Smoothness of finite-dimensional functional calculus












7












$begingroup$


Assume that $f:mathbb Rtomathbb R$ is continuous.
Given a real symmetric matrix $Aintext{Sym}(n)$, we can define $f(A)$ by applying $f$ to its spectrum. More explicitly,
$$ f(A):=sum f(lambda)P_lambda,qquad A=sumlambda P_lambda. $$
Here both sums are finite, and the second one is the decomposition of $A$ as a linear combination of orthogonal projections ($P_lambda$ is the projection onto the eigenspace for the eigenvalue $lambda$, so that $P_lambda P_{lambda'}=0$). Such decomposition exists and is unique by the spectral theorem.



I guess it is well known that $f:text{Sym}(n)totext{Sym}(n)$ is continuous.




Assuming $fin C^infty(mathbb R)$, is the induced map $f:text{Sym}(n)totext{Sym}(n)$ also smooth?




I think I can show that it is (Fréchet) differentiable everywhere, but I am wondering whether it is always $C^1$ or even $C^infty$.










share|cite|improve this question









$endgroup$

















    7












    $begingroup$


    Assume that $f:mathbb Rtomathbb R$ is continuous.
    Given a real symmetric matrix $Aintext{Sym}(n)$, we can define $f(A)$ by applying $f$ to its spectrum. More explicitly,
    $$ f(A):=sum f(lambda)P_lambda,qquad A=sumlambda P_lambda. $$
    Here both sums are finite, and the second one is the decomposition of $A$ as a linear combination of orthogonal projections ($P_lambda$ is the projection onto the eigenspace for the eigenvalue $lambda$, so that $P_lambda P_{lambda'}=0$). Such decomposition exists and is unique by the spectral theorem.



    I guess it is well known that $f:text{Sym}(n)totext{Sym}(n)$ is continuous.




    Assuming $fin C^infty(mathbb R)$, is the induced map $f:text{Sym}(n)totext{Sym}(n)$ also smooth?




    I think I can show that it is (Fréchet) differentiable everywhere, but I am wondering whether it is always $C^1$ or even $C^infty$.










    share|cite|improve this question









    $endgroup$















      7












      7








      7


      2



      $begingroup$


      Assume that $f:mathbb Rtomathbb R$ is continuous.
      Given a real symmetric matrix $Aintext{Sym}(n)$, we can define $f(A)$ by applying $f$ to its spectrum. More explicitly,
      $$ f(A):=sum f(lambda)P_lambda,qquad A=sumlambda P_lambda. $$
      Here both sums are finite, and the second one is the decomposition of $A$ as a linear combination of orthogonal projections ($P_lambda$ is the projection onto the eigenspace for the eigenvalue $lambda$, so that $P_lambda P_{lambda'}=0$). Such decomposition exists and is unique by the spectral theorem.



      I guess it is well known that $f:text{Sym}(n)totext{Sym}(n)$ is continuous.




      Assuming $fin C^infty(mathbb R)$, is the induced map $f:text{Sym}(n)totext{Sym}(n)$ also smooth?




      I think I can show that it is (Fréchet) differentiable everywhere, but I am wondering whether it is always $C^1$ or even $C^infty$.










      share|cite|improve this question









      $endgroup$




      Assume that $f:mathbb Rtomathbb R$ is continuous.
      Given a real symmetric matrix $Aintext{Sym}(n)$, we can define $f(A)$ by applying $f$ to its spectrum. More explicitly,
      $$ f(A):=sum f(lambda)P_lambda,qquad A=sumlambda P_lambda. $$
      Here both sums are finite, and the second one is the decomposition of $A$ as a linear combination of orthogonal projections ($P_lambda$ is the projection onto the eigenspace for the eigenvalue $lambda$, so that $P_lambda P_{lambda'}=0$). Such decomposition exists and is unique by the spectral theorem.



      I guess it is well known that $f:text{Sym}(n)totext{Sym}(n)$ is continuous.




      Assuming $fin C^infty(mathbb R)$, is the induced map $f:text{Sym}(n)totext{Sym}(n)$ also smooth?




      I think I can show that it is (Fréchet) differentiable everywhere, but I am wondering whether it is always $C^1$ or even $C^infty$.







      fa.functional-analysis real-analysis sp.spectral-theory






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      share|cite|improve this question











      share|cite|improve this question




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      asked 10 hours ago









      MizarMizar

      1,5981023




      1,5981023






















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          6












          $begingroup$

          Yes. The can be derived from the resolvent formalism.



          I'll just do the $C^1$ case and leave higher derivatives as an exercise - ask if it's not clear how to generalize. I am basically using formula (2.7) of "On differentiability of symmetric matrix valued functions", Alexander Shapiro, http://www.optimization-online.org/DB_HTML/2002/07/499.html. (I think there's a typo in the middle case of the display preceeding (2.7): $f(mu_j)/(mu_j-mu_k)$ should be $(f(mu_j)-f(mu_k))/(mu_j-mu_k).$) Shapiro's paper references "(cf., [4])", where [4] is the 600-page textbook "Perturbation Theory for Linear Operators" by T. Kato, but I don't know if that is helpful for this specific question.



          I will call the induced map $f^*$ to distinguish it from $f.$ I'll also call the dimension $p$ instead of $n.$



          It suffices to show $f^*$ is $C^1$ for matrices with eigenvalues in a given bounded interval $J.$ Approximate $f$ by polynomials $f_n$ such that $sup_{xin J}|f(x)-f_n(x)|to 0$ and $sup_{xin J}|f'(x)-f_n'(x)|to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:



          $$f_n^*(X) = frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1} dz$$
          where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $Hinmathrm{Sym}(p),$



          begin{align*}
          f_n^*(X+H)
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X-H)^{-1} dz\
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1}+f_n(z)(z I_p - X)^{-1}H(z I_p - X)^{-1} +dots dz\
          &= frac{1}{2pi i}int_C f_n(z)sum_{lambda}(z-lambda)^{-1}P_lambda +f_n(z)sum_{lambda_1,lambda_2}(z-lambda_1)^{-1}(z-lambda_2)^{-1}P_{lambda_1}HP_{lambda_2}+dots dz\
          &= f_n^*(X)+sum_{lambda_1,lambda_2} P_{lambda_1} H P_{lambda_2}int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
          end{align*}



          The second equality uses the Taylor expansion $$(A-H)^{-1}=A^{-1}+A^{-1}HA^{-1}+dots$$ with $A=z I_p-X.$
          The third equality uses $(zI_p - X)^{-1}=sum_lambda (z-lambda)^{-1} P_lambda.$ The fourth equality uses $int_C f_n(z)(z-lambda)^{-1}(z-mu)^{-1}dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$



          This gives a bound



          $$|Df^*_n(X)H| leq c_p|H|cdot sup_{xin J}|f'_n(x)-f_n'(x)|$$



          for some constant $c_p>0,$ where $|cdot|$ is any matrix norm. This shows that $f^*$ can be approximated arbitrarily well in the $C^1$ norm, which means it's $C^1.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
            $endgroup$
            – Mizar
            9 hours ago



















          0












          $begingroup$

          Yes. To show that f(A) is n-times differentiable at A=B, simply interpolate f by a polynomial P so that P and its derivatives up to order n agree with f on the spectrum of B. Clearly P(A) is n-times differentiable, and it isn't too much work to show that f and P have the same nth derivative.






          share|cite|improve this answer








          New contributor




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






          $endgroup$













          • $begingroup$
            Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
            $endgroup$
            – Mizar
            4 hours ago














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          2 Answers
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          2 Answers
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          active

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          6












          $begingroup$

          Yes. The can be derived from the resolvent formalism.



          I'll just do the $C^1$ case and leave higher derivatives as an exercise - ask if it's not clear how to generalize. I am basically using formula (2.7) of "On differentiability of symmetric matrix valued functions", Alexander Shapiro, http://www.optimization-online.org/DB_HTML/2002/07/499.html. (I think there's a typo in the middle case of the display preceeding (2.7): $f(mu_j)/(mu_j-mu_k)$ should be $(f(mu_j)-f(mu_k))/(mu_j-mu_k).$) Shapiro's paper references "(cf., [4])", where [4] is the 600-page textbook "Perturbation Theory for Linear Operators" by T. Kato, but I don't know if that is helpful for this specific question.



          I will call the induced map $f^*$ to distinguish it from $f.$ I'll also call the dimension $p$ instead of $n.$



          It suffices to show $f^*$ is $C^1$ for matrices with eigenvalues in a given bounded interval $J.$ Approximate $f$ by polynomials $f_n$ such that $sup_{xin J}|f(x)-f_n(x)|to 0$ and $sup_{xin J}|f'(x)-f_n'(x)|to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:



          $$f_n^*(X) = frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1} dz$$
          where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $Hinmathrm{Sym}(p),$



          begin{align*}
          f_n^*(X+H)
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X-H)^{-1} dz\
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1}+f_n(z)(z I_p - X)^{-1}H(z I_p - X)^{-1} +dots dz\
          &= frac{1}{2pi i}int_C f_n(z)sum_{lambda}(z-lambda)^{-1}P_lambda +f_n(z)sum_{lambda_1,lambda_2}(z-lambda_1)^{-1}(z-lambda_2)^{-1}P_{lambda_1}HP_{lambda_2}+dots dz\
          &= f_n^*(X)+sum_{lambda_1,lambda_2} P_{lambda_1} H P_{lambda_2}int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
          end{align*}



          The second equality uses the Taylor expansion $$(A-H)^{-1}=A^{-1}+A^{-1}HA^{-1}+dots$$ with $A=z I_p-X.$
          The third equality uses $(zI_p - X)^{-1}=sum_lambda (z-lambda)^{-1} P_lambda.$ The fourth equality uses $int_C f_n(z)(z-lambda)^{-1}(z-mu)^{-1}dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$



          This gives a bound



          $$|Df^*_n(X)H| leq c_p|H|cdot sup_{xin J}|f'_n(x)-f_n'(x)|$$



          for some constant $c_p>0,$ where $|cdot|$ is any matrix norm. This shows that $f^*$ can be approximated arbitrarily well in the $C^1$ norm, which means it's $C^1.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
            $endgroup$
            – Mizar
            9 hours ago
















          6












          $begingroup$

          Yes. The can be derived from the resolvent formalism.



          I'll just do the $C^1$ case and leave higher derivatives as an exercise - ask if it's not clear how to generalize. I am basically using formula (2.7) of "On differentiability of symmetric matrix valued functions", Alexander Shapiro, http://www.optimization-online.org/DB_HTML/2002/07/499.html. (I think there's a typo in the middle case of the display preceeding (2.7): $f(mu_j)/(mu_j-mu_k)$ should be $(f(mu_j)-f(mu_k))/(mu_j-mu_k).$) Shapiro's paper references "(cf., [4])", where [4] is the 600-page textbook "Perturbation Theory for Linear Operators" by T. Kato, but I don't know if that is helpful for this specific question.



          I will call the induced map $f^*$ to distinguish it from $f.$ I'll also call the dimension $p$ instead of $n.$



          It suffices to show $f^*$ is $C^1$ for matrices with eigenvalues in a given bounded interval $J.$ Approximate $f$ by polynomials $f_n$ such that $sup_{xin J}|f(x)-f_n(x)|to 0$ and $sup_{xin J}|f'(x)-f_n'(x)|to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:



          $$f_n^*(X) = frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1} dz$$
          where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $Hinmathrm{Sym}(p),$



          begin{align*}
          f_n^*(X+H)
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X-H)^{-1} dz\
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1}+f_n(z)(z I_p - X)^{-1}H(z I_p - X)^{-1} +dots dz\
          &= frac{1}{2pi i}int_C f_n(z)sum_{lambda}(z-lambda)^{-1}P_lambda +f_n(z)sum_{lambda_1,lambda_2}(z-lambda_1)^{-1}(z-lambda_2)^{-1}P_{lambda_1}HP_{lambda_2}+dots dz\
          &= f_n^*(X)+sum_{lambda_1,lambda_2} P_{lambda_1} H P_{lambda_2}int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
          end{align*}



          The second equality uses the Taylor expansion $$(A-H)^{-1}=A^{-1}+A^{-1}HA^{-1}+dots$$ with $A=z I_p-X.$
          The third equality uses $(zI_p - X)^{-1}=sum_lambda (z-lambda)^{-1} P_lambda.$ The fourth equality uses $int_C f_n(z)(z-lambda)^{-1}(z-mu)^{-1}dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$



          This gives a bound



          $$|Df^*_n(X)H| leq c_p|H|cdot sup_{xin J}|f'_n(x)-f_n'(x)|$$



          for some constant $c_p>0,$ where $|cdot|$ is any matrix norm. This shows that $f^*$ can be approximated arbitrarily well in the $C^1$ norm, which means it's $C^1.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
            $endgroup$
            – Mizar
            9 hours ago














          6












          6








          6





          $begingroup$

          Yes. The can be derived from the resolvent formalism.



          I'll just do the $C^1$ case and leave higher derivatives as an exercise - ask if it's not clear how to generalize. I am basically using formula (2.7) of "On differentiability of symmetric matrix valued functions", Alexander Shapiro, http://www.optimization-online.org/DB_HTML/2002/07/499.html. (I think there's a typo in the middle case of the display preceeding (2.7): $f(mu_j)/(mu_j-mu_k)$ should be $(f(mu_j)-f(mu_k))/(mu_j-mu_k).$) Shapiro's paper references "(cf., [4])", where [4] is the 600-page textbook "Perturbation Theory for Linear Operators" by T. Kato, but I don't know if that is helpful for this specific question.



          I will call the induced map $f^*$ to distinguish it from $f.$ I'll also call the dimension $p$ instead of $n.$



          It suffices to show $f^*$ is $C^1$ for matrices with eigenvalues in a given bounded interval $J.$ Approximate $f$ by polynomials $f_n$ such that $sup_{xin J}|f(x)-f_n(x)|to 0$ and $sup_{xin J}|f'(x)-f_n'(x)|to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:



          $$f_n^*(X) = frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1} dz$$
          where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $Hinmathrm{Sym}(p),$



          begin{align*}
          f_n^*(X+H)
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X-H)^{-1} dz\
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1}+f_n(z)(z I_p - X)^{-1}H(z I_p - X)^{-1} +dots dz\
          &= frac{1}{2pi i}int_C f_n(z)sum_{lambda}(z-lambda)^{-1}P_lambda +f_n(z)sum_{lambda_1,lambda_2}(z-lambda_1)^{-1}(z-lambda_2)^{-1}P_{lambda_1}HP_{lambda_2}+dots dz\
          &= f_n^*(X)+sum_{lambda_1,lambda_2} P_{lambda_1} H P_{lambda_2}int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
          end{align*}



          The second equality uses the Taylor expansion $$(A-H)^{-1}=A^{-1}+A^{-1}HA^{-1}+dots$$ with $A=z I_p-X.$
          The third equality uses $(zI_p - X)^{-1}=sum_lambda (z-lambda)^{-1} P_lambda.$ The fourth equality uses $int_C f_n(z)(z-lambda)^{-1}(z-mu)^{-1}dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$



          This gives a bound



          $$|Df^*_n(X)H| leq c_p|H|cdot sup_{xin J}|f'_n(x)-f_n'(x)|$$



          for some constant $c_p>0,$ where $|cdot|$ is any matrix norm. This shows that $f^*$ can be approximated arbitrarily well in the $C^1$ norm, which means it's $C^1.$






          share|cite|improve this answer









          $endgroup$



          Yes. The can be derived from the resolvent formalism.



          I'll just do the $C^1$ case and leave higher derivatives as an exercise - ask if it's not clear how to generalize. I am basically using formula (2.7) of "On differentiability of symmetric matrix valued functions", Alexander Shapiro, http://www.optimization-online.org/DB_HTML/2002/07/499.html. (I think there's a typo in the middle case of the display preceeding (2.7): $f(mu_j)/(mu_j-mu_k)$ should be $(f(mu_j)-f(mu_k))/(mu_j-mu_k).$) Shapiro's paper references "(cf., [4])", where [4] is the 600-page textbook "Perturbation Theory for Linear Operators" by T. Kato, but I don't know if that is helpful for this specific question.



          I will call the induced map $f^*$ to distinguish it from $f.$ I'll also call the dimension $p$ instead of $n.$



          It suffices to show $f^*$ is $C^1$ for matrices with eigenvalues in a given bounded interval $J.$ Approximate $f$ by polynomials $f_n$ such that $sup_{xin J}|f(x)-f_n(x)|to 0$ and $sup_{xin J}|f'(x)-f_n'(x)|to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:



          $$f_n^*(X) = frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1} dz$$
          where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $Hinmathrm{Sym}(p),$



          begin{align*}
          f_n^*(X+H)
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X-H)^{-1} dz\
          &= frac{1}{2pi i}int_C f_n(z)(z I_p - X)^{-1}+f_n(z)(z I_p - X)^{-1}H(z I_p - X)^{-1} +dots dz\
          &= frac{1}{2pi i}int_C f_n(z)sum_{lambda}(z-lambda)^{-1}P_lambda +f_n(z)sum_{lambda_1,lambda_2}(z-lambda_1)^{-1}(z-lambda_2)^{-1}P_{lambda_1}HP_{lambda_2}+dots dz\
          &= f_n^*(X)+sum_{lambda_1,lambda_2} P_{lambda_1} H P_{lambda_2}int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
          end{align*}



          The second equality uses the Taylor expansion $$(A-H)^{-1}=A^{-1}+A^{-1}HA^{-1}+dots$$ with $A=z I_p-X.$
          The third equality uses $(zI_p - X)^{-1}=sum_lambda (z-lambda)^{-1} P_lambda.$ The fourth equality uses $int_C f_n(z)(z-lambda)^{-1}(z-mu)^{-1}dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$



          This gives a bound



          $$|Df^*_n(X)H| leq c_p|H|cdot sup_{xin J}|f'_n(x)-f_n'(x)|$$



          for some constant $c_p>0,$ where $|cdot|$ is any matrix norm. This shows that $f^*$ can be approximated arbitrarily well in the $C^1$ norm, which means it's $C^1.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 10 hours ago









          DapDap

          91826




          91826








          • 1




            $begingroup$
            Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
            $endgroup$
            – Mizar
            9 hours ago














          • 1




            $begingroup$
            Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
            $endgroup$
            – Mizar
            9 hours ago








          1




          1




          $begingroup$
          Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
          $endgroup$
          – Mizar
          9 hours ago




          $begingroup$
          Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
          $endgroup$
          – Mizar
          9 hours ago











          0












          $begingroup$

          Yes. To show that f(A) is n-times differentiable at A=B, simply interpolate f by a polynomial P so that P and its derivatives up to order n agree with f on the spectrum of B. Clearly P(A) is n-times differentiable, and it isn't too much work to show that f and P have the same nth derivative.






          share|cite|improve this answer








          New contributor




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






          $endgroup$













          • $begingroup$
            Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
            $endgroup$
            – Mizar
            4 hours ago


















          0












          $begingroup$

          Yes. To show that f(A) is n-times differentiable at A=B, simply interpolate f by a polynomial P so that P and its derivatives up to order n agree with f on the spectrum of B. Clearly P(A) is n-times differentiable, and it isn't too much work to show that f and P have the same nth derivative.






          share|cite|improve this answer








          New contributor




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






          $endgroup$













          • $begingroup$
            Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
            $endgroup$
            – Mizar
            4 hours ago
















          0












          0








          0





          $begingroup$

          Yes. To show that f(A) is n-times differentiable at A=B, simply interpolate f by a polynomial P so that P and its derivatives up to order n agree with f on the spectrum of B. Clearly P(A) is n-times differentiable, and it isn't too much work to show that f and P have the same nth derivative.






          share|cite|improve this answer








          New contributor




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






          $endgroup$



          Yes. To show that f(A) is n-times differentiable at A=B, simply interpolate f by a polynomial P so that P and its derivatives up to order n agree with f on the spectrum of B. Clearly P(A) is n-times differentiable, and it isn't too much work to show that f and P have the same nth derivative.







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          answered 7 hours ago









          B ChinB Chin

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          • $begingroup$
            Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
            $endgroup$
            – Mizar
            4 hours ago




















          • $begingroup$
            Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
            $endgroup$
            – Mizar
            4 hours ago


















          $begingroup$
          Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
          $endgroup$
          – Mizar
          4 hours ago






          $begingroup$
          Yeah, I thought about this strategy, but it's not so clear why this gives $C^n$ regularity rather than just a Taylor approximation of degree $n$ at each $A$. (If one manages to infer such an approximation exists with coefficients depending continuously on $A$, then one could invoke this result: mathoverflow.net/questions/88501/converse-of-taylors-theorem)
          $endgroup$
          – Mizar
          4 hours ago




















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