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1

$begingroup$

First of all I want to thank you for the help you provide on this website! Whenever I had a hard time understanding things in math I visited this website and (nearly) allways found a hint or a solution to a problem I had.
But this time it seems that I have to ask a question myself, because I haven't found anything to solve a problem I found in a book I'm currently reading. Although it seems like a pretty basic question in functional calculus.

Setting of the Problem:
Let $T$ be a self adjoint (unbounded) linear operator in an Hilbert space $X$ with spectral family (resolution of the identity) $E$.
Let $f : mathbbR rightarrow mathbbC$ be a measurable function, then $f(T)$ can be defined by the spectral theorem:

$f(T)x := intlimits_mathbbR f(t) dE(t)x$ for all $x in D(f(T)) := left^2 d langle y, E(t) y rangle < infty right$

Problem: Assume $f$ is continous and real-valued, ie: $f : mathbbR rightarrow mathbbR$, then there holds:
$sigma(f(T)) = overlinef(sigma(T))$, where $f(A) := left t in A right$ for $A subseteq mathbbR$.
Moreover if $|f(t)| rightarrow 0$ for $|t| rightarrow infty$ then the upper equation holds without the closure.

My attempted solution: I know from functional calculus that
$sigma(f(T)) = left lambda in mathbbC Big $
and I have the feeling that this might be the solution but I can't see the relation between this and the stated problem.

I hope somebody has a solution or hint to this problem and is able to help out a poor little fellow who is relatively new to functional calculus. Thanks in advance,
GordonFreeman

Ps.: I apologize for any mistakes in my english. If somethings unclear because of my phrasing, please let me know.

functional-analysis spectral-theory functional-calculus

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asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

First of all I want to thank you for the help you provide on this website! Whenever I had a hard time understanding things in math I visited this website and (nearly) allways found a hint or a solution to a problem I had.
But this time it seems that I have to ask a question myself, because I haven't found anything to solve a problem I found in a book I'm currently reading. Although it seems like a pretty basic question in functional calculus.

Setting of the Problem:
Let $T$ be a self adjoint (unbounded) linear operator in an Hilbert space $X$ with spectral family (resolution of the identity) $E$.
Let $f : mathbbR rightarrow mathbbC$ be a measurable function, then $f(T)$ can be defined by the spectral theorem:

$f(T)x := intlimits_mathbbR f(t) dE(t)x$ for all $x in D(f(T)) := left^2 d langle y, E(t) y rangle < infty right$

Problem: Assume $f$ is continous and real-valued, ie: $f : mathbbR rightarrow mathbbR$, then there holds:
$sigma(f(T)) = overlinef(sigma(T))$, where $f(A) := left t in A right$ for $A subseteq mathbbR$.
Moreover if $|f(t)| rightarrow 0$ for $|t| rightarrow infty$ then the upper equation holds without the closure.

My attempted solution: I know from functional calculus that
$sigma(f(T)) = left lambda in mathbbC Big $
and I have the feeling that this might be the solution but I can't see the relation between this and the stated problem.

I hope somebody has a solution or hint to this problem and is able to help out a poor little fellow who is relatively new to functional calculus. Thanks in advance,
GordonFreeman

Ps.: I apologize for any mistakes in my english. If somethings unclear because of my phrasing, please let me know.

functional-analysis spectral-theory functional-calculus

share|cite|improve this question

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

First of all I want to thank you for the help you provide on this website! Whenever I had a hard time understanding things in math I visited this website and (nearly) allways found a hint or a solution to a problem I had.
But this time it seems that I have to ask a question myself, because I haven't found anything to solve a problem I found in a book I'm currently reading. Although it seems like a pretty basic question in functional calculus.

Setting of the Problem:
Let $T$ be a self adjoint (unbounded) linear operator in an Hilbert space $X$ with spectral family (resolution of the identity) $E$.
Let $f : mathbbR rightarrow mathbbC$ be a measurable function, then $f(T)$ can be defined by the spectral theorem:

$f(T)x := intlimits_mathbbR f(t) dE(t)x$ for all $x in D(f(T)) := left^2 d langle y, E(t) y rangle < infty right$

Problem: Assume $f$ is continous and real-valued, ie: $f : mathbbR rightarrow mathbbR$, then there holds:
$sigma(f(T)) = overlinef(sigma(T))$, where $f(A) := left t in A right$ for $A subseteq mathbbR$.
Moreover if $|f(t)| rightarrow 0$ for $|t| rightarrow infty$ then the upper equation holds without the closure.

My attempted solution: I know from functional calculus that
$sigma(f(T)) = left lambda in mathbbC Big $
and I have the feeling that this might be the solution but I can't see the relation between this and the stated problem.

I hope somebody has a solution or hint to this problem and is able to help out a poor little fellow who is relatively new to functional calculus. Thanks in advance,
GordonFreeman

Ps.: I apologize for any mistakes in my english. If somethings unclear because of my phrasing, please let me know.

functional-analysis spectral-theory functional-calculus

share|cite|improve this question

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63

New contributor

GordonFreeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 8:46

GordonFreemanGordonFreeman

63