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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      DivergenceForm

      685




      685






















          3 Answers
          3






          active

          oldest

          votes


















          1














          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





















                Your Answer





                StackExchange.ifUsing("editor", function () {
                return StackExchange.using("mathjaxEditing", function () {
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                });
                });
                }, "mathjax-editing");

                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "69"
                };
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function() {
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled) {
                StackExchange.using("snippets", function() {
                createEditor();
                });
                }
                else {
                createEditor();
                }
                });

                function createEditor() {
                StackExchange.prepareEditor({
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: true,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: 10,
                bindNavPrevention: true,
                postfix: "",
                imageUploader: {
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                },
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058707%2fbounded-function-of-compact-normal-operator-on-hilbert-space-is-normal%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1














                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














                  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












                    1








                    1






                    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 1 hour ago

























                    answered 1 hour ago









                    DisintegratingByParts

                    58.5k42579




                    58.5k42579























                        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 1 hour ago









                            Aweygan

                            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 1 hour ago









                                    mechanodroid

                                    26.2k62245




                                    26.2k62245






























                                        draft saved

                                        draft discarded




















































                                        Thanks for contributing an answer to Mathematics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid …



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.





                                        Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                        Please pay close attention to the following guidance:


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid …



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058707%2fbounded-function-of-compact-normal-operator-on-hilbert-space-is-normal%23new-answer', 'question_page');
                                        }
                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        -

                                        Popular posts from this blog

                                        Eastern Orthodox Church

                                        Image
                                        Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
                                        Read more

                                        Zagreb

                                        Image
                                        This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
                                        Read more

                                        Understanding the information contained in the Deep Space Network XML data?

                                        Image
                                        up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
                                        Read more