Is the function mapping a self-adjoint operator on a Hilbert space to the maximum of its spectrum differentiable?

Question

Let $H$ be a $\mathbb R$-Hilbert space. Is $$\left\{A\in\mathfrak L(H):A\text{ is self-adjoint}\right\}\to\mathbb R\;,\;\;\;A\mapsto\max\sigma(A)\tag1$$ differentiable? We know that this is true in the finite-dimensional case $H=\mathbb R^d$, $d\in\mathbb N$. If the claim as stated is not true in general, is there at least a similar result?

Answer 1

Define $A(t) = \pmatrix{ t & 0 \\ 0 & -t}$. Then $$ \max \sigma (A(t)) = |t|, $$ which seems not to be differentiable at zero.

Answer 2

The mapping is Lipschitz and from $$\max \sigma(A) = \sup\{(x,Ax) \;|\; \|x\| \le 1\}$$ it should follow that it is convex (as a supremum of linear functions (w.r.t. $A$)). Hence, it should be subdifferentiable and directionally differentiable.