Spectrum of $l^p$ multiplication operator (Brezis 6.17)
$begingroup$
I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.
Take $l^p(mathbb{R})$, for $1leq p leq infty$. For a fixed, bounded, real sequence $lambda_n$ define the multiplication operator $M: l^p to l^p$ by:
$$
M(x_1,x_2,x_3, cdots ) = (lambda_1 x_1 , lambda_2x_2, lambda_3x_3 ,cdots)
$$
i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $sigma(M)$.
First, if we have an eigenvalue $mu$, we have:
$$
mu x_1 = lambda_1 x_1
\
mu x_2 = lambda_2 x_2
\
mu x_3 = lambda_3x_3
\
vdots
$$
But $lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that:
$$
sigma(M) = {mu in mathbb{R} : (M - mu I) text{ is not invertible}}
$$
In general, if $M - mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by:
$$
left(frac{1}{lambda_1 - mu}y_1, frac{1}{lambda_2- mu}y_2,frac{1}{lambda_3 - mu}y_3 , cdots right)
$$
Thus, the operator is not invertible when $mu = lambda_k$ for some $lambda_k$. Thus, the spectrum is given by ${lambda_n : n in mathbb{N}}$.
functional-analysis operator-theory spectral-theory
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
$endgroup$
add a comment |
$begingroup$
I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.
Take $l^p(mathbb{R})$, for $1leq p leq infty$. For a fixed, bounded, real sequence $lambda_n$ define the multiplication operator $M: l^p to l^p$ by:
$$
M(x_1,x_2,x_3, cdots ) = (lambda_1 x_1 , lambda_2x_2, lambda_3x_3 ,cdots)
$$
i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $sigma(M)$.
First, if we have an eigenvalue $mu$, we have:
$$
mu x_1 = lambda_1 x_1
\
mu x_2 = lambda_2 x_2
\
mu x_3 = lambda_3x_3
\
vdots
$$
But $lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that:
$$
sigma(M) = {mu in mathbb{R} : (M - mu I) text{ is not invertible}}
$$
In general, if $M - mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by:
$$
left(frac{1}{lambda_1 - mu}y_1, frac{1}{lambda_2- mu}y_2,frac{1}{lambda_3 - mu}y_3 , cdots right)
$$
Thus, the operator is not invertible when $mu = lambda_k$ for some $lambda_k$. Thus, the spectrum is given by ${lambda_n : n in mathbb{N}}$.
functional-analysis operator-theory spectral-theory
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
$endgroup$
1
$begingroup$
How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 17:34
$begingroup$
Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
$endgroup$
– rubikscube09
Dec 8 '18 at 17:42
$begingroup$
Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:01
$begingroup$
I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
$endgroup$
– rubikscube09
Dec 8 '18 at 18:43
$begingroup$
Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:43
add a comment |
$begingroup$
I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.
Take $l^p(mathbb{R})$, for $1leq p leq infty$. For a fixed, bounded, real sequence $lambda_n$ define the multiplication operator $M: l^p to l^p$ by:
$$
M(x_1,x_2,x_3, cdots ) = (lambda_1 x_1 , lambda_2x_2, lambda_3x_3 ,cdots)
$$
i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $sigma(M)$.
First, if we have an eigenvalue $mu$, we have:
$$
mu x_1 = lambda_1 x_1
\
mu x_2 = lambda_2 x_2
\
mu x_3 = lambda_3x_3
\
vdots
$$
But $lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that:
$$
sigma(M) = {mu in mathbb{R} : (M - mu I) text{ is not invertible}}
$$
In general, if $M - mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by:
$$
left(frac{1}{lambda_1 - mu}y_1, frac{1}{lambda_2- mu}y_2,frac{1}{lambda_3 - mu}y_3 , cdots right)
$$
Thus, the operator is not invertible when $mu = lambda_k$ for some $lambda_k$. Thus, the spectrum is given by ${lambda_n : n in mathbb{N}}$.
functional-analysis operator-theory spectral-theory
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
$endgroup$
I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.
Take $l^p(mathbb{R})$, for $1leq p leq infty$. For a fixed, bounded, real sequence $lambda_n$ define the multiplication operator $M: l^p to l^p$ by:
$$
M(x_1,x_2,x_3, cdots ) = (lambda_1 x_1 , lambda_2x_2, lambda_3x_3 ,cdots)
$$
i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $sigma(M)$.
First, if we have an eigenvalue $mu$, we have:
$$
mu x_1 = lambda_1 x_1
\
mu x_2 = lambda_2 x_2
\
mu x_3 = lambda_3x_3
\
vdots
$$
But $lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that:
$$
sigma(M) = {mu in mathbb{R} : (M - mu I) text{ is not invertible}}
$$
In general, if $M - mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by:
$$
left(frac{1}{lambda_1 - mu}y_1, frac{1}{lambda_2- mu}y_2,frac{1}{lambda_3 - mu}y_3 , cdots right)
$$
Thus, the operator is not invertible when $mu = lambda_k$ for some $lambda_k$. Thus, the spectrum is given by ${lambda_n : n in mathbb{N}}$.
functional-analysis operator-theory spectral-theory
functional-analysis operator-theory spectral-theory
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
asked Dec 8 '18 at 17:33
rubikscube09rubikscube09
1,265719
1,265719
1
$begingroup$
How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 17:34
$begingroup$
Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
$endgroup$
– rubikscube09
Dec 8 '18 at 17:42
$begingroup$
Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:01
$begingroup$
I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
$endgroup$
– rubikscube09
Dec 8 '18 at 18:43
$begingroup$
Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:43
add a comment |
1
$begingroup$
How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 17:34
$begingroup$
Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
$endgroup$
– rubikscube09
Dec 8 '18 at 17:42
$begingroup$
Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:01
$begingroup$
I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
$endgroup$
– rubikscube09
Dec 8 '18 at 18:43
$begingroup$
Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:43
1
1
$begingroup$
How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 17:34
$begingroup$
How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 17:34
$begingroup$
Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
$endgroup$
– rubikscube09
Dec 8 '18 at 17:42
$begingroup$
Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
$endgroup$
– rubikscube09
Dec 8 '18 at 17:42
$begingroup$
Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:01
$begingroup$
Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:01
$begingroup$
I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
$endgroup$
– rubikscube09
Dec 8 '18 at 18:43
$begingroup$
I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
$endgroup$
– rubikscube09
Dec 8 '18 at 18:43
$begingroup$
Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:43
$begingroup$
Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
$endgroup$
– MisterRiemann
Dec 8 '18 at 18:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You will have eigenvectors wherever the sequence $lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m={n: lambda_n=lambda_m}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $nin E_m$ and zero otherwise, satisfies $Mx=lambda_m x$.
Thus ${lambda_n: ninmathbb N}subsetsigma(M)$. As the spectrum is always closed, $overline{{lambda_n: ninmathbb N}}subsetsigma(M)$.
Conversely, if $munotinoverline{{lambda_n: ninmathbb N}}$, there exists $delta>0$ such that $|mu-lambda_n|>delta>0$ for all $n$. Then $$left|frac1{mu-lambda_n}right|<frac1delta$$ for all $n$, which gives that the sequence ${1/(mu-lambda_n)}_n$ is bounded, and so $mu I-M$ is invertible. This shows that $overline{{lambda_n: ninmathbb N}}supsetsigma(M)$ and so
$$
sigma(M)=overline{{lambda_n: ninmathbb N}}
$$
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
$begingroup$
That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
add a comment |
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1 Answer
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$begingroup$
You will have eigenvectors wherever the sequence $lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m={n: lambda_n=lambda_m}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $nin E_m$ and zero otherwise, satisfies $Mx=lambda_m x$.
Thus ${lambda_n: ninmathbb N}subsetsigma(M)$. As the spectrum is always closed, $overline{{lambda_n: ninmathbb N}}subsetsigma(M)$.
Conversely, if $munotinoverline{{lambda_n: ninmathbb N}}$, there exists $delta>0$ such that $|mu-lambda_n|>delta>0$ for all $n$. Then $$left|frac1{mu-lambda_n}right|<frac1delta$$ for all $n$, which gives that the sequence ${1/(mu-lambda_n)}_n$ is bounded, and so $mu I-M$ is invertible. This shows that $overline{{lambda_n: ninmathbb N}}supsetsigma(M)$ and so
$$
sigma(M)=overline{{lambda_n: ninmathbb N}}
$$
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
$begingroup$
That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
add a comment |
$begingroup$
You will have eigenvectors wherever the sequence $lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m={n: lambda_n=lambda_m}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $nin E_m$ and zero otherwise, satisfies $Mx=lambda_m x$.
Thus ${lambda_n: ninmathbb N}subsetsigma(M)$. As the spectrum is always closed, $overline{{lambda_n: ninmathbb N}}subsetsigma(M)$.
Conversely, if $munotinoverline{{lambda_n: ninmathbb N}}$, there exists $delta>0$ such that $|mu-lambda_n|>delta>0$ for all $n$. Then $$left|frac1{mu-lambda_n}right|<frac1delta$$ for all $n$, which gives that the sequence ${1/(mu-lambda_n)}_n$ is bounded, and so $mu I-M$ is invertible. This shows that $overline{{lambda_n: ninmathbb N}}supsetsigma(M)$ and so
$$
sigma(M)=overline{{lambda_n: ninmathbb N}}
$$
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
$begingroup$
That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
add a comment |
$begingroup$
You will have eigenvectors wherever the sequence $lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m={n: lambda_n=lambda_m}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $nin E_m$ and zero otherwise, satisfies $Mx=lambda_m x$.
Thus ${lambda_n: ninmathbb N}subsetsigma(M)$. As the spectrum is always closed, $overline{{lambda_n: ninmathbb N}}subsetsigma(M)$.
Conversely, if $munotinoverline{{lambda_n: ninmathbb N}}$, there exists $delta>0$ such that $|mu-lambda_n|>delta>0$ for all $n$. Then $$left|frac1{mu-lambda_n}right|<frac1delta$$ for all $n$, which gives that the sequence ${1/(mu-lambda_n)}_n$ is bounded, and so $mu I-M$ is invertible. This shows that $overline{{lambda_n: ninmathbb N}}supsetsigma(M)$ and so
$$
sigma(M)=overline{{lambda_n: ninmathbb N}}
$$
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
You will have eigenvectors wherever the sequence $lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m={n: lambda_n=lambda_m}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $nin E_m$ and zero otherwise, satisfies $Mx=lambda_m x$.
Thus ${lambda_n: ninmathbb N}subsetsigma(M)$. As the spectrum is always closed, $overline{{lambda_n: ninmathbb N}}subsetsigma(M)$.
Conversely, if $munotinoverline{{lambda_n: ninmathbb N}}$, there exists $delta>0$ such that $|mu-lambda_n|>delta>0$ for all $n$. Then $$left|frac1{mu-lambda_n}right|<frac1delta$$ for all $n$, which gives that the sequence ${1/(mu-lambda_n)}_n$ is bounded, and so $mu I-M$ is invertible. This shows that $overline{{lambda_n: ninmathbb N}}supsetsigma(M)$ and so
$$
sigma(M)=overline{{lambda_n: ninmathbb N}}
$$
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 18:37
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
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Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
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– rubikscube09
Dec 8 '18 at 19:06
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If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
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– Martin Argerami
Dec 8 '18 at 19:14
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That makes sense. Thank you.
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– rubikscube09
Dec 8 '18 at 19:16
add a comment |
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Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
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That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
$begingroup$
Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $mu$ is not in the closure, there is no sequence of $lambda_n$ s converging to $mu$, hence there exist infinitely many $m$ such that $frac{1}{lambda_m - mu} < frac{1}{delta}$. However, how can we choose $N$ such that the above is true for all $n geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:06
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
$begingroup$
If $K$ is closed and $munotin K$, this means that $operatorname{dist}(mu,K)>delta$ for some $delta$. That is, $|mu-lambda|>delta$ for all $lambdain K$.
$endgroup$
– Martin Argerami
Dec 8 '18 at 19:14
$begingroup$
That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
$begingroup$
That makes sense. Thank you.
$endgroup$
– rubikscube09
Dec 8 '18 at 19:16
add a comment |
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How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $lambda_n$? Regarding the spectrum, what if $mu-lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-mu I$ then?
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– MisterRiemann
Dec 8 '18 at 17:34
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Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $mu - lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $lambda_n$ here) the quotients $(mu - lambda_n)^{-1}$ would still exist and hence we would still have an inverse.
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– rubikscube09
Dec 8 '18 at 17:42
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Assuming that it was invertible, it would give a bijection between $ell^p$ and $ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $epsilon, delta$ such that $epsilon < |lambda_n-mu| < delta$. Do you see what the spectrum must be then?
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– MisterRiemann
Dec 8 '18 at 18:01
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I believe I understand your point now. If $lambda_n to mu$ then we may have the issue that $frac{1}{lambda_n - mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $lambda_n$ converge to $mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized.
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– rubikscube09
Dec 8 '18 at 18:43
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Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion.
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– MisterRiemann
Dec 8 '18 at 18:43