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.



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


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



Smoothness of finite-dimensional functional calculus
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 $C^1.$


