T linear continuous function of normed spaces with $|T|<1$, then $|T^n|le |T|^n$
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 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
add a comment |
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 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
add a comment |
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 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
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 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
asked Nov 27 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
edited Nov 27 '18 at 22:11
AnalyticHarmony
asked Nov 27 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
asked Nov 27 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
asked Nov 27 '18 at 20:48
AnalyticHarmonyAnalyticHarmony
642313
642313
add a comment |
add a comment |
2 Answers
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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 '18 at 22:23
rubikscube09rubikscube09
1,174718
add a comment |
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 '18 at 20:52
jgonjgon
13.4k22041
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 at 21:24
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
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 '18 at 22:23
rubikscube09rubikscube09
1,174718
add a comment |
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 '18 at 22:23
rubikscube09rubikscube09
1,174718
add a comment |
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 '18 at 22:23
rubikscube09rubikscube09
1,174718
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 '18 at 22:23
rubikscube09rubikscube09
1,174718
answered Nov 27 '18 at 22:23
rubikscube09rubikscube09
1,174718
answered Nov 27 '18 at 22:23
rubikscube09rubikscube09
1,174718
answered Nov 27 '18 at 22:23
rubikscube09rubikscube09
1,174718
1,174718
add a comment |
add a comment |
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 '18 at 20:52
jgonjgon
13.4k22041
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 at 21:24
add a comment |
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 '18 at 20:52
jgonjgon
13.4k22041
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 at 21:24
add a comment |
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 '18 at 20:52
jgonjgon
13.4k22041
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 '18 at 20:52
jgonjgon
13.4k22041
answered Nov 27 '18 at 20:52
jgonjgon
13.4k22041
answered Nov 27 '18 at 20:52
jgonjgon
13.4k22041
answered Nov 27 '18 at 20:52
jgonjgon
13.4k22041
13.4k22041
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 at 21:24
add a comment |
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 at 21:24
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 at 22:07
And how can one prove that first inequality?
– AnalyticHarmony
Nov 27 '18 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 '18 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 '18 at 21:24
add a comment |
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