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Spectrum of an unbounded operator is not compact.

-2

$begingroup$

My problem:

Let $X=C[0,pi]$ and define an operator $T: D to X$ where $$D= x in X mid x',x'' in Xquad textand quad x(0)=x(pi)=0
$$defined by $T(x)=x''.$
Show that spectrum of $T$ is not compact.

My approach to this problem is that;

1. boundedness of the spectrum follows from the Neumann series expansion in $lambda$; the spectrum $sigma(T)$ is bounded by $|T|$.




2. The spectrum of an unbounded operator is in general a closed, possibly empty, subset of the complex plane.

If I show that $T$ is an unbounded then for the spectrum will non-compact if it is not bounded by statement(1) and by statement(2) i must have to show that spectrum of $T$ is non-empty.

My approach is right or not? If yes, then please tell me how can I show that spectrum of an unbounded operator is non-empty? Because the result for non-empty spectrum is hold bounded operators.

functional-analysis spectral-theory

share|cite|improve this question

edited Mar 27 at 17:48

khujista muhreen

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what have you tried?
  $endgroup$
  – supinf
  Mar 26 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $T$? Based on what you say here it could be that $T=0$, which certainly has compact spectrum.
  $endgroup$
  – David C. Ullrich
  Mar 26 at 16:54
  

  
  
  
  

add a comment | 

-2

$begingroup$

My problem:

Let $X=C[0,pi]$ and define an operator $T: D to X$ where $$D= x in X mid x',x'' in Xquad textand quad x(0)=x(pi)=0
$$defined by $T(x)=x''.$
Show that spectrum of $T$ is not compact.

My approach to this problem is that;

1. boundedness of the spectrum follows from the Neumann series expansion in $lambda$; the spectrum $sigma(T)$ is bounded by $|T|$.




2. The spectrum of an unbounded operator is in general a closed, possibly empty, subset of the complex plane.

If I show that $T$ is an unbounded then for the spectrum will non-compact if it is not bounded by statement(1) and by statement(2) i must have to show that spectrum of $T$ is non-empty.

My approach is right or not? If yes, then please tell me how can I show that spectrum of an unbounded operator is non-empty? Because the result for non-empty spectrum is hold bounded operators.

functional-analysis spectral-theory

share|cite|improve this question

edited Mar 27 at 17:48

khujista muhreen

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what have you tried?
  $endgroup$
  – supinf
  Mar 26 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $T$? Based on what you say here it could be that $T=0$, which certainly has compact spectrum.
  $endgroup$
  – David C. Ullrich
  Mar 26 at 16:54
  

  
  
  
  

add a comment | 

$begingroup$

My problem:

Let $X=C[0,pi]$ and define an operator $T: D to X$ where $$D= x in X mid x',x'' in Xquad textand quad x(0)=x(pi)=0
$$defined by $T(x)=x''.$
Show that spectrum of $T$ is not compact.

My approach to this problem is that;

1. boundedness of the spectrum follows from the Neumann series expansion in $lambda$; the spectrum $sigma(T)$ is bounded by $|T|$.




2. The spectrum of an unbounded operator is in general a closed, possibly empty, subset of the complex plane.

If I show that $T$ is an unbounded then for the spectrum will non-compact if it is not bounded by statement(1) and by statement(2) i must have to show that spectrum of $T$ is non-empty.

My approach is right or not? If yes, then please tell me how can I show that spectrum of an unbounded operator is non-empty? Because the result for non-empty spectrum is hold bounded operators.

functional-analysis spectral-theory

share|cite|improve this question

edited Mar 27 at 17:48

khujista muhreen

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

$endgroup$

share|cite|improve this question

edited Mar 27 at 17:48

khujista muhreen

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

share|cite|improve this question

edited Mar 27 at 17:48

khujista muhreen

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13

asked Mar 26 at 13:46

khujista muhreenkhujista muhreen

13