Closed operator, closed graph
$begingroup$
From a course based on Kreyszig's Introduction to Functional analysis.
Let $X neq {0}$ denote a complex normed vector space, and
assume that the operator $T : D(T) ⊂ X → X$ is closed. Let $λ ∈ C$. Show
that the operator $T − λI$ is closed. I have tried applying theorem $4.13-3$, but am unsure of how to choose a sequence and showing convergence.
real-analysis functional-analysis operator-theory closed-graph
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
$endgroup$
add a comment |
$begingroup$
From a course based on Kreyszig's Introduction to Functional analysis.
Let $X neq {0}$ denote a complex normed vector space, and
assume that the operator $T : D(T) ⊂ X → X$ is closed. Let $λ ∈ C$. Show
that the operator $T − λI$ is closed. I have tried applying theorem $4.13-3$, but am unsure of how to choose a sequence and showing convergence.
real-analysis functional-analysis operator-theory closed-graph
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
$endgroup$
add a comment |
$begingroup$
From a course based on Kreyszig's Introduction to Functional analysis.
Let $X neq {0}$ denote a complex normed vector space, and
assume that the operator $T : D(T) ⊂ X → X$ is closed. Let $λ ∈ C$. Show
that the operator $T − λI$ is closed. I have tried applying theorem $4.13-3$, but am unsure of how to choose a sequence and showing convergence.
real-analysis functional-analysis operator-theory closed-graph
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
$endgroup$
From a course based on Kreyszig's Introduction to Functional analysis.
Let $X neq {0}$ denote a complex normed vector space, and
assume that the operator $T : D(T) ⊂ X → X$ is closed. Let $λ ∈ C$. Show
that the operator $T − λI$ is closed. I have tried applying theorem $4.13-3$, but am unsure of how to choose a sequence and showing convergence.
real-analysis functional-analysis operator-theory closed-graph
real-analysis functional-analysis operator-theory closed-graph
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
edited Dec 4 '18 at 11:35
thaumoctopus
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
asked Dec 4 '18 at 10:57
thaumoctopusthaumoctopus
9519
9519
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that for any $lambdainmathbb{C}$, $D(T)=D(T-lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)to X$ and $F:D(F)to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)cap D(F)to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)in overline{G_{F+T}}$ then $exists$ a sequence ${(x_n,y_n)}_{ninmathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),forall ninmathbb{N}implies y_n=T(x_n)+F(x_n),forall ninmathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)to T(x)$, $F(x_n)to F(x)$ as $x_nto x$ as $nto infty$. Hence, $y_n=T(x_n)+F(x_n)to T(x)+F(x)=(T+F)(x)$ as $nto infty$, also $y_nto y$ is clear by definition so $y=(T+F)x$ gives $overline{G_{T+F}}subset G_{T+F}$. Hence our claim is proved.
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
$endgroup$
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
add a comment |
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$begingroup$
Note that for any $lambdainmathbb{C}$, $D(T)=D(T-lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)to X$ and $F:D(F)to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)cap D(F)to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)in overline{G_{F+T}}$ then $exists$ a sequence ${(x_n,y_n)}_{ninmathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),forall ninmathbb{N}implies y_n=T(x_n)+F(x_n),forall ninmathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)to T(x)$, $F(x_n)to F(x)$ as $x_nto x$ as $nto infty$. Hence, $y_n=T(x_n)+F(x_n)to T(x)+F(x)=(T+F)(x)$ as $nto infty$, also $y_nto y$ is clear by definition so $y=(T+F)x$ gives $overline{G_{T+F}}subset G_{T+F}$. Hence our claim is proved.
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
$endgroup$
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
add a comment |
$begingroup$
Note that for any $lambdainmathbb{C}$, $D(T)=D(T-lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)to X$ and $F:D(F)to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)cap D(F)to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)in overline{G_{F+T}}$ then $exists$ a sequence ${(x_n,y_n)}_{ninmathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),forall ninmathbb{N}implies y_n=T(x_n)+F(x_n),forall ninmathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)to T(x)$, $F(x_n)to F(x)$ as $x_nto x$ as $nto infty$. Hence, $y_n=T(x_n)+F(x_n)to T(x)+F(x)=(T+F)(x)$ as $nto infty$, also $y_nto y$ is clear by definition so $y=(T+F)x$ gives $overline{G_{T+F}}subset G_{T+F}$. Hence our claim is proved.
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
$endgroup$
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
add a comment |
$begingroup$
Note that for any $lambdainmathbb{C}$, $D(T)=D(T-lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)to X$ and $F:D(F)to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)cap D(F)to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)in overline{G_{F+T}}$ then $exists$ a sequence ${(x_n,y_n)}_{ninmathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),forall ninmathbb{N}implies y_n=T(x_n)+F(x_n),forall ninmathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)to T(x)$, $F(x_n)to F(x)$ as $x_nto x$ as $nto infty$. Hence, $y_n=T(x_n)+F(x_n)to T(x)+F(x)=(T+F)(x)$ as $nto infty$, also $y_nto y$ is clear by definition so $y=(T+F)x$ gives $overline{G_{T+F}}subset G_{T+F}$. Hence our claim is proved.
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
$endgroup$
Note that for any $lambdainmathbb{C}$, $D(T)=D(T-lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)to X$ and $F:D(F)to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)cap D(F)to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)in overline{G_{F+T}}$ then $exists$ a sequence ${(x_n,y_n)}_{ninmathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),forall ninmathbb{N}implies y_n=T(x_n)+F(x_n),forall ninmathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)to T(x)$, $F(x_n)to F(x)$ as $x_nto x$ as $nto infty$. Hence, $y_n=T(x_n)+F(x_n)to T(x)+F(x)=(T+F)(x)$ as $nto infty$, also $y_nto y$ is clear by definition so $y=(T+F)x$ gives $overline{G_{T+F}}subset G_{T+F}$. Hence our claim is proved.
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
answered Dec 4 '18 at 11:28
Sujit BhattacharyyaSujit Bhattacharyya
1,092418
1,092418
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
add a comment |
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
Thank you for a very detailed answer
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:30
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
welcome. For your case, we have the domain set zero $D(T)subset X={0}$ so $D(T)={0}$ so the result holds trivially.
$endgroup$
– Sujit Bhattacharyya
Dec 4 '18 at 11:32
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
$begingroup$
Yeah it was meant to say $neq$, edited it now :)
$endgroup$
– thaumoctopus
Dec 4 '18 at 11:35
add a comment |
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