Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
add a comment |
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
1
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
edited Nov 2 '17 at 14:10
amWhy
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
edited Nov 2 '17 at 14:10
amWhy
192k28224439
edited Nov 2 '17 at 14:10
amWhy
192k28224439
edited Nov 2 '17 at 14:10
amWhy
192k28224439
192k28224439
asked Nov 2 '17 at 8:24
NYRAHHH
35129
asked Nov 2 '17 at 8:24
NYRAHHH
35129
asked Nov 2 '17 at 8:24
NYRAHHH
35129
35129
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
1
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
1
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
1
1
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
1 Answer
1
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oldest
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This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
add a comment |
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This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
add a comment |
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
add a comment |
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
answered Dec 4 '18 at 1:57
Martin Argerami
124k1176174
124k1176174
add a comment |
add a comment |
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Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
– Harambe
Nov 2 '17 at 8:31
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
– Demophilus
Nov 2 '17 at 13:22
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
– Fedor Goncharov
Nov 2 '17 at 16:52
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
– Fedor Goncharov
Nov 2 '17 at 17:03
1
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
– mechanodroid
Nov 4 '17 at 10:53