Mixing
Equivalence of definitions for the approximate point spectrum
$begingroup$
Let $T: X rightarrow X$ be a continuous, linear operator on some Banach space $X$.
We defined the approximate point spectrum $APsigma(T)$ as the set
$$
{ lambda in mathbb{C} : lambda - T ;text{is not injective or}; text{Im}(lambda - T) ;text{is not closed in X} }.
$$
I want to show equivalence with the definition
$$
lambda in APsigma(T) :Leftrightarrow exists (x_n) subset X, Vert x_n Vert =1 ;text{with}; Vert lambda x_n - Tx_nVert rightarrow 0.
$$
What I have done: "$Leftarrow$".
What I need: Show that if $lambda -T$ is injective and $text{Im}(lambda - T)$ is not closed in $X$, then there is a sequence such as above.
functional-analysis spectral-theory
asked Jan 15 at 18:55
fpmoofpmoo
382113
$endgroup$
add a comment |
$begingroup$
Let $T: X rightarrow X$ be a continuous, linear operator on some Banach space $X$.
We defined the approximate point spectrum $APsigma(T)$ as the set
$$
{ lambda in mathbb{C} : lambda - T ;text{is not injective or}; text{Im}(lambda - T) ;text{is not closed in X} }.
$$
I want to show equivalence with the definition
$$
lambda in APsigma(T) :Leftrightarrow exists (x_n) subset X, Vert x_n Vert =1 ;text{with}; Vert lambda x_n - Tx_nVert rightarrow 0.
$$
What I have done: "$Leftarrow$".
What I need: Show that if $lambda -T$ is injective and $text{Im}(lambda - T)$ is not closed in $X$, then there is a sequence such as above.
functional-analysis spectral-theory
asked Jan 15 at 18:55
fpmoofpmoo
382113
$endgroup$
$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25
add a comment |
$begingroup$
Let $T: X rightarrow X$ be a continuous, linear operator on some Banach space $X$.
We defined the approximate point spectrum $APsigma(T)$ as the set
$$
{ lambda in mathbb{C} : lambda - T ;text{is not injective or}; text{Im}(lambda - T) ;text{is not closed in X} }.
$$
I want to show equivalence with the definition
$$
lambda in APsigma(T) :Leftrightarrow exists (x_n) subset X, Vert x_n Vert =1 ;text{with}; Vert lambda x_n - Tx_nVert rightarrow 0.
$$
What I have done: "$Leftarrow$".
What I need: Show that if $lambda -T$ is injective and $text{Im}(lambda - T)$ is not closed in $X$, then there is a sequence such as above.
functional-analysis spectral-theory
asked Jan 15 at 18:55
fpmoofpmoo
382113
$endgroup$
Let $T: X rightarrow X$ be a continuous, linear operator on some Banach space $X$.
We defined the approximate point spectrum $APsigma(T)$ as the set
$$
{ lambda in mathbb{C} : lambda - T ;text{is not injective or}; text{Im}(lambda - T) ;text{is not closed in X} }.
$$
I want to show equivalence with the definition
$$
lambda in APsigma(T) :Leftrightarrow exists (x_n) subset X, Vert x_n Vert =1 ;text{with}; Vert lambda x_n - Tx_nVert rightarrow 0.
$$
What I have done: "$Leftarrow$".
What I need: Show that if $lambda -T$ is injective and $text{Im}(lambda - T)$ is not closed in $X$, then there is a sequence such as above.
functional-analysis spectral-theory
functional-analysis spectral-theory
asked Jan 15 at 18:55
fpmoofpmoo
382113
asked Jan 15 at 18:55
fpmoofpmoo
382113
asked Jan 15 at 18:55
fpmoofpmoo
382113
asked Jan 15 at 18:55
fpmoofpmoo
382113
asked Jan 15 at 18:55
fpmoofpmoo
382113
382113
$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25
add a comment |
$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25
$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $lambda I-T$ is injective and that the range $mathcal{R}(lambda I -T)$ is not closed. Then $(lambda I-T)^{-1}$ cannot be bounded; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{lambda}$ on the closure $mathcal{R}(lambda I-T)^c$ of the range, and
$$
(lambda I-T)(lambda I-T)^{-1}=I
$$
would automatically extend (by continuity) to
$$
(lambda I-T)R_{lambda}x=x,;;; xinmathcal{R}(lambda I-T)^c.
$$
But that would contradict the fact that the range of $lambda I-T$ is not closed. So $(lambda I-T)^{-1}$ cannot be bounded, which implies the existence of a sequence of unit vectors ${ e_n }subset mathcal{R}(lambda I-T)$ such that $|(lambda I-T)^{-1}e_n|rightarrowinfty$. Then
$$
f_n= frac{1}{|(lambda I-T)^{-1}e_n|}(lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
(lambda I-T)f_n =frac{1}{|(lambda I-T)^{-1}e_n|}e_n rightarrow 0.
$$
Hence $lambda$ is in the approximate point spectrum of $T$.
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose that $lambda I-T$ is injective and that the range $mathcal{R}(lambda I -T)$ is not closed. Then $(lambda I-T)^{-1}$ cannot be bounded; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{lambda}$ on the closure $mathcal{R}(lambda I-T)^c$ of the range, and
$$
(lambda I-T)(lambda I-T)^{-1}=I
$$
would automatically extend (by continuity) to
$$
(lambda I-T)R_{lambda}x=x,;;; xinmathcal{R}(lambda I-T)^c.
$$
But that would contradict the fact that the range of $lambda I-T$ is not closed. So $(lambda I-T)^{-1}$ cannot be bounded, which implies the existence of a sequence of unit vectors ${ e_n }subset mathcal{R}(lambda I-T)$ such that $|(lambda I-T)^{-1}e_n|rightarrowinfty$. Then
$$
f_n= frac{1}{|(lambda I-T)^{-1}e_n|}(lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
(lambda I-T)f_n =frac{1}{|(lambda I-T)^{-1}e_n|}e_n rightarrow 0.
$$
Hence $lambda$ is in the approximate point spectrum of $T$.
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
Suppose that $lambda I-T$ is injective and that the range $mathcal{R}(lambda I -T)$ is not closed. Then $(lambda I-T)^{-1}$ cannot be bounded; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{lambda}$ on the closure $mathcal{R}(lambda I-T)^c$ of the range, and
$$
(lambda I-T)(lambda I-T)^{-1}=I
$$
would automatically extend (by continuity) to
$$
(lambda I-T)R_{lambda}x=x,;;; xinmathcal{R}(lambda I-T)^c.
$$
But that would contradict the fact that the range of $lambda I-T$ is not closed. So $(lambda I-T)^{-1}$ cannot be bounded, which implies the existence of a sequence of unit vectors ${ e_n }subset mathcal{R}(lambda I-T)$ such that $|(lambda I-T)^{-1}e_n|rightarrowinfty$. Then
$$
f_n= frac{1}{|(lambda I-T)^{-1}e_n|}(lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
(lambda I-T)f_n =frac{1}{|(lambda I-T)^{-1}e_n|}e_n rightarrow 0.
$$
Hence $lambda$ is in the approximate point spectrum of $T$.
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
Suppose that $lambda I-T$ is injective and that the range $mathcal{R}(lambda I -T)$ is not closed. Then $(lambda I-T)^{-1}$ cannot be bounded; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{lambda}$ on the closure $mathcal{R}(lambda I-T)^c$ of the range, and
$$
(lambda I-T)(lambda I-T)^{-1}=I
$$
would automatically extend (by continuity) to
$$
(lambda I-T)R_{lambda}x=x,;;; xinmathcal{R}(lambda I-T)^c.
$$
But that would contradict the fact that the range of $lambda I-T$ is not closed. So $(lambda I-T)^{-1}$ cannot be bounded, which implies the existence of a sequence of unit vectors ${ e_n }subset mathcal{R}(lambda I-T)$ such that $|(lambda I-T)^{-1}e_n|rightarrowinfty$. Then
$$
f_n= frac{1}{|(lambda I-T)^{-1}e_n|}(lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
(lambda I-T)f_n =frac{1}{|(lambda I-T)^{-1}e_n|}e_n rightarrow 0.
$$
Hence $lambda$ is in the approximate point spectrum of $T$.
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
Suppose that $lambda I-T$ is injective and that the range $mathcal{R}(lambda I -T)$ is not closed. Then $(lambda I-T)^{-1}$ cannot be bounded; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{lambda}$ on the closure $mathcal{R}(lambda I-T)^c$ of the range, and
$$
(lambda I-T)(lambda I-T)^{-1}=I
$$
would automatically extend (by continuity) to
$$
(lambda I-T)R_{lambda}x=x,;;; xinmathcal{R}(lambda I-T)^c.
$$
But that would contradict the fact that the range of $lambda I-T$ is not closed. So $(lambda I-T)^{-1}$ cannot be bounded, which implies the existence of a sequence of unit vectors ${ e_n }subset mathcal{R}(lambda I-T)$ such that $|(lambda I-T)^{-1}e_n|rightarrowinfty$. Then
$$
f_n= frac{1}{|(lambda I-T)^{-1}e_n|}(lambda I-T)^{-1}e_n
$$
is a sequence of unit vectors such that
$$
(lambda I-T)f_n =frac{1}{|(lambda I-T)^{-1}e_n|}e_n rightarrow 0.
$$
Hence $lambda$ is in the approximate point spectrum of $T$.
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
edited Jan 16 at 6:03
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 16 at 5:56
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
add a comment |
add a comment |
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$begingroup$
Did you mean to include the point spectrum in the approximate point spectrum?
$endgroup$
– DisintegratingByParts
Jan 16 at 4:24
$begingroup$
The point spectrum is included as ${ lambda : lambda - T ;text{not injective} }$ is a subset of $APsigma(T)$.
$endgroup$
– fpmoo
Jan 16 at 10:25