map $t\to (\p(x_0+th_1),h_2)$ is continuous in a neighborhood of $t=0$ for any $h_1,h_1\in H$.

A mapping $\p$ is strongly monotone in a Hilbert space $H$ if there exists a constant $k>0$ such that $$(\p(x_1)-\p(x_2),x_1-x_2)\ge k||x_1-x_2||^2,\q \forall x_1,x_2\in H.$$

If $F$ is monotone and hemicontinuous then the problem (\[pr1\]) is uniquely solvable. \[l1\]

[**Proof**]{} [**of Lemma \[l1\]**]{} According to [@deim p. 100] problem (\[pr1\]) is solvable if the operator $F_\ep$ is monotone, hemicontinuous and $||F_\ep(x)||\to\infty$ as $||x||\to\infty$. For sufficiently small $\delta>0$ from (\[pr1\]) one has: $$||F_\ep(x)||^2\ge
||F(x)||^2+\ep^2||x-z_0||^2-\f{1}{4\delta}||F(z_0)||^2-\delta||x-z_0||^2$$ $$\ge(\ep^2-\delta)||x-z_0||^2-C\to +\infty$$ as $||x||\to+\infty$. Here $C$ is a constant. Since $F_\ep$ is strongly monotone the solution to (\[pr1\]) is unique. Therefore Lemma  \[l1\] is proved. $\Box$

The result given by Lemma \[l1\] is well known and its proof is given for the convenience of the reader.

Let $\rightharpoonup$ denote weak convergence in $H$.

We say that $F$ is [*w*]{}-continuous if $x\rightharpoonup \xi$ implies $F(x)
\rightharpoonup F(\xi)$.

Suppose that $F$ is w-continuous, all the assumptions of Lemma  \[l1\] are satisfied, and there exists a unique solution $y$ to (\[noneq\]). Let $x(t)$ solve (\[pr1\]) for $\ep=\ep(t)$, and $\ep(t)$ tend to zero as $t\to +\infty$. Then $$\label{znorms}
\lim_{t\to+\infty}||x(t)-y||=0.$$ \[l2\]

[**Proof**]{} First let us show that $x(t)$ is 