$$\lim_{\stackrel{ z\longrightarrow0}{{
\sphericalangle}}}f'(\alpha\otimes1_n+zv)=f'(\alpha\otimes1_n)=
f'(\alpha)\otimes{\rm Id}_n$$ is a positive map on $M_n(\mathcal A)$.

\[Main\] Let $\mathcal A$ be a von Neumann algebra and $f\colon
H^+(\mathcal A)\to H^+(\mathcal A)$ be a noncommutative analytic map. Fix $\alpha=\alpha^*\in\mathcal A$.

1.  Assume that for any $v\in \mathcal A,
    v>0$ and any state $\varphi$ on $\mathcal A$, $$\label{5}
    \liminf_{z\to0,z\in\mathbb C^+}\frac{\varphi(\Im f(\alpha
    +zv))}{\Im z}<\infty.$$ Then there exists $c=c(v)\in\mathcal A$, $c>0$ such that $$\label{6}
    \lim_{y\downarrow0}\frac{\Im f(\alpha+iyv)}{y}=c$$ in the strong operator [(so)]{} topology. Moreover, $\displaystyle\lim_{\stackrel{z\longrightarrow0}{{\sphericalangle}}}
    f(\alpha+zv)=f(\alpha)$ exists, does not depend on $v$ and is selfadjoint. The limits $$\label{7}
    \lim_{\stackrel{ z\longrightarrow0}{{
    \sphericalangle}}}\Delta f(\alpha+zv_1,\alpha+zv_2)\quad\text{and}\quad
    \lim_{\stackrel{z\longrightarrow0}{{\sphericalangle}}}f'(\alpha+zv)$$ exist pointwise [wo]{} for any $v,v_1,v_2>0$, and $\displaystyle\lim_{\stackrel{ z\longrightarrow0}{{
    \sphericalangle}}}f'(\alpha+zv)(v)=c(v)$. All statements remain true for any $n\in\mathbb N$, $v,v_1,v_2>0$ in $M_n(\mathcal A)$ and $\alpha$ replaced by $\alpha\otimes 1_n$.

2.  Assume in addition to the hypothesis that for any $v,w>0$ in $\mathcal A$ and any state $\varphi$ on $\mathcal A$, the gradient of the two-variable complex function $\{(z,\zeta)\in\mathbb C^2\colon\Im(zv+\zeta w)>0\}\ni(z,\zeta)\mapsto\varphi(f(\alpha+zv+\zeta w))
    \in\mathbb C^+$ admits the limit $$\lim_{\stackrel{y_1,y_2\downarrow0}{(y_1,y_2)\in[0,1)^2\setminus\{(0,0)\}}}
    \left(\varphi(f'(\alpha+iy_1v+iy_2w)(v)),\varphi(f'(\alpha+iy_1v+iy_2w)(w))\right).$$ Then the limits are equal to each other, completely positive and do not depend on $v,v_1,v_2$.

3.  Assume that the pointwise [wo]{} limit $\displaystyle\lim_{y\downarrow0}f'(\alpha+iyv):=f'(\alpha)$ exists for any $v>0$, does not depend on $v$ and $f'(\alpha)$ is a completely bounded 