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Assume that $f:\mathbb R\to\mathbb R$ is continuous. Given a real symmetric matrix $A\in\text{Sym}(n)$, we can define $f(A)$ by applying $f$ to its spectrum. More explicitly, $$ f(A):=\sum f(\lambda)P_\lambda,\qquad A=\sum\lambda P_\lambda. $$ 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 $f:\text{Sym}(n)\to\text{Sym}(n)$ is continuous.

Assuming $f\in C^\infty(\mathbb R)$, is the induced map $f:\text{Sym}(n)\to\text{Sym}(n)$ 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 $C^\infty$.

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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 $\sup_{x\in J}|f(x)-f_n(x)|\to 0$ and $\sup_{x\in J}|f'(x)-f_n'(x)|\to 0.$ Since $f_n$ is analytic, $f^*_n$ can be evaluated using resolvents:

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

\begin{align*} f_n^*(X+H) &= \frac{1}{2\pi i}\int_C f_n(z)(z I_p - X-H)^{-1} dz\\ &= \frac{1}{2\pi 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}{2\pi 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(t\lambda_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 $\int_C f_n(z)(z-\lambda)^{-1}(z-\mu)^{-1}dz =\int_0^1 f'_n(t\lambda+(1-t)\mu)dt.$

This gives a bound

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

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 $C^1.$

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    $\begingroup$ Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-) $\endgroup$
    – Mizar
    Commented Apr 6, 2019 at 18:15
  • $\begingroup$ Higher derivative estimates follow from the formula $\sum_{j=0}^k\frac{f(\lambda_j)}{\prod_{\ell\neq j}(\lambda_j-\lambda_\ell)}=\frac{1}{k!|\Delta_k|}\int_{\Delta_k}f^{(k)}(\sum_jt_j\lambda_j)\,dt_0\cdots dt_k$, $\Delta_k$ being the standard simplex $\{t_j\ge 0,\sum t_j=1\}$ (assuming wlog the $\lambda_j$'s are distinct). $\endgroup$
    – Mizar
    Commented Apr 7, 2019 at 15:21
  • $\begingroup$ 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 $LHS=\sum_{j=1}^k\int_0^1\frac{f'(t_0\lambda_0+(1-t_0)\lambda_j)}{\prod_{\ell\neq j,\ell>0}(\lambda_j-\lambda_\ell)}$. We are done applying induction with $g:=f'(t_0\lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^{(k-1)}(z)=(1-t_0)^{k-1}f^{(k)}(t_0\lambda_0+(1-t_0)z)$. $\endgroup$
    – Mizar
    Commented Apr 7, 2019 at 15:25
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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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  • $\begingroup$ 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
    Commented Apr 6, 2019 at 23:00

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