Show that $forall y in X$ the equation $x= y + Tx$ has a unique solution
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Exercise :
Let $X$ be a Banach space, $T in B(X)$ which means that $T$ is a bounded linear operator $T : X to X$. We suppose that for all $y in X$ the series
$$sum_{n=1}^infty|T^ny|$$
converges. Show that for all $y in X$, the equation $x = y + Tx$ has a unique solution.
Attempt/Thoughts :
Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n to T$ as $nto infty$ over $B(X)$.
Since the given series is convergent, this means that :
$$Bigg|sum_{n=k+1}^infty|T^ny|Bigg| < epsilon $$
Also, since $T$ is a bounded operator, this means that :
$$|Ty| leq M |y|$$
I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !
real-analysis sequences-and-series functional-analysis operator-theory banach-spaces
asked Nov 17 at 10:58
Rebellos
12.5k21041
add a comment |
up vote
2
down vote
favorite
Exercise :
Let $X$ be a Banach space, $T in B(X)$ which means that $T$ is a bounded linear operator $T : X to X$. We suppose that for all $y in X$ the series
$$sum_{n=1}^infty|T^ny|$$
converges. Show that for all $y in X$, the equation $x = y + Tx$ has a unique solution.
Attempt/Thoughts :
Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n to T$ as $nto infty$ over $B(X)$.
Since the given series is convergent, this means that :
$$Bigg|sum_{n=k+1}^infty|T^ny|Bigg| < epsilon $$
Also, since $T$ is a bounded operator, this means that :
$$|Ty| leq M |y|$$
I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !
real-analysis sequences-and-series functional-analysis operator-theory banach-spaces
asked Nov 17 at 10:58
Rebellos
12.5k21041
Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Exercise :
Let $X$ be a Banach space, $T in B(X)$ which means that $T$ is a bounded linear operator $T : X to X$. We suppose that for all $y in X$ the series
$$sum_{n=1}^infty|T^ny|$$
converges. Show that for all $y in X$, the equation $x = y + Tx$ has a unique solution.
Attempt/Thoughts :
Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n to T$ as $nto infty$ over $B(X)$.
Since the given series is convergent, this means that :
$$Bigg|sum_{n=k+1}^infty|T^ny|Bigg| < epsilon $$
Also, since $T$ is a bounded operator, this means that :
$$|Ty| leq M |y|$$
I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !
real-analysis sequences-and-series functional-analysis operator-theory banach-spaces
asked Nov 17 at 10:58
Rebellos
12.5k21041
Exercise :
Let $X$ be a Banach space, $T in B(X)$ which means that $T$ is a bounded linear operator $T : X to X$. We suppose that for all $y in X$ the series
$$sum_{n=1}^infty|T^ny|$$
converges. Show that for all $y in X$, the equation $x = y + Tx$ has a unique solution.
Attempt/Thoughts :
Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n to T$ as $nto infty$ over $B(X)$.
Since the given series is convergent, this means that :
$$Bigg|sum_{n=k+1}^infty|T^ny|Bigg| < epsilon $$
Also, since $T$ is a bounded operator, this means that :
$$|Ty| leq M |y|$$
I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !
real-analysis sequences-and-series functional-analysis operator-theory banach-spaces
real-analysis sequences-and-series functional-analysis operator-theory banach-spaces
asked Nov 17 at 10:58
Rebellos
12.5k21041
asked Nov 17 at 10:58
Rebellos
12.5k21041
edited Nov 17 at 11:11
asked Nov 17 at 10:58
Rebellos
12.5k21041
asked Nov 17 at 10:58
Rebellos
12.5k21041
asked Nov 17 at 10:58
Rebellos
12.5k21041
12.5k21041
Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23
add a comment |
Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23
Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
|
show 3 more comments
up vote
3
down vote
accepted
Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
|
show 3 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
answered Nov 17 at 11:29
Calvin Khor
10.8k21437
10.8k21437
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
|
show 3 more comments
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
All in all it makes sense but, how do we think of $x$ as $x = (1-T)^{-1}y$ when $Tx$ is $T(x)$ ? Also, shouldn't it be $|T| <1$ or $|T|<1$ so that the geometric series hold ?
– Rebellos
Nov 17 at 11:33
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
@Rebellos Sorry, I don't understand your first question, the fact that $Tx = T(x)$ is purely notation. Can you try to rephrase? The convergence of the series for every $y$ implies this inequality for the operator norm of $T$. (You may find the inequality $|T^n| le |T|^n$ useful)
– Calvin Khor
Nov 17 at 11:40
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
The expression $Tx = T(x)$ means that $x$ is the argument of the operator $T$. How come you can "solve" for $x$ like it's an ordinary quation (as $Tcdot x$), this is what I am asking.
– Rebellos
Nov 17 at 11:43
1
1
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
@Rebellos Well, you shouldn't be able to without proof. This is that proof.
– Calvin Khor
Nov 17 at 11:44
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
I get the concept now. Sorry for the questions, but this is the first time during both our course and my self study, that I handle something like this, so it took me some time (and your explanation) to grasp over the though process !
– Rebellos
Nov 17 at 11:47
|
show 3 more comments
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Try to arrive at "$frac 1{1-T}=1+T+T^2+T^3+ldots$"
– Hagen von Eitzen
Nov 17 at 11:16
@HagenvonEitzen Hi, thanks a lot for your comment ! I see this is the geometric series in case of $|T| <1$. How would I get to a solution using this, though ?
– Rebellos
Nov 17 at 11:23