bounded. Indeed, it follows from (\[pr1\]) that $$F(x(t))-F(y)+\ep(t)(x(t)-y)=\ep(t)(z_0-y).$$ Therefore $$\label{3.6}
(F(x(t))-F(y), x(t)-y)+\ep(t)||x(t)-y||^2=\ep(t)(z_0-y,x(t)-y).$$ This and (\[frder\]) imply $$\label{toobad}
||x(t)-y||\le||z_0-y||.$$ Thus there exists a sequence $\{x(t_n)\}$, $t_n\to\infty$ as $n\to\infty$, which converges weakly to some element $\y\in H$. Let us show that $\y$ is the (unique) solution to problem (\[noneq\]). Since $F$ is [*w*]{}-continuous, $F(x(t_n))\rightharpoonup F(\y)$. Because of the weak lower semicontinuity of the norm in a Hilbert space one has: $$\label{wsol}
||F(\y)||\le\liminf_{n\to\infty}||F(x(t_n))||=\liminf_{n\to\infty}\ep(t_n)||x(t_n)-z_0||
=0.$$ The conclusion $F(\y)=0$ follows from (\[wsol\]) and can also be derived directly from (\[pr1\]) with $\ep=\ep(t)\to 0$ as $t\to +\infty$. By the uniqueness of the solution to equation (\[noneq\]) one concludes that $\y=y$. Let us show that the sequence $\{x(t_n)\}$ converges strongly to $y$. Indeed, from (\[3.6\]), (\[frder\]) and the relation $x(t_n)\rightharpoonup y$, one gets: $$\label{ineq1}
||x(t_n)-y||^2\le (z_0-y,x(t_n)-y)\to 0\t{ as }n\to\infty.$$ Thus $$\label{lim1}   
\lim_{n\to\infty}||x(t_n)-y||=0.$$ From (\[lim1\]) it follows by the standard argument that $x(t)\to y$ as $t\to\infty$. Lemma \[l2\] is proved. $\Box$

Assume that $F$ is continuously Fréchet differentiable, $\sup_{x\in H}||F'(x)||\le N_1$, and condition (\[frder\]) holds. If $\ep(t)$ is continuously differentiable, then the solution $x(t)$ to problem (\[pr1\]) with $\ep=\ep(t)$ is continuously differentiable in the strong sense and one has $$\label{est8}
||\dot{x}(t)||\le\f{|\dot{\ep}(t)|}{\ep(t)}||y-z_0||,\q t\in [0,+\infty).$$ \[l3\]

[**Proof**]{} Fréchet differentiability of 