Let $A$ be a self-adjoint operator defined on a Hilbert space $\mathcal{H}$. By the Spectral Mapping Theorem, we know that if $g \in C(\sigma(A))$ then $ \sigma(g(A)) = g(\sigma(A))$. Since $\lambda \in \sigma(A)$ iff there is a sequence $(x_k)_k \subseteq \mathcal{H}$ such that $\|x_k\|=1$ for all $n \in \mathbb{N}$ and $\|(A-\lambda)x_k\| \to 0$, I wonder if this implies that $\| (g(A)-g(\lambda))x_k\| \to 0$. I know that this is true if the sequence $x_k$ is constant, that is if $\lambda$ is an eigenvalue of $A$.
Could you help me, please?
Let $p_n$ be a sequence of polynomials convergent uniformly to $g$ on $\sigma(A).$ Then $$\|p_n(A)-g(A)\|=\max_{t\in \sigma(A)}|p_n(t)-g(t)|\underset{n\to \infty}{\longrightarrow}0$$ For $\lambda\in \sigma(A)$ and $\|x_k\|=1$ we get $$\|[g(A)-g(\lambda)I]x_k\|\le \\ \|[g(A)-p_n(A)]x_k\|+\|[p_n(A)-p_n(\lambda )I]x_n\|+|p_n(\lambda)-g(\lambda)|\\ \le 2\max_{t\in \sigma(A)}|p_n(t)-g(t)|+\|[p_n(A)-p_n(\lambda )I]x_k\|$$ For $\delta>0$ there is $n_0$ such that $$2\max_{t\in \sigma(A)}|p_{n_0}(t)-g(t)|<{\delta\over 2}$$ We have $$p_{n_0}(A)-p_{n_0}(\lambda)I=q_{n_0}(A)[A-\lambda I]$$ for a polynomial $q_{n_0}.$ Thus $$\|q_{n_0}(A)[A-\lambda I]x_k\|\le \|q_{n_0}(A)\|\,\|(A-\lambda I)x_k\|$$ There is $k_0$ such that for $k\ge k_0$ we have $$\|q_{n_0}(A)\|\,\|(A-\lambda I)x_k\|<{\delta\over 2}$$ Summarizing, for $k\ge k_0$ we obtain $$\|[g(A)-g(\lambda)I]x_k\|<\delta$$ which shows the claim in OP.