T linear continuous function of normed spaces with $|T|<1$, then $|T^n|le |T|^n$
up vote
2
down vote
favorite
Let $E$ be a Banach space (complete normed vector space) and $T:Eto
E$ linear and continuous with $|T|<1$, with the norm $|f|=sup_{xin S}|f(x)|$ where $S$ is the unit sphere $S = {xin E$ st. $|x|=1 }$. Then $|T^n|le |T|^n$ for any $nge 0$.
I'm trying by induction: for $n=0, |Id| = 1$. For $n+1$ I think I should use $|T cdot S|le |T||S|$ but I'm stuck in proving this last inequality.
Thanks in advance for any hints.
linear-algebra metric-spaces
asked Nov 27 at 20:48
AnalyticHarmony
642313
add a comment |
up vote
2
down vote
favorite
Let $E$ be a Banach space (complete normed vector space) and $T:Eto
E$ linear and continuous with $|T|<1$, with the norm $|f|=sup_{xin S}|f(x)|$ where $S$ is the unit sphere $S = {xin E$ st. $|x|=1 }$. Then $|T^n|le |T|^n$ for any $nge 0$.
I'm trying by induction: for $n=0, |Id| = 1$. For $n+1$ I think I should use $|T cdot S|le |T||S|$ but I'm stuck in proving this last inequality.
Thanks in advance for any hints.
linear-algebra metric-spaces
asked Nov 27 at 20:48
AnalyticHarmony
642313
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $E$ be a Banach space (complete normed vector space) and $T:Eto
E$ linear and continuous with $|T|<1$, with the norm $|f|=sup_{xin S}|f(x)|$ where $S$ is the unit sphere $S = {xin E$ st. $|x|=1 }$. Then $|T^n|le |T|^n$ for any $nge 0$.
I'm trying by induction: for $n=0, |Id| = 1$. For $n+1$ I think I should use $|T cdot S|le |T||S|$ but I'm stuck in proving this last inequality.
Thanks in advance for any hints.
linear-algebra metric-spaces
asked Nov 27 at 20:48
AnalyticHarmony
642313
Let $E$ be a Banach space (complete normed vector space) and $T:Eto
E$ linear and continuous with $|T|<1$, with the norm $|f|=sup_{xin S}|f(x)|$ where $S$ is the unit sphere $S = {xin E$ st. $|x|=1 }$. Then $|T^n|le |T|^n$ for any $nge 0$.
I'm trying by induction: for $n=0, |Id| = 1$. For $n+1$ I think I should use $|T cdot S|le |T||S|$ but I'm stuck in proving this last inequality.
Thanks in advance for any hints.
linear-algebra metric-spaces
linear-algebra metric-spaces
asked Nov 27 at 20:48
AnalyticHarmony
642313
asked Nov 27 at 20:48
AnalyticHarmony
642313
edited Nov 27 at 22:11
asked Nov 27 at 20:48
AnalyticHarmony
642313
asked Nov 27 at 20:48
AnalyticHarmony
642313
asked Nov 27 at 20:48
AnalyticHarmony
642313
642313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
To prove the first inequality:
$$
|TS| = sup_{|x| = 1}|TSx| leq|T| sup_{|x| = 1}|Sx| leq |T||S|
$$
This follows from the fact that for any $x$:
$$
|Tx| = left|T left(|x|frac{x}{|x|}right)right| = |x|left|T left(frac{x}{|x|}right)right| leq |x|sup_{y = 1} |Ty| = |x||T|
$$
And this result uses the fact that $frac{x}{|x|}$ is a unit (norm 1) vector.
answered Nov 27 at 22:23
rubikscube09
1,169717
add a comment |
up vote
2
down vote
Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.
To complete it, just observe the following
$$|T^{n+1}|=|T^nT|le |T^n||T|le |T|^n|T|=|T|^{n+1},$$
where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.
answered Nov 27 at 20:52
jgon
12k21840
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
To prove the first inequality:
$$
|TS| = sup_{|x| = 1}|TSx| leq|T| sup_{|x| = 1}|Sx| leq |T||S|
$$
This follows from the fact that for any $x$:
$$
|Tx| = left|T left(|x|frac{x}{|x|}right)right| = |x|left|T left(frac{x}{|x|}right)right| leq |x|sup_{y = 1} |Ty| = |x||T|
$$
And this result uses the fact that $frac{x}{|x|}$ is a unit (norm 1) vector.
answered Nov 27 at 22:23
rubikscube09
1,169717
add a comment |
up vote
4
down vote
accepted
To prove the first inequality:
$$
|TS| = sup_{|x| = 1}|TSx| leq|T| sup_{|x| = 1}|Sx| leq |T||S|
$$
This follows from the fact that for any $x$:
$$
|Tx| = left|T left(|x|frac{x}{|x|}right)right| = |x|left|T left(frac{x}{|x|}right)right| leq |x|sup_{y = 1} |Ty| = |x||T|
$$
And this result uses the fact that $frac{x}{|x|}$ is a unit (norm 1) vector.
answered Nov 27 at 22:23
rubikscube09
1,169717
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
To prove the first inequality:
$$
|TS| = sup_{|x| = 1}|TSx| leq|T| sup_{|x| = 1}|Sx| leq |T||S|
$$
This follows from the fact that for any $x$:
$$
|Tx| = left|T left(|x|frac{x}{|x|}right)right| = |x|left|T left(frac{x}{|x|}right)right| leq |x|sup_{y = 1} |Ty| = |x||T|
$$
And this result uses the fact that $frac{x}{|x|}$ is a unit (norm 1) vector.
answered Nov 27 at 22:23
rubikscube09
1,169717
To prove the first inequality:
$$
|TS| = sup_{|x| = 1}|TSx| leq|T| sup_{|x| = 1}|Sx| leq |T||S|
$$
This follows from the fact that for any $x$:
$$
|Tx| = left|T left(|x|frac{x}{|x|}right)right| = |x|left|T left(frac{x}{|x|}right)right| leq |x|sup_{y = 1} |Ty| = |x||T|
$$
And this result uses the fact that $frac{x}{|x|}$ is a unit (norm 1) vector.
answered Nov 27 at 22:23
rubikscube09
1,169717
answered Nov 27 at 22:23
rubikscube09
1,169717
answered Nov 27 at 22:23
rubikscube09
1,169717
answered Nov 27 at 22:23
rubikscube09
1,169717
1,169717
add a comment |
add a comment |
up vote
2
down vote
Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.
To complete it, just observe the following
$$|T^{n+1}|=|T^nT|le |T^n||T|le |T|^n|T|=|T|^{n+1},$$
where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.
answered Nov 27 at 20:52
jgon
12k21840
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
add a comment |
up vote
2
down vote
Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.
To complete it, just observe the following
$$|T^{n+1}|=|T^nT|le |T^n||T|le |T|^n|T|=|T|^{n+1},$$
where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.
answered Nov 27 at 20:52
jgon
12k21840
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
add a comment |
up vote
2
down vote
up vote
2
down vote
Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.
To complete it, just observe the following
$$|T^{n+1}|=|T^nT|le |T^n||T|le |T|^n|T|=|T|^{n+1},$$
where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.
answered Nov 27 at 20:52
jgon
12k21840
Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.
To complete it, just observe the following
$$|T^{n+1}|=|T^nT|le |T^n||T|le |T|^n|T|=|T|^{n+1},$$
where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.
answered Nov 27 at 20:52
jgon
12k21840
answered Nov 27 at 20:52
jgon
12k21840
answered Nov 27 at 20:52
jgon
12k21840
answered Nov 27 at 20:52
jgon
12k21840
12k21840
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
add a comment |
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 at 22:07
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
@AnalyticHarmony Ah sorry, I misunderstood your question, I see you've edited and rubikscube09 has given a good answer.
– jgon
Nov 28 at 21:24
add a comment |
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