(\[lap\]) follows from (\[znorms\]), inequality (\[lap\]) and the triangle inequality: $$||z(t)-y||\le ||z(t)-x(t)||+||x(t)-y||.$$ Theorem \[t3\] is proved. $\Box$

[**Examples.**]{}

[**1.**]{} Let $\ep(t)=\ep_0(t_0+t)^{-\nu}$, $\ep_0$, $t_0$ and $\nu$ are positive constants. Then $C_\ep=\f{\nu}{t_0}$ and condition (\[rhoc\]) is satisfied if $\nu\in (0,1]$ and $t_0>\nu$.

[**2.**]{} If $\ep(t)=\frac {\ep_0}{\log(t_0+t)}$, then $C_\ep=\f{1}{t_0\log t_0}$ and condition (\[rhoc\]) is satisfied if $t_0\log t_0>1$.

Note that if $\ep(t)=\ep_0e^{-\nu t}$ then condition (\[rhoc\]) is not satisfied.

[*Let the assumptions of Theorem \[t3\] hold. Suppose also that the following inequality holds: $$\label{pr11}
(F(h),h-y)\ge c||h-y||^{1+a},\q a>0.$$ Then for the solution $z(t)$ to problem (\[caupr\]) the following estimate holds: $$\label{pr12}  
||z(t)-y||=O\left(\ep^{\f{1}{a}}(t)\right).$$* ]{} \[prop1\]

[**Proof**]{} Denote $||x(t)-y||:=\varrho(t)$. Since $F(y)=0$, inequality (\[3.6\]) implies $$\label{pr13}
c\varrho^{1+a}(t)+\ep(t)\varrho^2(t)\le \ep(t)||z_0-y||\varrho(t)$$ and $\varrho(t) \to 0$ as $t\to +\infty$. This inequality can be reduced to $$\label{pr14}
c\varrho^{a}(t)+\ep(t)\varrho(t)\le \ep(t)||z_0-y||.$$ Thus $\varrho^{a}(t)\le \f{||z_0-y||}{c} \ep(t)$, and $$\label{pr15}
 ||x(t)-y||\le \left( \f{||z_0-y||}{c}\right)^{\f{1}{a}}\ep^{\f{1}{a}}(t).$$ Combining this estimate with estimate (\[lap\]) for $||z(t)-x(t)||$ one completes the proof. $\Box$

[**Example.**]{} In the case of a scalar function $f(h)$ and even integer $a>0$ the estimate $f(h)(h-y)\ge c|h-y|^{1+a}$ means that $f(h)=(h-y)^{a}g(h)$, where $g(h)\ge c>0$, and hence $y$ is a zero of the multiplicity $a$ for $f$.

[*Let all the assumptions of Theorem \[t3\] hold and there exists $v\in H$ such that $$\label{pr21}
