Show that $forall y in X$ the equation $x= y + Tx$ has a unique solution
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
12.9k21041
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
12.9k21041
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
12.9k21041
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
12.9k21041
asked Nov 17 at 10:58
Rebellos
12.9k21041
edited Nov 17 at 11:11
asked Nov 17 at 10:58
Rebellos
12.9k21041
asked Nov 17 at 10:58
Rebellos
12.9k21041
asked Nov 17 at 10:58
Rebellos
12.9k21041
12.9k21041
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
10.9k21437
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
10.9k21437
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
10.9k21437
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
10.9k21437
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
10.9k21437
answered Nov 17 at 11:29
Calvin Khor
10.9k21437
answered Nov 17 at 11:29
Calvin Khor
10.9k21437
answered Nov 17 at 11:29
Calvin Khor
10.9k21437
10.9k21437
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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002213%2fshow-that-forall-y-in-x-the-equation-x-y-tx-has-a-unique-solution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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