Mixing
Proving compactness of $A:l_2to l_2::Ax=(0,x_1,frac12x_2,…,frac1nx_n,…)$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Problem: Show that the operator $A:l_2to l_2::Ax=(0,x_1,frac12x_2,...,frac1nx_n,...)$ is compact but has no pontual spectrum.
My solution: $(0,x_1,frac12x_2,...,frac1nx_n,...)=(lambda x_1,lambda x_2,...,lambda x_n,...)$
Since $lambda x_1=0implies lambda=0$, then the $ker(A-lambda)=0$.
My problem lies with compactness.
If we consider a sequence in $l_2$ like $x_n$,that converges to some $x$. Consider $epsilon>0$ so that $||x_m-x||<epsilon$ for $m>NinmathbbN$. Then $||A(x_m)-A(x)||^2=||A(x_m-x)||^2=sum_limitsn=1^infty |frac1n(x_m_i-x_i)|^2<sum_limitsn=1^infty |frac1n^2(fracepsilon^frac1n2)^n|<cepsilon<delta:::delta>0$ Therefore we can find a subsequence that converges proving that $A$ is compact.
Question:
Is my proof right? If not. How should I prove compactness?
Thanks in advance!
functional-analysis
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
add a comment |Â
up vote
2
down vote
favorite
Problem: Show that the operator $A:l_2to l_2::Ax=(0,x_1,frac12x_2,...,frac1nx_n,...)$ is compact but has no pontual spectrum.
My solution: $(0,x_1,frac12x_2,...,frac1nx_n,...)=(lambda x_1,lambda x_2,...,lambda x_n,...)$
Since $lambda x_1=0implies lambda=0$, then the $ker(A-lambda)=0$.
My problem lies with compactness.
If we consider a sequence in $l_2$ like $x_n$,that converges to some $x$. Consider $epsilon>0$ so that $||x_m-x||<epsilon$ for $m>NinmathbbN$. Then $||A(x_m)-A(x)||^2=||A(x_m-x)||^2=sum_limitsn=1^infty |frac1n(x_m_i-x_i)|^2<sum_limitsn=1^infty |frac1n^2(fracepsilon^frac1n2)^n|<cepsilon<delta:::delta>0$ Therefore we can find a subsequence that converges proving that $A$ is compact.
Question:
Is my proof right? If not. How should I prove compactness?
Thanks in advance!
functional-analysis
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
2
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem: Show that the operator $A:l_2to l_2::Ax=(0,x_1,frac12x_2,...,frac1nx_n,...)$ is compact but has no pontual spectrum.
My solution: $(0,x_1,frac12x_2,...,frac1nx_n,...)=(lambda x_1,lambda x_2,...,lambda x_n,...)$
Since $lambda x_1=0implies lambda=0$, then the $ker(A-lambda)=0$.
My problem lies with compactness.
If we consider a sequence in $l_2$ like $x_n$,that converges to some $x$. Consider $epsilon>0$ so that $||x_m-x||<epsilon$ for $m>NinmathbbN$. Then $||A(x_m)-A(x)||^2=||A(x_m-x)||^2=sum_limitsn=1^infty |frac1n(x_m_i-x_i)|^2<sum_limitsn=1^infty |frac1n^2(fracepsilon^frac1n2)^n|<cepsilon<delta:::delta>0$ Therefore we can find a subsequence that converges proving that $A$ is compact.
Question:
Is my proof right? If not. How should I prove compactness?
Thanks in advance!
functional-analysis
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
Problem: Show that the operator $A:l_2to l_2::Ax=(0,x_1,frac12x_2,...,frac1nx_n,...)$ is compact but has no pontual spectrum.
My solution: $(0,x_1,frac12x_2,...,frac1nx_n,...)=(lambda x_1,lambda x_2,...,lambda x_n,...)$
Since $lambda x_1=0implies lambda=0$, then the $ker(A-lambda)=0$.
My problem lies with compactness.
If we consider a sequence in $l_2$ like $x_n$,that converges to some $x$. Consider $epsilon>0$ so that $||x_m-x||<epsilon$ for $m>NinmathbbN$. Then $||A(x_m)-A(x)||^2=||A(x_m-x)||^2=sum_limitsn=1^infty |frac1n(x_m_i-x_i)|^2<sum_limitsn=1^infty |frac1n^2(fracepsilon^frac1n2)^n|<cepsilon<delta:::delta>0$ Therefore we can find a subsequence that converges proving that $A$ is compact.
Question:
Is my proof right? If not. How should I prove compactness?
Thanks in advance!
functional-analysis
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
edited Jul 28 at 20:25
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
asked Jul 28 at 18:49
Pedro Gomes
1,3052618
1,3052618
2
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07
add a comment |Â
2
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07
2
2
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating).
answered Jul 29 at 1:47
Martin Argerami
115k1071164
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
add a comment |Â
up vote
1
down vote
To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,frac12x_2,dots,frac1nx_n, 0, dots)$$ is bounded.
First, we show bounded.
$$||A_n x||^2=sum_j=1^n |frac1jx_j|^2 leq frac11 sum_j=1^n |x_j|^2 leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A.
$$||(A-A_n) x||^2=sum_j=1^infty |frac1jx_j|^2 - sum_j=1^n|frac1jx_j|^2$$
$$=sum_j=n+1^infty|frac1jx_j|^2 leq frac1(n+1)^2sum_j=n+1^infty|x_j|^2 = frac1(n+1)^2 ||x||$$
So that as we take $nrightarrow infty$, $||(A-A_n) x||^2 rightarrow 0$
Finally to show that the operator has no point spectrum.
To be in the point spectrum, we must have that $$R_lambda = (A-lambda I)^-1$$ does not exist.
First take $lambda neq 0$
We have that $$T_lambda x = 0 rightarrow (A-lambda I)x= Ax-lambda x =0$$
$$rightarrow (0,frac12x_1,...,frac1n,...)-lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of
$$x_1=0$$ and $$x_j=frac1j-1 frac1lambdax_n-1$$ for $jgeq 2$.
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_lambda =0$ and so $T_lambda^-1=R_lambda$ exists and $lambda$ is not in the point spectrum.
Now take, $lambda =0$. This means that $T_lambda = Ax = 0$ implies that $x=0$ again so that $lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
answered Jul 29 at 7:16
MathIsHard
1,122415
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating).
answered Jul 29 at 1:47
Martin Argerami
115k1071164
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
add a comment |Â
up vote
1
down vote
Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating).
answered Jul 29 at 1:47
Martin Argerami
115k1071164
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating).
answered Jul 29 at 1:47
Martin Argerami
115k1071164
Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating).
answered Jul 29 at 1:47
Martin Argerami
115k1071164
answered Jul 29 at 1:47
Martin Argerami
115k1071164
answered Jul 29 at 1:47
Martin Argerami
115k1071164
answered Jul 29 at 1:47
Martin Argerami
115k1071164
115k1071164
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
add a comment |Â
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
Why? Could you please explain me why my proof is wrong in more detail? Thanks in advance!
– Pedro Gomes
Jul 29 at 17:24
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
You prove that if $x_nto x$, then $Ax_nto Ax$. That's continuity. It has nothing to do with compactness.
– Martin Argerami
Jul 30 at 2:17
add a comment |Â
up vote
1
down vote
To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,frac12x_2,dots,frac1nx_n, 0, dots)$$ is bounded.
First, we show bounded.
$$||A_n x||^2=sum_j=1^n |frac1jx_j|^2 leq frac11 sum_j=1^n |x_j|^2 leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A.
$$||(A-A_n) x||^2=sum_j=1^infty |frac1jx_j|^2 - sum_j=1^n|frac1jx_j|^2$$
$$=sum_j=n+1^infty|frac1jx_j|^2 leq frac1(n+1)^2sum_j=n+1^infty|x_j|^2 = frac1(n+1)^2 ||x||$$
So that as we take $nrightarrow infty$, $||(A-A_n) x||^2 rightarrow 0$
Finally to show that the operator has no point spectrum.
To be in the point spectrum, we must have that $$R_lambda = (A-lambda I)^-1$$ does not exist.
First take $lambda neq 0$
We have that $$T_lambda x = 0 rightarrow (A-lambda I)x= Ax-lambda x =0$$
$$rightarrow (0,frac12x_1,...,frac1n,...)-lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of
$$x_1=0$$ and $$x_j=frac1j-1 frac1lambdax_n-1$$ for $jgeq 2$.
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_lambda =0$ and so $T_lambda^-1=R_lambda$ exists and $lambda$ is not in the point spectrum.
Now take, $lambda =0$. This means that $T_lambda = Ax = 0$ implies that $x=0$ again so that $lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
answered Jul 29 at 7:16
MathIsHard
1,122415
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
 |Â
show 1 more comment
up vote
1
down vote
To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,frac12x_2,dots,frac1nx_n, 0, dots)$$ is bounded.
First, we show bounded.
$$||A_n x||^2=sum_j=1^n |frac1jx_j|^2 leq frac11 sum_j=1^n |x_j|^2 leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A.
$$||(A-A_n) x||^2=sum_j=1^infty |frac1jx_j|^2 - sum_j=1^n|frac1jx_j|^2$$
$$=sum_j=n+1^infty|frac1jx_j|^2 leq frac1(n+1)^2sum_j=n+1^infty|x_j|^2 = frac1(n+1)^2 ||x||$$
So that as we take $nrightarrow infty$, $||(A-A_n) x||^2 rightarrow 0$
Finally to show that the operator has no point spectrum.
To be in the point spectrum, we must have that $$R_lambda = (A-lambda I)^-1$$ does not exist.
First take $lambda neq 0$
We have that $$T_lambda x = 0 rightarrow (A-lambda I)x= Ax-lambda x =0$$
$$rightarrow (0,frac12x_1,...,frac1n,...)-lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of
$$x_1=0$$ and $$x_j=frac1j-1 frac1lambdax_n-1$$ for $jgeq 2$.
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_lambda =0$ and so $T_lambda^-1=R_lambda$ exists and $lambda$ is not in the point spectrum.
Now take, $lambda =0$. This means that $T_lambda = Ax = 0$ implies that $x=0$ again so that $lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
answered Jul 29 at 7:16
MathIsHard
1,122415
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,frac12x_2,dots,frac1nx_n, 0, dots)$$ is bounded.
First, we show bounded.
$$||A_n x||^2=sum_j=1^n |frac1jx_j|^2 leq frac11 sum_j=1^n |x_j|^2 leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A.
$$||(A-A_n) x||^2=sum_j=1^infty |frac1jx_j|^2 - sum_j=1^n|frac1jx_j|^2$$
$$=sum_j=n+1^infty|frac1jx_j|^2 leq frac1(n+1)^2sum_j=n+1^infty|x_j|^2 = frac1(n+1)^2 ||x||$$
So that as we take $nrightarrow infty$, $||(A-A_n) x||^2 rightarrow 0$
Finally to show that the operator has no point spectrum.
To be in the point spectrum, we must have that $$R_lambda = (A-lambda I)^-1$$ does not exist.
First take $lambda neq 0$
We have that $$T_lambda x = 0 rightarrow (A-lambda I)x= Ax-lambda x =0$$
$$rightarrow (0,frac12x_1,...,frac1n,...)-lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of
$$x_1=0$$ and $$x_j=frac1j-1 frac1lambdax_n-1$$ for $jgeq 2$.
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_lambda =0$ and so $T_lambda^-1=R_lambda$ exists and $lambda$ is not in the point spectrum.
Now take, $lambda =0$. This means that $T_lambda = Ax = 0$ implies that $x=0$ again so that $lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
answered Jul 29 at 7:16
MathIsHard
1,122415
To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,frac12x_2,dots,frac1nx_n, 0, dots)$$ is bounded.
First, we show bounded.
$$||A_n x||^2=sum_j=1^n |frac1jx_j|^2 leq frac11 sum_j=1^n |x_j|^2 leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A.
$$||(A-A_n) x||^2=sum_j=1^infty |frac1jx_j|^2 - sum_j=1^n|frac1jx_j|^2$$
$$=sum_j=n+1^infty|frac1jx_j|^2 leq frac1(n+1)^2sum_j=n+1^infty|x_j|^2 = frac1(n+1)^2 ||x||$$
So that as we take $nrightarrow infty$, $||(A-A_n) x||^2 rightarrow 0$
Finally to show that the operator has no point spectrum.
To be in the point spectrum, we must have that $$R_lambda = (A-lambda I)^-1$$ does not exist.
First take $lambda neq 0$
We have that $$T_lambda x = 0 rightarrow (A-lambda I)x= Ax-lambda x =0$$
$$rightarrow (0,frac12x_1,...,frac1n,...)-lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of
$$x_1=0$$ and $$x_j=frac1j-1 frac1lambdax_n-1$$ for $jgeq 2$.
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_lambda =0$ and so $T_lambda^-1=R_lambda$ exists and $lambda$ is not in the point spectrum.
Now take, $lambda =0$. This means that $T_lambda = Ax = 0$ implies that $x=0$ again so that $lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
answered Jul 29 at 7:16
MathIsHard
1,122415
answered Jul 29 at 7:16
MathIsHard
1,122415
answered Jul 29 at 7:16
MathIsHard
1,122415
answered Jul 29 at 7:16
MathIsHard
1,122415
1,122415
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
 |Â
show 1 more comment
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
1
1
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
Thanks for your answer! That is precisely where my doubt lies. You used the following theorem if I am not mistaken:"Theorem (Sequence of compact linear operators). Let ($T_n$) be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If ($T_n$) is uniformly operator convergent, say, $||Tn - T||to 0 $(cf. Sec. 4.9), then the limit operator $T$ is compact."
– Pedro Gomes
Jul 29 at 13:01
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
However I tried to prove it using the definition of compactness where a "sequence must have convergent subsequence". Where does my argument fails? Thanks in advance!
– Pedro Gomes
Jul 29 at 13:04
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Hello. Your first comment is indeed what I did. I agree with what uniquesolution commented above. We can only assume that the sequence is bounded. Not that it is convergent.
– MathIsHard
Jul 29 at 16:46
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
Furthermore, the theorem you are trying to use I believe is the compactness criterion that states that if T is a linear operator, then T is compact iff it maps every bounded sequence $(x_n)$ into a sequence $(Tx_n)$ which has a convergent subsequence. You may be able to argue that the $frac1n$ coefficients multiplied by bounded $(x_n)$ will converge to 0. However, I am not certain that this will work as I have not seen a proof done in this way. I always see people use the finite rank operators.
– MathIsHard
Jul 29 at 16:57
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
That is exactly my problem. I was told to prove that when an linear operator is bounded it is compact. And my Professor told me to use the sequence definition and use continuity of the afore-mentioned operator(since it is linear and bounded hence continuous). Why does everyone seem to talk of a bounded sequence? Thanks!
– Pedro Gomes
Jul 29 at 17:16
 |Â
show 1 more comment
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865496%2fproving-compactness-of-al-2-to-l-2-ax-0-x-1-frac12x-2-frac1n%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
In order to prove that $A$ is compact, you must show that for every bounded sequence $x_n$ in $l_2$, the sequence $Ax_n$ has a norm-convergent subsequence. What you did is to assume that $x_n$ is convergent. You can only assume it is bounded.
– uniquesolution
Jul 28 at 21:07