# Smoothness of finite-dimensional functional calculus

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 $$C^\infty$$.

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 $$C^1.$$

• Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-) – Mizar Apr 6 '19 at 18:15
• 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). – Mizar Apr 7 '19 at 15:21
• 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 $g^{(k-1)}(z)=(1-t_0)^{k-1}f^{(k)}(t_0\lambda_0+(1-t_0)z)$. – Mizar Apr 7 '19 at 15:25

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.

• 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) – Mizar Apr 6 '19 at 23:00