Smoothness of finite-dimensional functional calculus The 2019 Stack Overflow Developer Survey Results Are In“Converse” of Taylor's theoremHow “generalized eigenvalues” combine into producing the spectral measure?Can be this operator extended to an unbounded self-adjoint operator ? Resonance of Schrödinger operatorFinding the spectrum of the composition of a projection with a multiplication operatorSchrodinger's equation via Spectral TheoremIs there an asymptotic bound for this oscillatory integral?If $H$ is the closure of the set of solenoidal smooth vecor fields in $L^2$ and $P_H$ denote the orthogonal projection onto $H$, then $P_HH_0^1⊆H_0^1$Gaps in the spectrum of Laplace-Beltrami operatorsPerturbation theory compact operatorIf $A$ is a dissipative self-adjoint operator with spectral decomposition $(H_λ)$, then $e^tAx$ tends to the projection of $x$ onto $H_0$ as $t→∞$
Smoothness of finite-dimensional functional calculus
The 2019 Stack Overflow Developer Survey Results Are In“Converse” of Taylor's theoremHow “generalized eigenvalues” combine into producing the spectral measure?Can be this operator extended to an unbounded self-adjoint operator ? Resonance of Schrödinger operatorFinding the spectrum of the composition of a projection with a multiplication operatorSchrodinger's equation via Spectral TheoremIs there an asymptotic bound for this oscillatory integral?If $H$ is the closure of the set of solenoidal smooth vecor fields in $L^2$ and $P_H$ denote the orthogonal projection onto $H$, then $P_HH_0^1⊆H_0^1$Gaps in the spectrum of Laplace-Beltrami operatorsPerturbation theory compact operatorIf $A$ is a dissipative self-adjoint operator with spectral decomposition $(H_λ)$, then $e^tAx$ tends to the projection of $x$ onto $H_0$ as $t→∞$
$begingroup$
Assume that $f:mathbb Rtomathbb R$ is continuous.
Given a real symmetric matrix $AintextSym(n)$, 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 $f:textSym(n)totextSym(n)$ is continuous.
Assuming $fin C^infty(mathbb R)$, is the induced map $f:textSym(n)totextSym(n)$ 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
$endgroup$
add a comment |
$begingroup$
Assume that $f:mathbb Rtomathbb R$ is continuous.
Given a real symmetric matrix $AintextSym(n)$, 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 $f:textSym(n)totextSym(n)$ is continuous.
Assuming $fin C^infty(mathbb R)$, is the induced map $f:textSym(n)totextSym(n)$ 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
$endgroup$
add a comment |
$begingroup$
Assume that $f:mathbb Rtomathbb R$ is continuous.
Given a real symmetric matrix $AintextSym(n)$, 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 $f:textSym(n)totextSym(n)$ is continuous.
Assuming $fin C^infty(mathbb R)$, is the induced map $f:textSym(n)totextSym(n)$ 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
$endgroup$
Assume that $f:mathbb Rtomathbb R$ is continuous.
Given a real symmetric matrix $AintextSym(n)$, 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 $f:textSym(n)totextSym(n)$ is continuous.
Assuming $fin C^infty(mathbb R)$, is the induced map $f:textSym(n)totextSym(n)$ 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
fa.functional-analysis real-analysis sp.spectral-theory
asked Apr 6 at 17:05
MizarMizar
1,6581024
1,6581024
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$f_n^*(X) = frac12pi iint_C f_n(z)(z I_p - X)^-1 dz$$
where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $HinmathrmSym(p),$
beginalign*
f_n^*(X+H)
&= frac12pi iint_C f_n(z)(z I_p - X-H)^-1 dz\
&= frac12pi iint_C f_n(z)(z I_p - X)^-1+f_n(z)(z I_p - X)^-1H(z I_p - X)^-1 +dots dz\
&= frac12pi iint_C f_n(z)sum_lambda(z-lambda)^-1P_lambda +f_n(z)sum_lambda_1,lambda_2(z-lambda_1)^-1(z-lambda_2)^-1P_lambda_1HP_lambda_2+dots dz\
&= f_n^*(X)+sum_lambda_1,lambda_2 P_lambda_1 H P_lambda_2int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
endalign*
The second equality uses the Taylor expansion $$(A-H)^-1=A^-1+A^-1HA^-1+dots$$ 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 $int_C f_n(z)(z-lambda)^-1(z-mu)^-1dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$
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 $C^1.$
$endgroup$
1
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
add a comment |
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
Apr 6 at 23:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$f_n^*(X) = frac12pi iint_C f_n(z)(z I_p - X)^-1 dz$$
where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $HinmathrmSym(p),$
beginalign*
f_n^*(X+H)
&= frac12pi iint_C f_n(z)(z I_p - X-H)^-1 dz\
&= frac12pi iint_C f_n(z)(z I_p - X)^-1+f_n(z)(z I_p - X)^-1H(z I_p - X)^-1 +dots dz\
&= frac12pi iint_C f_n(z)sum_lambda(z-lambda)^-1P_lambda +f_n(z)sum_lambda_1,lambda_2(z-lambda_1)^-1(z-lambda_2)^-1P_lambda_1HP_lambda_2+dots dz\
&= f_n^*(X)+sum_lambda_1,lambda_2 P_lambda_1 H P_lambda_2int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
endalign*
The second equality uses the Taylor expansion $$(A-H)^-1=A^-1+A^-1HA^-1+dots$$ 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 $int_C f_n(z)(z-lambda)^-1(z-mu)^-1dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$
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 $C^1.$
$endgroup$
1
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
add a comment |
$begingroup$
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:
$$f_n^*(X) = frac12pi iint_C f_n(z)(z I_p - X)^-1 dz$$
where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $HinmathrmSym(p),$
beginalign*
f_n^*(X+H)
&= frac12pi iint_C f_n(z)(z I_p - X-H)^-1 dz\
&= frac12pi iint_C f_n(z)(z I_p - X)^-1+f_n(z)(z I_p - X)^-1H(z I_p - X)^-1 +dots dz\
&= frac12pi iint_C f_n(z)sum_lambda(z-lambda)^-1P_lambda +f_n(z)sum_lambda_1,lambda_2(z-lambda_1)^-1(z-lambda_2)^-1P_lambda_1HP_lambda_2+dots dz\
&= f_n^*(X)+sum_lambda_1,lambda_2 P_lambda_1 H P_lambda_2int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
endalign*
The second equality uses the Taylor expansion $$(A-H)^-1=A^-1+A^-1HA^-1+dots$$ 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 $int_C f_n(z)(z-lambda)^-1(z-mu)^-1dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$
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 $C^1.$
$endgroup$
1
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
add a comment |
$begingroup$
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:
$$f_n^*(X) = frac12pi iint_C f_n(z)(z I_p - X)^-1 dz$$
where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $HinmathrmSym(p),$
beginalign*
f_n^*(X+H)
&= frac12pi iint_C f_n(z)(z I_p - X-H)^-1 dz\
&= frac12pi iint_C f_n(z)(z I_p - X)^-1+f_n(z)(z I_p - X)^-1H(z I_p - X)^-1 +dots dz\
&= frac12pi iint_C f_n(z)sum_lambda(z-lambda)^-1P_lambda +f_n(z)sum_lambda_1,lambda_2(z-lambda_1)^-1(z-lambda_2)^-1P_lambda_1HP_lambda_2+dots dz\
&= f_n^*(X)+sum_lambda_1,lambda_2 P_lambda_1 H P_lambda_2int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
endalign*
The second equality uses the Taylor expansion $$(A-H)^-1=A^-1+A^-1HA^-1+dots$$ 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 $int_C f_n(z)(z-lambda)^-1(z-mu)^-1dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$
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 $C^1.$
$endgroup$
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:
$$f_n^*(X) = frac12pi iint_C f_n(z)(z I_p - X)^-1 dz$$
where $C$ is an anticlockwise circle in the complex plane with $J$ in its interior. For $HinmathrmSym(p),$
beginalign*
f_n^*(X+H)
&= frac12pi iint_C f_n(z)(z I_p - X-H)^-1 dz\
&= frac12pi iint_C f_n(z)(z I_p - X)^-1+f_n(z)(z I_p - X)^-1H(z I_p - X)^-1 +dots dz\
&= frac12pi iint_C f_n(z)sum_lambda(z-lambda)^-1P_lambda +f_n(z)sum_lambda_1,lambda_2(z-lambda_1)^-1(z-lambda_2)^-1P_lambda_1HP_lambda_2+dots dz\
&= f_n^*(X)+sum_lambda_1,lambda_2 P_lambda_1 H P_lambda_2int_0^1 f'_n(tlambda_1+(1-t)lambda_2)+dots dt
endalign*
The second equality uses the Taylor expansion $$(A-H)^-1=A^-1+A^-1HA^-1+dots$$ 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 $int_C f_n(z)(z-lambda)^-1(z-mu)^-1dz =int_0^1 f'_n(tlambda+(1-t)mu)dt.$
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 $C^1.$
answered Apr 6 at 17:25
DapDap
94826
94826
1
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
add a comment |
1
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days 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
Apr 6 at 18:15
$begingroup$
Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-)
$endgroup$
– Mizar
Apr 6 at 18:15
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
Higher derivative estimates follow from the formula $sum_j=0^kfracf(lambda_j)prod_ellneq j(lambda_j-lambda_ell)=frac1int_Delta_kf^(k)(sum_jt_jlambda_j),dt_0cdots dt_k$, $Delta_k$ being the standard simplex $t_jge 0,sum t_j=1$ (assuming wlog the $lambda_j$'s are distinct).
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
$begingroup$
The formula, in turn, is easy to prove by induction: we can subtract $f(lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=sum_j=1^kint_0^1fracf'(t_0lambda_0+(1-t_0)lambda_j)prod_ellneq j,ell>0(lambda_j-lambda_ell)$. We are done applying induction with $g:=f'(t_0lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^(k-1)(z)=(1-t_0)^k-1f^(k)(t_0lambda_0+(1-t_0)z)$.
$endgroup$
– Mizar
2 days ago
add a comment |
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
Apr 6 at 23:00
add a comment |
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
Apr 6 at 23:00
add a comment |
$begingroup$
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered Apr 6 at 20:36
B ChinB Chin
1
1
New contributor
New contributor
$begingroup$
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
Apr 6 at 23:00
add a comment |
$begingroup$
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
Apr 6 at 23:00
$begingroup$
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
Apr 6 at 23:00
$begingroup$
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
Apr 6 at 23:00
add a comment |
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-fa.functional-analysis, real-analysis, sp.spectral-theory