Unique solution for $x = y + Tx$ if $T(x_1,x_2,dots) = (frac{1}{2}x_2,frac{1}{3}x_3,dots)$
up vote
2
down vote
favorite
Exercise :
Let $T:ell^infty to ell^infty$ be an operator such that :
$$T(x_1,x_2,dots) = bigg(frac{1}{2}x_2,frac{1}{3}x_3,dotsbigg)$$
Show that for all $y in ell^infty$, the equation
$$x = y + Tx$$
has a unique solution.
Attempt :
I have proved that $T$ is a linear operator. Now, $ell^infty$ is the space defined as :
$$ell^infty = {x =(x_n) : |x|< infty} quad |x| :=sup|x_n|$$
From the definition of the norm over $ell^infty$, we can observe that
$$|T(x_1,x_2,dots)|<|(x_1,x_2,dots)|$$
This means that there exists an $M<1$, such that :
$$|T(x_1,x_2,dots)|leq M|(x_1,x_2,dots)|$$
Thus, $T$ is a bounded linear operator $T in B(ell^infty)$ with $|T| leq M <1$.
Now, it is
$$x = y + Tx Leftrightarrow x-Tx = y Leftrightarrow(1-T)x=y$$
where $1$ is the identity operator.
But $ell^infty$ is a Banach space and since $|T| <1$, then it is :
$$(1-T)^{-1}=sum_{n=0}^infty T^n Leftrightarrow (1-T)^{-1}y=sum_{n=0}^infty T^ny$$
Thus $x = sum_{n=0}^infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y in ell^infty$.
Question : Is my approach correct and rigorous enough ?
real-analysis sequences-and-series functional-analysis operator-theory
|
show 1 more comment
up vote
2
down vote
favorite
Exercise :
Let $T:ell^infty to ell^infty$ be an operator such that :
$$T(x_1,x_2,dots) = bigg(frac{1}{2}x_2,frac{1}{3}x_3,dotsbigg)$$
Show that for all $y in ell^infty$, the equation
$$x = y + Tx$$
has a unique solution.
Attempt :
I have proved that $T$ is a linear operator. Now, $ell^infty$ is the space defined as :
$$ell^infty = {x =(x_n) : |x|< infty} quad |x| :=sup|x_n|$$
From the definition of the norm over $ell^infty$, we can observe that
$$|T(x_1,x_2,dots)|<|(x_1,x_2,dots)|$$
This means that there exists an $M<1$, such that :
$$|T(x_1,x_2,dots)|leq M|(x_1,x_2,dots)|$$
Thus, $T$ is a bounded linear operator $T in B(ell^infty)$ with $|T| leq M <1$.
Now, it is
$$x = y + Tx Leftrightarrow x-Tx = y Leftrightarrow(1-T)x=y$$
where $1$ is the identity operator.
But $ell^infty$ is a Banach space and since $|T| <1$, then it is :
$$(1-T)^{-1}=sum_{n=0}^infty T^n Leftrightarrow (1-T)^{-1}y=sum_{n=0}^infty T^ny$$
Thus $x = sum_{n=0}^infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y in ell^infty$.
Question : Is my approach correct and rigorous enough ?
real-analysis sequences-and-series functional-analysis operator-theory
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
1
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
1
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Exercise :
Let $T:ell^infty to ell^infty$ be an operator such that :
$$T(x_1,x_2,dots) = bigg(frac{1}{2}x_2,frac{1}{3}x_3,dotsbigg)$$
Show that for all $y in ell^infty$, the equation
$$x = y + Tx$$
has a unique solution.
Attempt :
I have proved that $T$ is a linear operator. Now, $ell^infty$ is the space defined as :
$$ell^infty = {x =(x_n) : |x|< infty} quad |x| :=sup|x_n|$$
From the definition of the norm over $ell^infty$, we can observe that
$$|T(x_1,x_2,dots)|<|(x_1,x_2,dots)|$$
This means that there exists an $M<1$, such that :
$$|T(x_1,x_2,dots)|leq M|(x_1,x_2,dots)|$$
Thus, $T$ is a bounded linear operator $T in B(ell^infty)$ with $|T| leq M <1$.
Now, it is
$$x = y + Tx Leftrightarrow x-Tx = y Leftrightarrow(1-T)x=y$$
where $1$ is the identity operator.
But $ell^infty$ is a Banach space and since $|T| <1$, then it is :
$$(1-T)^{-1}=sum_{n=0}^infty T^n Leftrightarrow (1-T)^{-1}y=sum_{n=0}^infty T^ny$$
Thus $x = sum_{n=0}^infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y in ell^infty$.
Question : Is my approach correct and rigorous enough ?
real-analysis sequences-and-series functional-analysis operator-theory
Exercise :
Let $T:ell^infty to ell^infty$ be an operator such that :
$$T(x_1,x_2,dots) = bigg(frac{1}{2}x_2,frac{1}{3}x_3,dotsbigg)$$
Show that for all $y in ell^infty$, the equation
$$x = y + Tx$$
has a unique solution.
Attempt :
I have proved that $T$ is a linear operator. Now, $ell^infty$ is the space defined as :
$$ell^infty = {x =(x_n) : |x|< infty} quad |x| :=sup|x_n|$$
From the definition of the norm over $ell^infty$, we can observe that
$$|T(x_1,x_2,dots)|<|(x_1,x_2,dots)|$$
This means that there exists an $M<1$, such that :
$$|T(x_1,x_2,dots)|leq M|(x_1,x_2,dots)|$$
Thus, $T$ is a bounded linear operator $T in B(ell^infty)$ with $|T| leq M <1$.
Now, it is
$$x = y + Tx Leftrightarrow x-Tx = y Leftrightarrow(1-T)x=y$$
where $1$ is the identity operator.
But $ell^infty$ is a Banach space and since $|T| <1$, then it is :
$$(1-T)^{-1}=sum_{n=0}^infty T^n Leftrightarrow (1-T)^{-1}y=sum_{n=0}^infty T^ny$$
Thus $x = sum_{n=0}^infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y in ell^infty$.
Question : Is my approach correct and rigorous enough ?
real-analysis sequences-and-series functional-analysis operator-theory
real-analysis sequences-and-series functional-analysis operator-theory
edited Nov 17 at 18:01
asked Nov 17 at 17:54
Rebellos
12.5k21041
12.5k21041
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
1
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
1
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47
|
show 1 more comment
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
1
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
1
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
1
1
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
1
1
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You are correct and the proof looks sufficiently rigorous. However, $|T|_text{op}$ can be computed exactly. That is, $|T|_text{op}=dfrac12$. To show this, let $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
Note that the equality holds for $z=(0,1,0,0,0,ldots)$. This implies $|T|_{text{op}}= dfrac{1}{2}$.
You can write an explicit solution $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are correct and the proof looks sufficiently rigorous. However, $|T|_text{op}$ can be computed exactly. That is, $|T|_text{op}=dfrac12$. To show this, let $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
Note that the equality holds for $z=(0,1,0,0,0,ldots)$. This implies $|T|_{text{op}}= dfrac{1}{2}$.
You can write an explicit solution $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
add a comment |
up vote
1
down vote
accepted
You are correct and the proof looks sufficiently rigorous. However, $|T|_text{op}$ can be computed exactly. That is, $|T|_text{op}=dfrac12$. To show this, let $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
Note that the equality holds for $z=(0,1,0,0,0,ldots)$. This implies $|T|_{text{op}}= dfrac{1}{2}$.
You can write an explicit solution $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are correct and the proof looks sufficiently rigorous. However, $|T|_text{op}$ can be computed exactly. That is, $|T|_text{op}=dfrac12$. To show this, let $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
Note that the equality holds for $z=(0,1,0,0,0,ldots)$. This implies $|T|_{text{op}}= dfrac{1}{2}$.
You can write an explicit solution $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
You are correct and the proof looks sufficiently rigorous. However, $|T|_text{op}$ can be computed exactly. That is, $|T|_text{op}=dfrac12$. To show this, let $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
Note that the equality holds for $z=(0,1,0,0,0,ldots)$. This implies $|T|_{text{op}}= dfrac{1}{2}$.
You can write an explicit solution $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
answered Nov 17 at 18:13
Batominovski
31.8k23190
31.8k23190
add a comment |
add a comment |
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%2f3002622%2funique-solution-for-x-y-tx-if-tx-1-x-2-dots-frac12x-2-frac1%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
You are correct and the proof looks sufficiently rigorous. However, $|T|$ can be computed exactly. That is, $|T|=dfrac12$.
– Batominovski
Nov 17 at 18:01
@Batominovski I guess that stems from the fact that $T(x_n) = frac{x_{n+1}}{n+1}$ ?
– Rebellos
Nov 17 at 18:04
Yes, see the answer below.
– Batominovski
Nov 17 at 18:14
1
Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first.
– Batominovski
Nov 17 at 18:18
1
The thing is there are some cases where you have an operator $T$ such that $big|T(x)big|< M|x|$ for all $xneq 0$, but it turns out that $|T|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $mathcal{T}$ in that example satisfies $big|mathcal{T}(f)big|<|f|$ for all nonzero $finmathcal{C}$, but $|mathcal{T}|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $fequiv 0$).
– Batominovski
Nov 17 at 19:47