Bounding || zx - yz || given that || x - y || < M in a Banach algebra.
$begingroup$
Let $ X $ be an Banach algebra (not necessarily commutative), and let $ x, y, z in X $. Suppose that $ | x - y | < M $.
I want to bound $ | zx - yz | $ in terms of $ M $ by writing $ zx - yz $ as a product of terms including the term $ x - y $, but I don't see how to do so (or if it is even possible).
If it is possible, I am assuming it is just an elementary factoring trick that I am blanking out on.
Any help would be appreciated. Thanks!
functional-analysis banach-algebras
asked Jan 21 at 18:26
LMWLMW
1207
$endgroup$
add a comment |
$begingroup$
Let $ X $ be an Banach algebra (not necessarily commutative), and let $ x, y, z in X $. Suppose that $ | x - y | < M $.
I want to bound $ | zx - yz | $ in terms of $ M $ by writing $ zx - yz $ as a product of terms including the term $ x - y $, but I don't see how to do so (or if it is even possible).
If it is possible, I am assuming it is just an elementary factoring trick that I am blanking out on.
Any help would be appreciated. Thanks!
functional-analysis banach-algebras
asked Jan 21 at 18:26
LMWLMW
1207
$endgroup$
$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39
add a comment |
$begingroup$
Let $ X $ be an Banach algebra (not necessarily commutative), and let $ x, y, z in X $. Suppose that $ | x - y | < M $.
I want to bound $ | zx - yz | $ in terms of $ M $ by writing $ zx - yz $ as a product of terms including the term $ x - y $, but I don't see how to do so (or if it is even possible).
If it is possible, I am assuming it is just an elementary factoring trick that I am blanking out on.
Any help would be appreciated. Thanks!
functional-analysis banach-algebras
asked Jan 21 at 18:26
LMWLMW
1207
$endgroup$
Let $ X $ be an Banach algebra (not necessarily commutative), and let $ x, y, z in X $. Suppose that $ | x - y | < M $.
I want to bound $ | zx - yz | $ in terms of $ M $ by writing $ zx - yz $ as a product of terms including the term $ x - y $, but I don't see how to do so (or if it is even possible).
If it is possible, I am assuming it is just an elementary factoring trick that I am blanking out on.
Any help would be appreciated. Thanks!
functional-analysis banach-algebras
functional-analysis banach-algebras
asked Jan 21 at 18:26
LMWLMW
1207
asked Jan 21 at 18:26
LMWLMW
1207
edited Jan 21 at 18:39
LMW
asked Jan 21 at 18:26
LMWLMW
1207
asked Jan 21 at 18:26
LMWLMW
1207
asked Jan 21 at 18:26
LMWLMW
1207
1207
$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39
add a comment |
$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39
$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39
add a comment |
1 Answer
1
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$begingroup$
You can't in general. If you pick $y=x$ then you have $|x-y|<M$ for all $M$, while $|zx-yz|=|[z,x]|$ may be arbitrary (pick a ring of matrices for example).
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
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
$begingroup$
You can't in general. If you pick $y=x$ then you have $|x-y|<M$ for all $M$, while $|zx-yz|=|[z,x]|$ may be arbitrary (pick a ring of matrices for example).
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
add a comment |
$begingroup$
You can't in general. If you pick $y=x$ then you have $|x-y|<M$ for all $M$, while $|zx-yz|=|[z,x]|$ may be arbitrary (pick a ring of matrices for example).
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
add a comment |
$begingroup$
You can't in general. If you pick $y=x$ then you have $|x-y|<M$ for all $M$, while $|zx-yz|=|[z,x]|$ may be arbitrary (pick a ring of matrices for example).
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
$endgroup$
You can't in general. If you pick $y=x$ then you have $|x-y|<M$ for all $M$, while $|zx-yz|=|[z,x]|$ may be arbitrary (pick a ring of matrices for example).
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
answered Jan 21 at 18:56
Jose BroxJose Brox
3,15711128
3,15711128
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
add a comment |
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
$begingroup$
@LMW You didn't like my answer?
$endgroup$
– Jose Brox
Jan 24 at 22:09
add a comment |
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$begingroup$
For $M = 0$ you do not have a chance.
$endgroup$
– Paul Frost
Jan 21 at 18:37
$begingroup$
Whoops. I should have wrote strict inequality like in the title. I'll edit it now. Thanks for the catch.
$endgroup$
– LMW
Jan 21 at 18:39