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Pre-composing a Closed operator by a bounded operator

0

$begingroup$

A linear map ( not necessarily bounded ) between normed linear spaces is called a closed operator if its graph is closed. Suppose $X$ is a n.l.s and $Y, Z$ are Banach spaces. Let $A : X_0 ⊆ X → Y$ be a closed operator and $B ∈ B(Z, X)$ such that $Ran(B) ⊆ X_0$ . Prove that $AB ∈ B(Z, Y )$

Since, $Z,Y$ both are given to be Banach, I was willing to apply Closed graph Theorem to $AB$ . Let $(x,y)$ be a limit point of the Graph, then Enough to show that $AB(x)=y$ . Consider any sequence $(x_n,ABx_n_n ge 1$ in the Graph such that it converges to $(x,y)$ i.e. $x_n to x$ and $ABx_n to y$ . Since , B is given to be continuous, $x_n to x implies B(x_n) to B(x)= z $ (say) .

So all we need to show is $Az=y$ . Here I am stuck. How to use the fact that $A$ admits a closed graph?

Thanks in advance for help!

functional-analysis closed-graph

share|cite|improve this question

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

$endgroup$

add a comment | 

0

$begingroup$

A linear map ( not necessarily bounded ) between normed linear spaces is called a closed operator if its graph is closed. Suppose $X$ is a n.l.s and $Y, Z$ are Banach spaces. Let $A : X_0 ⊆ X → Y$ be a closed operator and $B ∈ B(Z, X)$ such that $Ran(B) ⊆ X_0$ . Prove that $AB ∈ B(Z, Y )$

Since, $Z,Y$ both are given to be Banach, I was willing to apply Closed graph Theorem to $AB$ . Let $(x,y)$ be a limit point of the Graph, then Enough to show that $AB(x)=y$ . Consider any sequence $(x_n,ABx_n_n ge 1$ in the Graph such that it converges to $(x,y)$ i.e. $x_n to x$ and $ABx_n to y$ . Since , B is given to be continuous, $x_n to x implies B(x_n) to B(x)= z $ (say) .

So all we need to show is $Az=y$ . Here I am stuck. How to use the fact that $A$ admits a closed graph?

Thanks in advance for help!

functional-analysis closed-graph

share|cite|improve this question

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

$endgroup$

add a comment | 

$begingroup$

A linear map ( not necessarily bounded ) between normed linear spaces is called a closed operator if its graph is closed. Suppose $X$ is a n.l.s and $Y, Z$ are Banach spaces. Let $A : X_0 ⊆ X → Y$ be a closed operator and $B ∈ B(Z, X)$ such that $Ran(B) ⊆ X_0$ . Prove that $AB ∈ B(Z, Y )$

Since, $Z,Y$ both are given to be Banach, I was willing to apply Closed graph Theorem to $AB$ . Let $(x,y)$ be a limit point of the Graph, then Enough to show that $AB(x)=y$ . Consider any sequence $(x_n,ABx_n_n ge 1$ in the Graph such that it converges to $(x,y)$ i.e. $x_n to x$ and $ABx_n to y$ . Since , B is given to be continuous, $x_n to x implies B(x_n) to B(x)= z $ (say) .

So all we need to show is $Az=y$ . Here I am stuck. How to use the fact that $A$ admits a closed graph?

Thanks in advance for help!

functional-analysis closed-graph

share|cite|improve this question

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

$endgroup$

share|cite|improve this question

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

share|cite|improve this question

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

edited Mar 28 at 4:07

Saad

20.3k92352

edited Mar 28 at 4:07

Saad

20.3k92352

asked Mar 28 at 3:16

reflexivereflexive

1,178625

asked Mar 28 at 3:16

reflexivereflexive

1,178625

1 Answer
1

active

oldest

votes

0

$begingroup$

Note that $(Bx_n, ABx_n)$ is a sequence in the graph of $A$. You know $Bx_n to Bx$ and $ABx_n to y$ so, since the graph of A is closed, $Bx in D(A)$ and $ABx =y$. This implies that $x in D(AB)$ and $ABx =y$ which is what we needed to show that $AB$ has closed graph.

share|cite|improve this answer

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

$endgroup$

add a comment | 

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0

$begingroup$

Note that $(Bx_n, ABx_n)$ is a sequence in the graph of $A$. You know $Bx_n to Bx$ and $ABx_n to y$ so, since the graph of A is closed, $Bx in D(A)$ and $ABx =y$. This implies that $x in D(AB)$ and $ABx =y$ which is what we needed to show that $AB$ has closed graph.

share|cite|improve this answer

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

$endgroup$

add a comment | 

0

$begingroup$

Note that $(Bx_n, ABx_n)$ is a sequence in the graph of $A$. You know $Bx_n to Bx$ and $ABx_n to y$ so, since the graph of A is closed, $Bx in D(A)$ and $ABx =y$. This implies that $x in D(AB)$ and $ABx =y$ which is what we needed to show that $AB$ has closed graph.

share|cite|improve this answer

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

$endgroup$

add a comment | 

$begingroup$

Note that $(Bx_n, ABx_n)$ is a sequence in the graph of $A$. You know $Bx_n to Bx$ and $ABx_n to y$ so, since the graph of A is closed, $Bx in D(A)$ and $ABx =y$. This implies that $x in D(AB)$ and $ABx =y$ which is what we needed to show that $AB$ has closed graph.

share|cite|improve this answer

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

$endgroup$

share|cite|improve this answer

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930

answered Mar 28 at 3:54

Rhys SteeleRhys Steele

7,5551930