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 13 hours ago
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 13 hours ago
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
13 hours ago
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
13 hours ago
$begingroup$
@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
12 hours ago
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 13 hours ago
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 13 hours ago
MarkusMarkus
37018
asked 13 hours ago
MarkusMarkus
37018
asked 13 hours ago
MarkusMarkus
37018
asked 13 hours ago
MarkusMarkus
37018
asked 13 hours ago
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
13 hours ago
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
13 hours ago
$begingroup$
@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
12 hours ago
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
13 hours ago
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
13 hours ago
$begingroup$
@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
12 hours ago
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
13 hours ago
$begingroup$
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
13 hours ago
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
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
@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
12 hours ago
$begingroup$
@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
12 hours ago
add a comment |
1 Answer
1
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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 13 hours ago
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
12 hours 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
7 hours 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 13 hours ago
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
12 hours 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
7 hours 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 13 hours ago
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
12 hours 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
7 hours 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 13 hours ago
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 13 hours ago
Bill JohnsonBill Johnson
24.4k371117
answered 13 hours ago
Bill JohnsonBill Johnson
24.4k371117
answered 13 hours ago
Bill JohnsonBill Johnson
24.4k371117
answered 13 hours ago
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
12 hours 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
7 hours 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
12 hours 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
7 hours 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
12 hours ago
$begingroup$
Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"?
$endgroup$
– Fedor Petrov
12 hours 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
7 hours 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
7 hours 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
13 hours ago
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
13 hours ago
$begingroup$
@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
12 hours ago