Quasinilpotent , non-compact operators
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked Mar 22 at 16:48
MarkusMarkus
37018
$endgroup$
add a comment |
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked Mar 22 at 16:48
MarkusMarkus
37018
$endgroup$
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05
add a comment |
5
5
5
$begingroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked Mar 22 at 16:48
MarkusMarkus
37018
$endgroup$
If $X$ is a separable Banach space and $(epsilon_n)downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
fa.functional-analysis banach-spaces operator-theory hilbert-spaces
asked Mar 22 at 16:48
MarkusMarkus
37018
asked Mar 22 at 16:48
MarkusMarkus
37018
asked Mar 22 at 16:48
MarkusMarkus
37018
asked Mar 22 at 16:48
MarkusMarkus
37018
asked Mar 22 at 16:48
MarkusMarkus
37018
37018
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05
add a comment |
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05
2
2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
2
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05
add a comment |
1 Answer
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7
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On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
add a comment |
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
add a comment |
7
7
7
$begingroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
$endgroup$
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
answered Mar 22 at 17:03
Bill JohnsonBill Johnson
24.4k371117
24.4k371117
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
add a comment |
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
3
3
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
Mar 22 at 17:35
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
$begingroup$
Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular.
$endgroup$
– Bill Johnson
2 days ago
add a comment |
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2
$begingroup$
What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+frac{1}{2})$ if $x<frac{1}{2}$ and $(Tf)(x)=0$ elswhere?
$endgroup$
– András Bátkai
Mar 22 at 16:59
2
$begingroup$
@AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.)
$endgroup$
– Jochen Glueck
Mar 22 at 17:02
$begingroup$
@AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(epsilon_n)$.
$endgroup$
– Jochen Glueck
Mar 22 at 17:05