Mixing
If $A$ is invertible and $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
Let $mathcal{A}$ be a unital Banach algebra and define $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n}$.
It is possible to show that $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n} = inf_{n rightarrow +infty} | A^n |^{1/n}$.
I need to show that if $A$ is invertible and $B$ is such that $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
What I did is the following: consider the element $C = A^{-1} + A^{-1}sum_{n=1}^{+infty}(BA^{-1})^n$. This is known as the Neumann series.
If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.
I tried to show if $r^{text{Gelf}}(BA^{-1}) < 1$, then $| BA^{-1} | < 1$, but I did not succeed (and maybe this isn't even true).
I would appreciate very much if someone could help me!
functional-analysis banach-algebras
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
add a comment |
Let $mathcal{A}$ be a unital Banach algebra and define $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n}$.
It is possible to show that $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n} = inf_{n rightarrow +infty} | A^n |^{1/n}$.
I need to show that if $A$ is invertible and $B$ is such that $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
What I did is the following: consider the element $C = A^{-1} + A^{-1}sum_{n=1}^{+infty}(BA^{-1})^n$. This is known as the Neumann series.
If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.
I tried to show if $r^{text{Gelf}}(BA^{-1}) < 1$, then $| BA^{-1} | < 1$, but I did not succeed (and maybe this isn't even true).
I would appreciate very much if someone could help me!
functional-analysis banach-algebras
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
add a comment |
Let $mathcal{A}$ be a unital Banach algebra and define $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n}$.
It is possible to show that $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n} = inf_{n rightarrow +infty} | A^n |^{1/n}$.
I need to show that if $A$ is invertible and $B$ is such that $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
What I did is the following: consider the element $C = A^{-1} + A^{-1}sum_{n=1}^{+infty}(BA^{-1})^n$. This is known as the Neumann series.
If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.
I tried to show if $r^{text{Gelf}}(BA^{-1}) < 1$, then $| BA^{-1} | < 1$, but I did not succeed (and maybe this isn't even true).
I would appreciate very much if someone could help me!
functional-analysis banach-algebras
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
Let $mathcal{A}$ be a unital Banach algebra and define $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n}$.
It is possible to show that $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n} = inf_{n rightarrow +infty} | A^n |^{1/n}$.
I need to show that if $A$ is invertible and $B$ is such that $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
What I did is the following: consider the element $C = A^{-1} + A^{-1}sum_{n=1}^{+infty}(BA^{-1})^n$. This is known as the Neumann series.
If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.
I tried to show if $r^{text{Gelf}}(BA^{-1}) < 1$, then $| BA^{-1} | < 1$, but I did not succeed (and maybe this isn't even true).
I would appreciate very much if someone could help me!
functional-analysis banach-algebras
functional-analysis banach-algebras
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
edited Nov 20 '18 at 18:18
Aweygan
13.5k21441
13.5k21441
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
asked Nov 20 '18 at 17:41
Luísa Borsato
1,496315
1,496315
add a comment |
add a comment |
1 Answer
1
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oldest
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HINT:
Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and
$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$
Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
add a comment |
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1 Answer
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1 Answer
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HINT:
Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and
$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$
Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
add a comment |
HINT:
Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and
$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$
Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
add a comment |
HINT:
Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and
$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$
Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
HINT:
Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and
$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$
Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
edited Nov 20 '18 at 18:30
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
answered Nov 20 '18 at 17:57
Aweygan
13.5k21441
13.5k21441
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
add a comment |
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
Yes, thank you!
– Luísa Borsato
Nov 20 '18 at 18:03
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
You're welcome. Glad to help!
– Aweygan
Nov 20 '18 at 18:11
1
1
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
Thank you very much for your side note! :)
– Luísa Borsato
Nov 20 '18 at 18:17
1
1
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
You're welcome! I was curious and thought I'd share what I found.
– Aweygan
Nov 20 '18 at 18:20
add a comment |
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