Orthogonally invariant functions of symmetric matrices

Definition of $g$. For $\lambda=(\lambda_1\ge\dots\ge\lambda_n)$ put
$$g(\lambda)=f\bigl(\operatorname{diag}(\lambda_1,\dots,\lambda_n)\bigr).$$

Continuity. $g$ is the composition of the continuous map $\lambda\mapsto\operatorname{diag}(\lambda)$ with $f$, so it is continuous.

$f=g\circ\lambda$. Let $S\in\mathrm{Sym}_n$. By the spectral theorem there is $Q\in O(n)$ with $S=Q\operatorname{diag}(\lambda(S))Q^T$, where we order the eigenvalues decreasingly. By invariance,
$$f(S)=f\bigl(\operatorname{diag}(\lambda(S))\bigr)=g(\lambda(S)).$$

Remark (converse and continuity of the spectrum). Conversely, every continuous $g$ gives a continuous invariant $f=g\circ\lambda$, because the ordered eigenvalues depend continuously on $S$: by Weyl's inequality $|\lambda_i(S)-\lambda_i(S')|\le\|S-S'\|$ (operator norm). Examples: $\operatorname{tr}S=\sum\lambda_i$, $\det S=\prod\lambda_i$, $\|S\|_F^2=\sum\lambda_i^2$ and $\lambda_1(S)$ are all invariant functions. Lesson. An invariant function is determined by its restriction to a "slice" (here the diagonal matrices).