:50
Davide GiraudoDavide Giraudo
127k16151264
127k16151264
add a comment |
add a comment |
$begingroup$
Take a sequence ${x_n}$ with $|x_n|=1$ such that $|A_nTx_n-ATx_n|ge|A_nT-AT|-frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=|A_ny_n-Ay_n|to0$. Let ${s_{n_k}}$ be any subsequence. Since $T$ is compact, the sequence ${y_{n_k}}$ is relatively compact, so there is a further subsequence ${y_{n_{k_ell}}}$ such that $y_{n_{k_ell}}to y$. Then
$$s_{n_{k_ell}}le|A_{n_{k_ell}}(y_{n_{k_ell}}-y)|+|(A_{n_{k_ell}}-A)y|+|A(y_{n_{k_ell}}-y)|to0$$
where the first term can be seen to converge to zero since $|A_{n_{k_ell}}|le c$. Thus for any subsequence ${s_{n_k}}$, there is a further subsequence converging to zero, so in particular ${s_n}$ cannot have a subsequence bounded away from zero. This implies $s_nto0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
answered Oct 18 '17 at 13:49
JasonJason
12k11232
$endgroup$
add a comment |
$begingroup$
Take a sequence ${x_n}$ with $|x_n|=1$ such that $|A_nTx_n-ATx_n|ge|A_nT-AT|-frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=|A_ny_n-Ay_n|to0$. Let ${s_{n_k}}$ be any subsequence. Since $T$ is compact, the sequence ${y_{n_k}}$ is relatively compact, so there is a further subsequence ${y_{n_{k_ell}}}$ such that $y_{n_{k_ell}}to y$. Then
$$s_{n_{k_ell}}le|A_{n_{k_ell}}(y_{n_{k_ell}}-y)|+|(A_{n_{k_ell}}-A)y|+|A(y_{n_{k_ell}}-y)|to0$$
where the first term can be seen to converge to zero since $|A_{n_{k_ell}}|le c$. Thus for any subsequence ${s_{n_k}}$, there is a further subsequence converging to zero, so in particular ${s_n}$ cannot have a subsequence bounded away from zero. This implies $s_nto0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
answered Oct 18 '17 at 13:49
JasonJason
12k11232
$endgroup$
add a comment |
$begingroup$
Take a sequence ${x_n}$ with $|x_n|=1$ such that $|A_nTx_n-ATx_n|ge|A_nT-AT|-frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=|A_ny_n-Ay_n|to0$. Let ${s_{n_k}}$ be any subsequence. Since $T$ is compact, the sequence ${y_{n_k}}$ is relatively compact, so there is a further subsequence ${y_{n_{k_ell}}}$ such that $y_{n_{k_ell}}to y$. Then
$$s_{n_{k_ell}}le|A_{n_{k_ell}}(y_{n_{k_ell}}-y)|+|(A_{n_{k_ell}}-A)y|+|A(y_{n_{k_ell}}-y)|to0$$
where the first term can be seen to converge to zero since $|A_{n_{k_ell}}|le c$. Thus for any subsequence ${s_{n_k}}$, there is a further subsequence converging to zero, so in particular ${s_n}$ cannot have a subsequence bounded away from zero. This implies $s_nto0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
answered Oct 18 '17 at 13:49
JasonJason
12k11232
$endgroup$
Take a sequence ${x_n}$ with $|x_n|=1$ such that $|A_nTx_n-ATx_n|ge|A_nT-AT|-frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=|A_ny_n-Ay_n|to0$. Let ${s_{n_k}}$ be any subsequence. Since $T$ is compact, the sequence ${y_{n_k}}$ is relatively compact, so there is a further subsequence ${y_{n_{k_ell}}}$ such that $y_{n_{k_ell}}to y$. Then
$$s_{n_{k_ell}}le|A_{n_{k_ell}}(y_{n_{k_ell}}-y)|+|(A_{n_{k_ell}}-A)y|+|A(y_{n_{k_ell}}-y)|to0$$
where the first term can be seen to converge to zero since $|A_{n_{k_ell}}|le c$. Thus for any subsequence ${s_{n_k}}$, there is a further subsequence converging to zero, so in particular ${s_n}$ cannot have a subsequence bounded away from zero. This implies $s_nto0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
answered Oct 18 '17 at 13:49
JasonJason
12k11232
answered Oct 18 '17 at 13:49
JasonJason
12k11232
answered Oct 18 '17 at 13:49
JasonJason
12k11232
answered Oct 18 '17 at 13:49
JasonJason
12k11232
12k11232
add a comment |
add a comment |
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$begingroup$
I think it's worth to mention that the assumption $|A_n|_{mathrm{op}}$ is superfluous as the Banach-Steinhaus Theorem already guarantees uniform boundedness of the sequence $A_n$.
$endgroup$
– el_tenedor
Jan 16 at 15:31