Mixing
proof about self-adjoint operators $T,P$
$begingroup$
Prove that for self-adjoint operators $T,P in B(H)$ where $H$ is a Hilbert space ($B(H)$ is the space of all bounded functions from $H to H$) and $T leq P$ then $|T|leq |P|$ .
So by assumption $Tleq P$ so $langle Tx,x rangle leq langle Px,x rangle$ for every $x in H$. So then the norm is just the supremum over all $xin H$ where $|x| = 1$ and this holds because the inequality was true for every $xin H$. Is this correct? I feel like it is quite obvious.
functional-analysis operator-theory hilbert-spaces self-adjoint-operators
asked Feb 1 at 17:35
mandellamandella
782522
$endgroup$
add a comment |
$begingroup$
Prove that for self-adjoint operators $T,P in B(H)$ where $H$ is a Hilbert space ($B(H)$ is the space of all bounded functions from $H to H$) and $T leq P$ then $|T|leq |P|$ .
So by assumption $Tleq P$ so $langle Tx,x rangle leq langle Px,x rangle$ for every $x in H$. So then the norm is just the supremum over all $xin H$ where $|x| = 1$ and this holds because the inequality was true for every $xin H$. Is this correct? I feel like it is quite obvious.
functional-analysis operator-theory hilbert-spaces self-adjoint-operators
asked Feb 1 at 17:35
mandellamandella
782522
$endgroup$
1
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56
add a comment |
$begingroup$
Prove that for self-adjoint operators $T,P in B(H)$ where $H$ is a Hilbert space ($B(H)$ is the space of all bounded functions from $H to H$) and $T leq P$ then $|T|leq |P|$ .
So by assumption $Tleq P$ so $langle Tx,x rangle leq langle Px,x rangle$ for every $x in H$. So then the norm is just the supremum over all $xin H$ where $|x| = 1$ and this holds because the inequality was true for every $xin H$. Is this correct? I feel like it is quite obvious.
functional-analysis operator-theory hilbert-spaces self-adjoint-operators
asked Feb 1 at 17:35
mandellamandella
782522
$endgroup$
Prove that for self-adjoint operators $T,P in B(H)$ where $H$ is a Hilbert space ($B(H)$ is the space of all bounded functions from $H to H$) and $T leq P$ then $|T|leq |P|$ .
So by assumption $Tleq P$ so $langle Tx,x rangle leq langle Px,x rangle$ for every $x in H$. So then the norm is just the supremum over all $xin H$ where $|x| = 1$ and this holds because the inequality was true for every $xin H$. Is this correct? I feel like it is quite obvious.
functional-analysis operator-theory hilbert-spaces self-adjoint-operators
functional-analysis operator-theory hilbert-spaces self-adjoint-operators
asked Feb 1 at 17:35
mandellamandella
782522
asked Feb 1 at 17:35
mandellamandella
782522
asked Feb 1 at 17:35
mandellamandella
782522
asked Feb 1 at 17:35
mandellamandella
782522
asked Feb 1 at 17:35
mandellamandella
782522
782522
1
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56
add a comment |
1
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56
1
1
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $Ole Tle P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $|T|le |P|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $lambda = |P|$ be the maximal element of the spectrum $sigma(P)$. Since $sigma(P)subset [0,lambda]$, we have $Ole Tle Ple lambda I$. We find that $
sigma(T)subset [0,lambda],
$ which implies that $|T|le lambda = |P|$.
answered Feb 1 at 18:00
SongSong
18.6k21651
$endgroup$
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $Ole Tle P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $|T|le |P|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $lambda = |P|$ be the maximal element of the spectrum $sigma(P)$. Since $sigma(P)subset [0,lambda]$, we have $Ole Tle Ple lambda I$. We find that $
sigma(T)subset [0,lambda],
$ which implies that $|T|le lambda = |P|$.
answered Feb 1 at 18:00
SongSong
18.6k21651
$endgroup$
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
add a comment |
$begingroup$
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $Ole Tle P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $|T|le |P|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $lambda = |P|$ be the maximal element of the spectrum $sigma(P)$. Since $sigma(P)subset [0,lambda]$, we have $Ole Tle Ple lambda I$. We find that $
sigma(T)subset [0,lambda],
$ which implies that $|T|le lambda = |P|$.
answered Feb 1 at 18:00
SongSong
18.6k21651
$endgroup$
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
add a comment |
$begingroup$
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $Ole Tle P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $|T|le |P|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $lambda = |P|$ be the maximal element of the spectrum $sigma(P)$. Since $sigma(P)subset [0,lambda]$, we have $Ole Tle Ple lambda I$. We find that $
sigma(T)subset [0,lambda],
$ which implies that $|T|le lambda = |P|$.
answered Feb 1 at 18:00
SongSong
18.6k21651
$endgroup$
We need additional assumption to make your statement as true. A very obvious counterexample is $T=-2I$, $P=I$.
If we make additional assumption that $Ole Tle P$, i.e. $T,P$ are non-negative elements of $B(H)$, then it is true that $|T|le |P|$. Here's a spectral argument that can show this. Note that the norm is equal to the spectral radius for every self-adjoint operator. Let $lambda = |P|$ be the maximal element of the spectrum $sigma(P)$. Since $sigma(P)subset [0,lambda]$, we have $Ole Tle Ple lambda I$. We find that $
sigma(T)subset [0,lambda],
$ which implies that $|T|le lambda = |P|$.
answered Feb 1 at 18:00
SongSong
18.6k21651
answered Feb 1 at 18:00
SongSong
18.6k21651
answered Feb 1 at 18:00
SongSong
18.6k21651
answered Feb 1 at 18:00
SongSong
18.6k21651
18.6k21651
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
add a comment |
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
$begingroup$
I understand that we need the additional assumption, and I also like your proof. I was just wondering if my proof is ok under the assumption that they are non-negative?
$endgroup$
– mandella
Feb 2 at 14:37
1
1
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
@mandella Your proof is very correct!
$endgroup$
– Song
Feb 2 at 16:47
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
$begingroup$
Thank you very much!
$endgroup$
– mandella
Feb 2 at 20:03
add a comment |
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1
$begingroup$
That's correct.
$endgroup$
– Stefan Lafon
Feb 1 at 17:56