Bounded function of compact normal operator on Hilbert space is normal
$begingroup$
Let $H$ be a Hilbert space and consider a compact normal linear operator $A:H to H$. Moreover, let $f$ be a bounded function on the spectrum $sigma(A)$ of $A$ and consider the operator $f(A)$ in the sense of functional calculus.
I want to show that $f(A)$ is also a normal operator. I know that a bounded linear operator $B:H to H$ is normal if and only if
$$ ||Bx||=||B^star x|| quad forall x in H,$$
which might be useful in this context.
functional-analysis operator-theory spectral-theory functional-calculus
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space and consider a compact normal linear operator $A:H to H$. Moreover, let $f$ be a bounded function on the spectrum $sigma(A)$ of $A$ and consider the operator $f(A)$ in the sense of functional calculus.
I want to show that $f(A)$ is also a normal operator. I know that a bounded linear operator $B:H to H$ is normal if and only if
$$ ||Bx||=||B^star x|| quad forall x in H,$$
which might be useful in this context.
functional-analysis operator-theory spectral-theory functional-calculus
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space and consider a compact normal linear operator $A:H to H$. Moreover, let $f$ be a bounded function on the spectrum $sigma(A)$ of $A$ and consider the operator $f(A)$ in the sense of functional calculus.
I want to show that $f(A)$ is also a normal operator. I know that a bounded linear operator $B:H to H$ is normal if and only if
$$ ||Bx||=||B^star x|| quad forall x in H,$$
which might be useful in this context.
functional-analysis operator-theory spectral-theory functional-calculus
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
$endgroup$
Let $H$ be a Hilbert space and consider a compact normal linear operator $A:H to H$. Moreover, let $f$ be a bounded function on the spectrum $sigma(A)$ of $A$ and consider the operator $f(A)$ in the sense of functional calculus.
I want to show that $f(A)$ is also a normal operator. I know that a bounded linear operator $B:H to H$ is normal if and only if
$$ ||Bx||=||B^star x|| quad forall x in H,$$
which might be useful in this context.
functional-analysis operator-theory spectral-theory functional-calculus
functional-analysis operator-theory spectral-theory functional-calculus
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
asked Jan 1 at 18:03
DivergenceFormDivergenceForm
685
685
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcal{H}$ with eigenvalue $lambda_n$ such that
$$
A = sum_{n} lambda_n P_n, ;; I=sum_{n}P_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_{n}f(lambda_n)P_n$ is normal because $f(A)^*=sum_n overline{f(lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_{n}|f(lambda_n)|^2P_n=sum_{n}|overline{f(lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
begin{align}
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overline{lambda_n}|^2|P_nx|^2 = |f(A)^*x|^2
end{align}
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
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$begingroup$
The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
$endgroup$
add a comment |
$begingroup$
For any $f in B(sigma(A))$ we have $f(A) in C^*(A)$.
Since $f mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = overline{f}(A) in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcal{H}$ with eigenvalue $lambda_n$ such that
$$
A = sum_{n} lambda_n P_n, ;; I=sum_{n}P_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_{n}f(lambda_n)P_n$ is normal because $f(A)^*=sum_n overline{f(lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_{n}|f(lambda_n)|^2P_n=sum_{n}|overline{f(lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
begin{align}
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overline{lambda_n}|^2|P_nx|^2 = |f(A)^*x|^2
end{align}
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcal{H}$ with eigenvalue $lambda_n$ such that
$$
A = sum_{n} lambda_n P_n, ;; I=sum_{n}P_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_{n}f(lambda_n)P_n$ is normal because $f(A)^*=sum_n overline{f(lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_{n}|f(lambda_n)|^2P_n=sum_{n}|overline{f(lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
begin{align}
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overline{lambda_n}|^2|P_nx|^2 = |f(A)^*x|^2
end{align}
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcal{H}$ with eigenvalue $lambda_n$ such that
$$
A = sum_{n} lambda_n P_n, ;; I=sum_{n}P_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_{n}f(lambda_n)P_n$ is normal because $f(A)^*=sum_n overline{f(lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_{n}|f(lambda_n)|^2P_n=sum_{n}|overline{f(lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
begin{align}
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overline{lambda_n}|^2|P_nx|^2 = |f(A)^*x|^2
end{align}
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcal{H}$ with eigenvalue $lambda_n$ such that
$$
A = sum_{n} lambda_n P_n, ;; I=sum_{n}P_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_{n}f(lambda_n)P_n$ is normal because $f(A)^*=sum_n overline{f(lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_{n}|f(lambda_n)|^2P_n=sum_{n}|overline{f(lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
begin{align}
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overline{lambda_n}|^2|P_nx|^2 = |f(A)^*x|^2
end{align}
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
edited Jan 1 at 21:08
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Jan 1 at 20:57
DisintegratingByPartsDisintegratingByParts
59.5k42580
59.5k42580
add a comment |
add a comment |
$begingroup$
The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
$endgroup$
add a comment |
$begingroup$
The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
$endgroup$
add a comment |
$begingroup$
The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
$endgroup$
The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
answered Jan 1 at 20:44
AweyganAweygan
14.4k21442
14.4k21442
add a comment |
add a comment |
$begingroup$
For any $f in B(sigma(A))$ we have $f(A) in C^*(A)$.
Since $f mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = overline{f}(A) in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
$endgroup$
add a comment |
$begingroup$
For any $f in B(sigma(A))$ we have $f(A) in C^*(A)$.
Since $f mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = overline{f}(A) in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
$endgroup$
add a comment |
$begingroup$
For any $f in B(sigma(A))$ we have $f(A) in C^*(A)$.
Since $f mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = overline{f}(A) in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
$endgroup$
For any $f in B(sigma(A))$ we have $f(A) in C^*(A)$.
Since $f mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = overline{f}(A) in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
answered Jan 1 at 21:09
mechanodroidmechanodroid
28k62447
28k62447
add a comment |
add a comment |
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