Bounded function of compact normal operator on Hilbert space is normal












4














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







share|cite|improve this question



























    4














    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







    share|cite|improve this question

























      4












      4








      4







      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







      share|cite|improve this question













      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






      share|cite|improve this question













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      asked Jan 1 at 18:03









      DivergenceFormDivergenceForm

      685




      685






















          3 Answers
          3






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          2














          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}






          share|cite|improve this answer































            1














            The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
            $$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$






            share|cite|improve this answer





























              1














              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.






              share|cite|improve this answer





















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                3 Answers
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                active

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                3 Answers
                3






                active

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                active

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                active

                oldest

                votes









                2














                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}






                share|cite|improve this answer




























                  2














                  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}






                  share|cite|improve this answer


























                    2












                    2








                    2






                    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}






                    share|cite|improve this answer














                    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}







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 1 at 21:08

























                    answered Jan 1 at 20:57









                    DisintegratingByPartsDisintegratingByParts

                    58.7k42579




                    58.7k42579























                        1














                        The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
                        $$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$






                        share|cite|improve this answer


























                          1














                          The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
                          $$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$






                          share|cite|improve this answer
























                            1












                            1








                            1






                            The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
                            $$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$






                            share|cite|improve this answer












                            The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
                            $$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 1 at 20:44









                            AweyganAweygan

                            13.5k21441




                            13.5k21441























                                1














                                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.






                                share|cite|improve this answer


























                                  1














                                  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.






                                  share|cite|improve this answer
























                                    1












                                    1








                                    1






                                    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.






                                    share|cite|improve this answer












                                    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.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 1 at 21:09









                                    mechanodroidmechanodroid

                                    27k62446




                                    27k62446






























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