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Example Not convergent uniformly
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Are there Hilbert space $H$ and self-adjoint operator $T$ s.t., $||T||=1$, $(Tx,x)geq 0 (forall x in H)$ and $T^n$ not convergent uniformly?
　
I proved $T^n$ convergent strongly by using spectrum theory. But I can't prove it convergent uniformly, so I guess counterexample exists.
operator-theory hilbert-spaces uniform-convergence strong-convergence
asked Jul 20 at 18:01
B.T.O
250111
add a comment |Â
up vote
0
down vote
favorite
Are there Hilbert space $H$ and self-adjoint operator $T$ s.t., $||T||=1$, $(Tx,x)geq 0 (forall x in H)$ and $T^n$ not convergent uniformly?
　
I proved $T^n$ convergent strongly by using spectrum theory. But I can't prove it convergent uniformly, so I guess counterexample exists.
operator-theory hilbert-spaces uniform-convergence strong-convergence
asked Jul 20 at 18:01
B.T.O
250111
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Are there Hilbert space $H$ and self-adjoint operator $T$ s.t., $||T||=1$, $(Tx,x)geq 0 (forall x in H)$ and $T^n$ not convergent uniformly?
　
I proved $T^n$ convergent strongly by using spectrum theory. But I can't prove it convergent uniformly, so I guess counterexample exists.
operator-theory hilbert-spaces uniform-convergence strong-convergence
asked Jul 20 at 18:01
B.T.O
250111
Are there Hilbert space $H$ and self-adjoint operator $T$ s.t., $||T||=1$, $(Tx,x)geq 0 (forall x in H)$ and $T^n$ not convergent uniformly?
　
I proved $T^n$ convergent strongly by using spectrum theory. But I can't prove it convergent uniformly, so I guess counterexample exists.
operator-theory hilbert-spaces uniform-convergence strong-convergence
asked Jul 20 at 18:01
B.T.O
250111
asked Jul 20 at 18:01
B.T.O
250111
asked Jul 20 at 18:01
B.T.O
250111
asked Jul 20 at 18:01
B.T.O
250111
250111
add a comment |Â
add a comment |Â
2 Answers
2
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up vote
1
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accepted
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H to H$ as $T = operatornamediag(alpha_n)_n$, that is $Tx = sum_n=1^infty alpha_n langle x, e_nrangle e_n$ where $(alpha_n)_n$ is a bounded sequence of scalars.
We have $|T| = sup_ninmathbbN |alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $alpha_n ge 0, forall n in mathbbN$.
Notice that $T^k = operatornamediag(alpha^k_n)_n$ so $$T^ke_n = alpha_n^k e_n$$
and this converges to $0$ when $ktoinfty$ if and only if $|alpha_n| < 1$.
Thus if $|alpha_n| < 1, forall n in mathbbN$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k to 0$ uniformly if and only if $$|T|^k to 0$$ if and only if $|T| < 1$.
So just take a sequence with $sup_ninmathbbN |alpha_n| ge 1$, such as the one @daw suggests: $alpha_n = fracnn+1$.
The only other option for the existence of the strong limit is that for some $ninmathbbN$ we have $alpha_n = 1$.
In that case the candidate for the limit is $operatornamediag(beta_n)_n$ where $beta_n =begincases 1,textif alpha_n = 1\
0, textif |alpha_n| < 1endcases$, and $T^k to operatornamediag(beta_n)_n$ uniformly if and only if $$sup_alpha_n |alpha_n| < 1$$
answered Jul 22 at 13:11
mechanodroid
22.2k52041
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
add a comment |Â
up vote
1
down vote
What about $H=l^2$ and $T$ defined by
$$
(Tx)_n = fracnn+1x_n.
$$
Then for the unit vector $e_k$ we have
$$
(T^m-T^n)e_k = left(left(frackk+1right)^m-left(frackk+1right)^nright)e_k,
$$
and $|(T^m-T^n)e_k|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
edited Jul 22 at 15:56
mechanodroid
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
Why the downvote?
– daw
Jul 22 at 14:53
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H to H$ as $T = operatornamediag(alpha_n)_n$, that is $Tx = sum_n=1^infty alpha_n langle x, e_nrangle e_n$ where $(alpha_n)_n$ is a bounded sequence of scalars.
We have $|T| = sup_ninmathbbN |alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $alpha_n ge 0, forall n in mathbbN$.
Notice that $T^k = operatornamediag(alpha^k_n)_n$ so $$T^ke_n = alpha_n^k e_n$$
and this converges to $0$ when $ktoinfty$ if and only if $|alpha_n| < 1$.
Thus if $|alpha_n| < 1, forall n in mathbbN$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k to 0$ uniformly if and only if $$|T|^k to 0$$ if and only if $|T| < 1$.
So just take a sequence with $sup_ninmathbbN |alpha_n| ge 1$, such as the one @daw suggests: $alpha_n = fracnn+1$.
The only other option for the existence of the strong limit is that for some $ninmathbbN$ we have $alpha_n = 1$.
In that case the candidate for the limit is $operatornamediag(beta_n)_n$ where $beta_n =begincases 1,textif alpha_n = 1\
0, textif |alpha_n| < 1endcases$, and $T^k to operatornamediag(beta_n)_n$ uniformly if and only if $$sup_alpha_n |alpha_n| < 1$$
answered Jul 22 at 13:11
mechanodroid
22.2k52041
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
add a comment |Â
up vote
1
down vote
accepted
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H to H$ as $T = operatornamediag(alpha_n)_n$, that is $Tx = sum_n=1^infty alpha_n langle x, e_nrangle e_n$ where $(alpha_n)_n$ is a bounded sequence of scalars.
We have $|T| = sup_ninmathbbN |alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $alpha_n ge 0, forall n in mathbbN$.
Notice that $T^k = operatornamediag(alpha^k_n)_n$ so $$T^ke_n = alpha_n^k e_n$$
and this converges to $0$ when $ktoinfty$ if and only if $|alpha_n| < 1$.
Thus if $|alpha_n| < 1, forall n in mathbbN$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k to 0$ uniformly if and only if $$|T|^k to 0$$ if and only if $|T| < 1$.
So just take a sequence with $sup_ninmathbbN |alpha_n| ge 1$, such as the one @daw suggests: $alpha_n = fracnn+1$.
The only other option for the existence of the strong limit is that for some $ninmathbbN$ we have $alpha_n = 1$.
In that case the candidate for the limit is $operatornamediag(beta_n)_n$ where $beta_n =begincases 1,textif alpha_n = 1\
0, textif |alpha_n| < 1endcases$, and $T^k to operatornamediag(beta_n)_n$ uniformly if and only if $$sup_alpha_n |alpha_n| < 1$$
answered Jul 22 at 13:11
mechanodroid
22.2k52041
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H to H$ as $T = operatornamediag(alpha_n)_n$, that is $Tx = sum_n=1^infty alpha_n langle x, e_nrangle e_n$ where $(alpha_n)_n$ is a bounded sequence of scalars.
We have $|T| = sup_ninmathbbN |alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $alpha_n ge 0, forall n in mathbbN$.
Notice that $T^k = operatornamediag(alpha^k_n)_n$ so $$T^ke_n = alpha_n^k e_n$$
and this converges to $0$ when $ktoinfty$ if and only if $|alpha_n| < 1$.
Thus if $|alpha_n| < 1, forall n in mathbbN$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k to 0$ uniformly if and only if $$|T|^k to 0$$ if and only if $|T| < 1$.
So just take a sequence with $sup_ninmathbbN |alpha_n| ge 1$, such as the one @daw suggests: $alpha_n = fracnn+1$.
The only other option for the existence of the strong limit is that for some $ninmathbbN$ we have $alpha_n = 1$.
In that case the candidate for the limit is $operatornamediag(beta_n)_n$ where $beta_n =begincases 1,textif alpha_n = 1\
0, textif |alpha_n| < 1endcases$, and $T^k to operatornamediag(beta_n)_n$ uniformly if and only if $$sup_alpha_n |alpha_n| < 1$$
answered Jul 22 at 13:11
mechanodroid
22.2k52041
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H to H$ as $T = operatornamediag(alpha_n)_n$, that is $Tx = sum_n=1^infty alpha_n langle x, e_nrangle e_n$ where $(alpha_n)_n$ is a bounded sequence of scalars.
We have $|T| = sup_ninmathbbN |alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $alpha_n ge 0, forall n in mathbbN$.
Notice that $T^k = operatornamediag(alpha^k_n)_n$ so $$T^ke_n = alpha_n^k e_n$$
and this converges to $0$ when $ktoinfty$ if and only if $|alpha_n| < 1$.
Thus if $|alpha_n| < 1, forall n in mathbbN$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k to 0$ uniformly if and only if $$|T|^k to 0$$ if and only if $|T| < 1$.
So just take a sequence with $sup_ninmathbbN |alpha_n| ge 1$, such as the one @daw suggests: $alpha_n = fracnn+1$.
The only other option for the existence of the strong limit is that for some $ninmathbbN$ we have $alpha_n = 1$.
In that case the candidate for the limit is $operatornamediag(beta_n)_n$ where $beta_n =begincases 1,textif alpha_n = 1\
0, textif |alpha_n| < 1endcases$, and $T^k to operatornamediag(beta_n)_n$ uniformly if and only if $$sup_alpha_n |alpha_n| < 1$$
answered Jul 22 at 13:11
mechanodroid
22.2k52041
edited Jul 22 at 15:54
answered Jul 22 at 13:11
mechanodroid
22.2k52041
answered Jul 22 at 13:11
mechanodroid
22.2k52041
answered Jul 22 at 13:11
mechanodroid
22.2k52041
22.2k52041
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
add a comment |Â
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
$alpha_n=1$ also leads to convergence ($T=id$)
– daw
Jul 22 at 14:52
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
@daw Thanks, I agree.
– mechanodroid
Jul 22 at 15:54
add a comment |Â
up vote
1
down vote
What about $H=l^2$ and $T$ defined by
$$
(Tx)_n = fracnn+1x_n.
$$
Then for the unit vector $e_k$ we have
$$
(T^m-T^n)e_k = left(left(frackk+1right)^m-left(frackk+1right)^nright)e_k,
$$
and $|(T^m-T^n)e_k|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
edited Jul 22 at 15:56
mechanodroid
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
Why the downvote?
– daw
Jul 22 at 14:53
add a comment |Â
up vote
1
down vote
What about $H=l^2$ and $T$ defined by
$$
(Tx)_n = fracnn+1x_n.
$$
Then for the unit vector $e_k$ we have
$$
(T^m-T^n)e_k = left(left(frackk+1right)^m-left(frackk+1right)^nright)e_k,
$$
and $|(T^m-T^n)e_k|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
edited Jul 22 at 15:56
mechanodroid
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
Why the downvote?
– daw
Jul 22 at 14:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What about $H=l^2$ and $T$ defined by
$$
(Tx)_n = fracnn+1x_n.
$$
Then for the unit vector $e_k$ we have
$$
(T^m-T^n)e_k = left(left(frackk+1right)^m-left(frackk+1right)^nright)e_k,
$$
and $|(T^m-T^n)e_k|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
edited Jul 22 at 15:56
mechanodroid
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
What about $H=l^2$ and $T$ defined by
$$
(Tx)_n = fracnn+1x_n.
$$
Then for the unit vector $e_k$ we have
$$
(T^m-T^n)e_k = left(left(frackk+1right)^m-left(frackk+1right)^nright)e_k,
$$
and $|(T^m-T^n)e_k|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
edited Jul 22 at 15:56
mechanodroid
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
edited Jul 22 at 15:56
mechanodroid
22.2k52041
edited Jul 22 at 15:56
mechanodroid
22.2k52041
edited Jul 22 at 15:56
mechanodroid
22.2k52041
22.2k52041
answered Jul 22 at 7:42
daw
21.9k1542
answered Jul 22 at 7:42
daw
21.9k1542
answered Jul 22 at 7:42
daw
21.9k1542
21.9k1542
Why the downvote?
– daw
Jul 22 at 14:53
add a comment |Â
Why the downvote?
– daw
Jul 22 at 14:53
Why the downvote?
– daw
Jul 22 at 14:53
Why the downvote?
– daw
Jul 22 at 14:53
add a comment |Â
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