Why is the largest eigenvalue Lipschitz continuous and not differentiable? Let
$$
A:\mathbb R^n\to \mathbb R^{n\times n}
$$
where $A(x)$ is symmetric for any $x=(x_1,..,x_n)$. $$A(x) = A_0+x_1A_1+x_2A_2+...x_nA_n$$ and all $A$ is positive semidefinite.
Consider
$$
\underset{x\in\mathbb R^n}{\rm minimize}~ \lambda_{\rm max}(A(x))
$$
I would like to show that $\lambda_{\rm max}(A(x))$ is Lipschitz continuous but not differentiable.
Attempt:
It is Lipschitz since $\lambda_{\rm max}(A(x))$ is equal to $\|A(x)\|_2$ (is this true?).
Does the non-differentiability also follow from this fact?
 A: We'll show that the $A\mapsto ( \lambda_1, \ldots, \lambda_n)$ from hermitian $n\times n$ matrices to the eigenvalues in decreasing order is Lipschitz. ( One can choose specific norms on both sides but that won't affect the statement).
First, let us show that if $A \preceq B$ ( the semidefinite order, meaning $B- A \succeq 0$ ) then the $k$th eigenvalue of $A$ is $\le $ the $k$-th eigenvalue of $B$, that is $\lambda_k(A) \le \lambda_k(B)$, for all $1 \le k \le n$. Indeed, there exists an $n-k+1$-dimensional subspace $V_1$ of $\mathbb{C}^n$ such that for every vector $x$ in that subspace we have $\langle A x, x \rangle \ge \lambda_k(A) \cdot ||x||^2$, and there exists a $k$-dimensional subspace $V_2$ of $\mathbb{C}^n$ such that for every vector $x$ in that subspace we have $\langle B x, x \rangle \le \lambda_k(B) \cdot ||x||^2$. These two subspaces have a non-zero intersection, let $0 \ne x \in V_1 \cap V_2$. We have 
$$\lambda_k(A) \cdot ||x||^2 \le \langle A x, x \rangle  \le \langle B x, x \rangle \le \lambda_k(B) \cdot ||x||^2$$ and so $\lambda_k(A) \le \lambda_k(B)$.
Now, recall an important ( not hard to prove ) result : if $C$ is an $n\times n$ hermitian matrix so that $c_{ii} > (\ge) \sum_{j, j\ne i} | c_{ij}|$ for all $1 \le i \le n $ ( diagonally dominant) then $C \succ (\succeq) 0$. 
Consider the norm $C \mapsto |C| = \max ( |c_{ij}|)$ on hermitian matrices. From the above we conclude that if $|C| \le \frac{\epsilon}{n}$ then 
$\epsilon \cdot I \pm C \succeq 0$.  As a consequence, if $|A-B| \le \frac{\epsilon}{n}$ then $\epsilon \cdot I \pm (A-B) \succeq 0$. From $\epsilon \cdot I + A \succeq B$ we get, using the preceding, $\epsilon + \lambda_k(A) \ge \lambda_k(B)$. Similarly $\epsilon + \lambda_k(B) \ge \lambda_k(A)$. Therefore
$$|\lambda_k(A) - \lambda_k(B)| \le n \cdot |A-B|$$
for all $1 \le k \le n$.
${\bf Added:}$ The constant $n$ is the best possible. Indeed, consider the $n\times n$ matrix  $A$ with all entries $1$. $A$ has $n$ as eigenvalue  ( eigenvector all entries $1$) and $|A| = |A-0| = 1$. 
As for the non-dfferentiability : the matrix$t \left(\matrix{t&0\\0& -t}\right)$ has largest eigenvalue $|t|$.
A: I was looking for a reference that I could cite. There is a theorem of Hoffman and Wielandt that says (the norm is the Frobenius norm):

Theorem 1. If $A$ and $B$ are normal matrices with eigenvalues $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ respectively, then there exists a suitable numbering of the eigenvalues such that $\sum_i \lvert \alpha_i - \beta_i \rvert^2 \leq \lVert A - B \rVert_F^2$.

For a Hermitian matrix $A$, the “suitable numbering” is such that the eigenvalues are ordered monotonously:

Remark 2. Although the arrangement of the eigenvalues mentioned in Theorem 1 is difficult, in general, to describe more explicitly, it is easy in the special case that $A$ is Hermitian. Then a “best” arrangement is $\alpha_1 \geq \ldots \geq \alpha_n$ and $\operatorname{Re}\beta_1 \geq \ldots \geq \operatorname{Re}\beta_n$.

See also The Wielandt-Hoffman theorem in Wilkinson (paragraph 48, page 104). There the version for real symmetric matrices is given.
Let $\lambda_k(\cdot)$ map a Hermitian matrix to its $k$-th largest eigenvalue and let $A$, $B$ be Hermitian. We obtain from the above, for every $1 \leq k \leq n$,
$$
    \lvert \lambda_k(A) - \lambda_k(B) \rvert^2 \leq\sum_i \lvert \lambda_i(A) - \lambda_i(B) \rvert^2 \leq  \lVert A - B \rVert_F^2,
$$
from which Lipschitz continuity with Lipschitz constant 1 follows. Note that this does not contradict the “Added…” remark by orangeskid, because different norms are used in the two answers.
Reference


*

*Hoffman, A. J.; Wielandt, H. W.: The variation of the spectrum of a normal matrix. Duke Math. J. 20 (1953), no. 1, 37–39. doi:10.1215/S0012-7094-53-02004-3. https://projecteuclid.org/euclid.dmj/1077465062

*Wilkinson, J. H.: The Algebraic Eigenvalue Problem (1988)

