$F$ implies hemicontinuity of $F$. Therefore problem (\[pr1\]) with $\ep=\ep(t)$ is uniquely solvable. The differentiability of $x(t)$ with respect to $t$ follows from the implicit function theorem \[2\]. To derive (\[est8\]) one differentiates equation (\[pr1\]) and uses the estimate $\left\Vert\left[F'(x(t))+\ep(t)I\right]^{-1}\right\Vert\le\f{1}{\ep(t)}$. The result is: $$\label{deriv}
||\dot{x}(t)||=|\dot{\ep}(t)|\cdot ||[F'(x(t))+\ep(t)I]^{-1}(x(t)-z_0)||\le 
\f{|\dot\ep(t)|}{\ep(t)}||x(t)-z_0||.$$ Here we have used the estimate $$\label{3.12}
||x(t)-z_0||\leq ||y-z_0||,$$ which can be derived from (\[pr1\]) similarly to the derivation of (\[toobad\]). Thus estimate (\[est8\]) follows from (\[deriv\]) and (\[3.12\]). $\Box$

Assume that $\ep=\ep(t)>0$, $F$ is twice Fréchet differentiable, condition (\[frder\]) holds, and $$\label{N}
||F'(x)||\le N_1, \q ||F''(x)||\le N_2\q  \forall x\in H.$$ Then for the operator $\P$ defined by (\[rchn\]) and $x(t)$, the solution to (\[pr1\]) with $\ep=\ep(t)$, estimate (\[semon\]) holds with $$\label{sigma}
\a(t)\equiv 0,\q \g(t)\equiv 1,\t{ and } \s(t):=\f{N_2}{2\ep(t)}.$$ \[l4\]

[**Proof**]{} Since $x(t)$ is the solution to (\[pr1\]) applying Taylor’s formula one gets: $$(\P(h,t),h-x(t))=-([F'(h)+\ep(t)I]^{-1}[F(h)-F(x(t))+\ep(t)(h-x(t))],h-x(t))$$ $$\leq - \Biggl([F'(h)+\ep(t)I]^{-1}[F'(h)(h-x(t))+\ep(t)(h-x(t))],h-x(t)\Biggl)+
\f{N_2||h-x(t)||^3}{2\ep(t)}$$ $$\label{comp}
= -||h-x(t)||^2+\f{N_2||h-x(t)||^3}{2\ep(t)}.$$ From (\[comp\]) and (\[semon\]) the conclusion of Lemma \[l4\] follows. $\Box$

Let us state the main result of this section.

Assume:

1.  problem (\[noneq\]) has a unique solution $y$;

2.  $F$ is w-continuous, twice Fréchet differentiable and inequalities (\[frder\]), (\[N\]) hold;

3.  $\ep(t)>0$ is continuously differentiable and monotonically tends to $0$, $$\label{rhoc}
    