Let $f : \mathbb{R} \to \mathbb{R}$ be a Schwartz function. Let $I$ be the identity operator on some separable Hilbert space $\mathcal{H}$ and $A$ be an unbounded self-adjoint operator on $\mathcal{H}$.
Then, $I+xA$ is self-adjoint for all $x \in \mathbb{R}$ and $f(I+xA)$ is a bounded operator on $\mathcal{H}$. However, is there any continuity or smoothness available for $f(I+xA)$ with respect to $x$?
For example, as $x \to 0$, does $f(I+xA)$ converge to $f(I)$ in any sense?
Yes, $f(I+xA)$ converges to $f(I)$ in the strong operator topology, meaning that $f(I+xA)\xi \to f(I)\xi $, for every vector $\xi $ in $\mathcal H$.
The fact that $f$ is a Schwartz function is not necessary and here is a proof that works for every bounded continuous function $f$. Write $$ A=\int_{\mathbb R} \lambda \,dP(\lambda ) $$ by the spectral Theorem, so that $$ f(I+xA) = \int_{\mathbb R} f(1+x\lambda ) \,dP(\lambda ). $$ Next observe that $f(1+x\lambda ) \to f(1)$, as $x\to 0$, pointwise on the variable $\lambda $. Writing $$ g_x(\lambda ) = f(1+x\lambda ) - f(1) $$ we then have that $g_x(\lambda )\to 0$, as $x\to 0$, pointwise on $\lambda $. Therefore, for every $\xi $ in $\mathcal H$, one has that $$ \|g_x(A)\xi \|^2 = \langle g_x(A)\xi , g_x(A)\xi \rangle = \langle g_x(A)^2\xi , \xi \rangle = \int_{\mathbb R} g_x(\lambda )^2 \,d\langle P(\lambda )\xi , \xi \rangle , $$ which converges to zero by the Lebesgue Dominated Convergence Theorem, and the fact that $g_x(\lambda )^2$ is uniformly bounded.