$T_n rightarrow T$ then we have $||T|| le liminf(||T_n||)$
$begingroup$
I know how to show that a cauchy sequence of linear continuous operators $T_n:X rightarrow Y$ has a limit that is also such an operator(if Y is a Banach space), but I found this relation here too $||T|| le liminf(||T_n||)$ and I just don't know where this liminf comes into play. Does anybody here know?
calculus real-analysis analysis functional-analysis
edited Jan 6 at 14:02
A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
I know how to show that a cauchy sequence of linear continuous operators $T_n:X rightarrow Y$ has a limit that is also such an operator(if Y is a Banach space), but I found this relation here too $||T|| le liminf(||T_n||)$ and I just don't know where this liminf comes into play. Does anybody here know?
calculus real-analysis analysis functional-analysis
edited Jan 6 at 14:02
A.Γ.
22.9k32656
$endgroup$
1
$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53
add a comment |
$begingroup$
I know how to show that a cauchy sequence of linear continuous operators $T_n:X rightarrow Y$ has a limit that is also such an operator(if Y is a Banach space), but I found this relation here too $||T|| le liminf(||T_n||)$ and I just don't know where this liminf comes into play. Does anybody here know?
calculus real-analysis analysis functional-analysis
edited Jan 6 at 14:02
A.Γ.
22.9k32656
$endgroup$
I know how to show that a cauchy sequence of linear continuous operators $T_n:X rightarrow Y$ has a limit that is also such an operator(if Y is a Banach space), but I found this relation here too $||T|| le liminf(||T_n||)$ and I just don't know where this liminf comes into play. Does anybody here know?
calculus real-analysis analysis functional-analysis
calculus real-analysis analysis functional-analysis
edited Jan 6 at 14:02
A.Γ.
22.9k32656
edited Jan 6 at 14:02
A.Γ.
22.9k32656
edited Jan 6 at 14:02
A.Γ.
22.9k32656
edited Jan 6 at 14:02
A.Γ.
22.9k32656
edited Jan 6 at 14:02
A.Γ.
22.9k32656
22.9k32656
asked Feb 7 '14 at 15:44
user66906
1
$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53
add a comment |
1
$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53
1
1
$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$lVert TrVert = lim_{ntoinfty} lVert T_nrVert$$
by the continuity of the norm ($leftlvert lVert TrVert - lVert SrVertrightrvert leqslant lVert T-SrVert$).
You get the $liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $Xto Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = liminflimits_{ntoinfty} lVert T_nrVert < infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$lVert TrVert leqslant M.$$
For, let $x in X$ with $lVert xrVert leqslant 1$. For every $varepsilon > 0$, there are infinitely many $n_k$ with $lVert T_{n_k}rVert < M + varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = lim_{ktoinfty} T_{n_k}x.$$
But $lVert T_{n_k}xrVert < M+varepsilon$ for all $k$, and hence $lVert TxrVert leqslant M+varepsilon$. That holds for all $x$ with $lVert xrVert leqslant 1$, so
$$lVert TrVert = sup_{lVert xrVert leqslant 1} lVert TxrVert leqslant M + varepsilon.$$
That holds for all $varepsilon > 0$, hence $lVert TrVert leqslant M$.
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
$endgroup$
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
add a comment |
Your Answer
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$begingroup$
If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$lVert TrVert = lim_{ntoinfty} lVert T_nrVert$$
by the continuity of the norm ($leftlvert lVert TrVert - lVert SrVertrightrvert leqslant lVert T-SrVert$).
You get the $liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $Xto Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = liminflimits_{ntoinfty} lVert T_nrVert < infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$lVert TrVert leqslant M.$$
For, let $x in X$ with $lVert xrVert leqslant 1$. For every $varepsilon > 0$, there are infinitely many $n_k$ with $lVert T_{n_k}rVert < M + varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = lim_{ktoinfty} T_{n_k}x.$$
But $lVert T_{n_k}xrVert < M+varepsilon$ for all $k$, and hence $lVert TxrVert leqslant M+varepsilon$. That holds for all $x$ with $lVert xrVert leqslant 1$, so
$$lVert TrVert = sup_{lVert xrVert leqslant 1} lVert TxrVert leqslant M + varepsilon.$$
That holds for all $varepsilon > 0$, hence $lVert TrVert leqslant M$.
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
$endgroup$
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
add a comment |
$begingroup$
If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$lVert TrVert = lim_{ntoinfty} lVert T_nrVert$$
by the continuity of the norm ($leftlvert lVert TrVert - lVert SrVertrightrvert leqslant lVert T-SrVert$).
You get the $liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $Xto Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = liminflimits_{ntoinfty} lVert T_nrVert < infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$lVert TrVert leqslant M.$$
For, let $x in X$ with $lVert xrVert leqslant 1$. For every $varepsilon > 0$, there are infinitely many $n_k$ with $lVert T_{n_k}rVert < M + varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = lim_{ktoinfty} T_{n_k}x.$$
But $lVert T_{n_k}xrVert < M+varepsilon$ for all $k$, and hence $lVert TxrVert leqslant M+varepsilon$. That holds for all $x$ with $lVert xrVert leqslant 1$, so
$$lVert TrVert = sup_{lVert xrVert leqslant 1} lVert TxrVert leqslant M + varepsilon.$$
That holds for all $varepsilon > 0$, hence $lVert TrVert leqslant M$.
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
$endgroup$
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
add a comment |
$begingroup$
If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$lVert TrVert = lim_{ntoinfty} lVert T_nrVert$$
by the continuity of the norm ($leftlvert lVert TrVert - lVert SrVertrightrvert leqslant lVert T-SrVert$).
You get the $liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $Xto Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = liminflimits_{ntoinfty} lVert T_nrVert < infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$lVert TrVert leqslant M.$$
For, let $x in X$ with $lVert xrVert leqslant 1$. For every $varepsilon > 0$, there are infinitely many $n_k$ with $lVert T_{n_k}rVert < M + varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = lim_{ktoinfty} T_{n_k}x.$$
But $lVert T_{n_k}xrVert < M+varepsilon$ for all $k$, and hence $lVert TxrVert leqslant M+varepsilon$. That holds for all $x$ with $lVert xrVert leqslant 1$, so
$$lVert TrVert = sup_{lVert xrVert leqslant 1} lVert TxrVert leqslant M + varepsilon.$$
That holds for all $varepsilon > 0$, hence $lVert TrVert leqslant M$.
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
$endgroup$
If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$lVert TrVert = lim_{ntoinfty} lVert T_nrVert$$
by the continuity of the norm ($leftlvert lVert TrVert - lVert SrVertrightrvert leqslant lVert T-SrVert$).
You get the $liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $Xto Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = liminflimits_{ntoinfty} lVert T_nrVert < infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$lVert TrVert leqslant M.$$
For, let $x in X$ with $lVert xrVert leqslant 1$. For every $varepsilon > 0$, there are infinitely many $n_k$ with $lVert T_{n_k}rVert < M + varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = lim_{ktoinfty} T_{n_k}x.$$
But $lVert T_{n_k}xrVert < M+varepsilon$ for all $k$, and hence $lVert TxrVert leqslant M+varepsilon$. That holds for all $x$ with $lVert xrVert leqslant 1$, so
$$lVert TrVert = sup_{lVert xrVert leqslant 1} lVert TxrVert leqslant M + varepsilon.$$
That holds for all $varepsilon > 0$, hence $lVert TrVert leqslant M$.
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
edited Feb 8 '14 at 10:26
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
answered Feb 7 '14 at 16:12
Daniel FischerDaniel Fischer
174k17171289
174k17171289
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
add a comment |
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
1
1
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
$begingroup$
Not following the part about pointwise convergence. Does it imply that there are operators such that $lim T_n x = Tx$ but $lim |T_n| neq |T|$? Any good examples?
$endgroup$
– Jeff
Mar 24 '14 at 4:01
2
2
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
Yes, we can have pointwise convergent sequences of operators with $lVert lim T_nrVert < liminf lVert T_nrVert$, for example on $ell^p$ spaces ($1 leqslant p < infty$), we can take $T_n(x) = x_ncdot e_1$. Then $T_n(x) to 0$ for all $x$, but $lVert T_nrVert = 1$ for all $n$.
$endgroup$
– Daniel Fischer
Mar 24 '14 at 10:24
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
$begingroup$
You also get the liminf for weak convergence, since the norm is weakly sequentially lower semicontinuous. You might also find this question interesting: math.stackexchange.com/questions/417349/…
$endgroup$
– anonymous
Apr 23 '14 at 20:31
add a comment |
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$begingroup$
If the sequence is a Cauchy sequence in the norm, you have $lVert TrVert = lim lVert T_nrVert$. You get the $liminf$ if the convergence is not norm-convergence, but for example pointwise convergence (and both spaces should be Banach, although, if you know that the limit exists everywhere, you need not require $Y$ to be complete).
$endgroup$
– Daniel Fischer
Feb 7 '14 at 15:48
$begingroup$
and how exactly do you see then that it is the liminf?
$endgroup$
– user66906
Feb 7 '14 at 15:53