Continuity of maximum eigenvalue function Let $f:S^n_{+}(\mathbb{R})\rightarrow \mathbb{R}_+$ be defined as $$f(A)=\lambda_{\max}(A),$$ where $S^n_{+}(\mathbb{R})$ is set of all positive definite $n\times n$ matrices.
What can we say about the continuity and differentiability of $f$?
Intuitively, I feel that it is continuous because if we perturb a matrix, its eigenvalues do not change abruptly. But I don't know how to approach it.
 A: It is both continuous and differentiable (under some conditions).
Continuity : Since $f(A)=\max\{\lambda_1(A),\ldots,\lambda_n(A)\}$, it is enough to prove that each $\lambda_i(A)$ is continuous. In fact, since the $\lambda_i$'s are the zeros of $\chi_A$ and that $A\mapsto\chi_A$ is continuous, it suffices to show that if $\alpha_1(P)\leqslant\ldots\leqslant\alpha_n(P)$ are the roots of a monic polynomial $P\in\mathbb{R}[X]$ of degree $n$ then each $\alpha_i$ is continuous. Suppose $P,Q$ are such polynomials and let $\varepsilon>0$. Take $\delta>0$ such that $|Q(\alpha_i(P))|\leqslant\frac{\varepsilon^n}{2}$ as soon as $\|P-Q\|_{\infty}\leqslant\delta$ (which is possible by continuity of $Q\mapsto Q(\alpha_i(P))$), then
$$ \prod_{j=1}^n|\alpha_i(P)-\alpha_j(Q)|=|Q(\alpha_i(P))|\leqslant\frac{\varepsilon^n}{2} $$
therefore there is at least one $j\in[\![1,n]\!]$ such that $|\alpha_i(P)-\alpha_j(Q)|\leqslant\varepsilon$. If you take $\varepsilon<\frac{1}{2}\min_{\alpha_i(P)\neq\alpha_j(P)}|\alpha_i(P)-\alpha_j(P)|$ then in fact $\alpha_j(Q)=\alpha_i(Q)$ so that $|\alpha_i(P)-\alpha_i(Q)|\leqslant\varepsilon$.
Differentiability : Suppose that there exists a unique $i$ such that $\lambda_i(A)=\lambda_{\max}(A)$, then $f$ is differentiable. See : Derivative of spectral norm of symmetric matrix
