Norm limit of sequence of orthogonal projections on Hilbert space “contractive”
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Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?
operator-theory hilbert-spaces projection
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
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add a comment |
$begingroup$
Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?
operator-theory hilbert-spaces projection
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
$endgroup$
add a comment |
$begingroup$
Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?
operator-theory hilbert-spaces projection
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
$endgroup$
Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?
operator-theory hilbert-spaces projection
operator-theory hilbert-spaces projection
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
asked Dec 11 '18 at 8:55
worldreporter14worldreporter14
31318
31318
add a comment |
add a comment |
1 Answer
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If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
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– worldreporter14
Dec 11 '18 at 11:01
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
add a comment |
$begingroup$
If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
add a comment |
$begingroup$
If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
edited Dec 11 '18 at 9:15
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Dec 11 '18 at 8:58
Kavi Rama MurthyKavi Rama Murthy
59k42161
59k42161
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
add a comment |
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
$begingroup$
Thanks for the answer! Can we at least say that $C notin (0, 1)$?
$endgroup$
– worldreporter14
Dec 11 '18 at 11:01
add a comment |
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