also firmly nonexpansive is nonexpansive.

It is very important to note that, similar to the case of nonexpansive mappings, any enriched nonexpansive mapping is [*continuous*]{}.

\[ex1\]

Let $X=\left[\dfrac{1}{2},2\right]$ be endowed with the usual norm and $T:X\rightarrow X$ be defined by $Tx=\dfrac{1}{x}$, for all $x\in \left[\dfrac{1}{2},2\right]$. Then

\(i) $T$ is Lipschitz continuous with Lipschitz constant $L=4$ and $T$ is not nonexpansive;

\(ii) $T$ is a $3/2$-enriched nonexpansive mapping.

\(iii) $Fix\,(T)=\{1\}$ but $T$ is not quasi-nonexpansive.

\(i) Assume $T$ is nonexpansive. Then $$|Tx-Ty|\leq |x-y|,\forall x,y \in X,$$ which, for $x=1$ and $y=1/2$, leads to the contradiction $1\leq 1/2$.

\(ii) The enriched nonexpansive condition reduces in this case to $$\left|b(x-y)+\frac{1}{x}-\frac{1}{y}\right|\leq (b+1) |x-y| \Leftrightarrow \left|b-\frac{1}{xy}\right|\cdot |x-y|\leq (b+1) \cdot |x-y|.$$ It easy to check that, for any $b\geq 3/2$, we have $$\left|b-\frac{1}{xy}\right|\leq b+1, \forall x,y \in \left[\dfrac{1}{2},2\right],$$ which proves that $T$ is a $3/2$-enriched nonexpansive mapping.

\(iii) Assume $T$ is quasinonexpansive. Then we must have $$\left|\frac{1}{x}-1\right |\leq |x-1|, \forall x \in \left[\dfrac{1}{2},2\right].$$ Just take $x=\dfrac{1}{2}$ to reach the contradiction $1\leq \dfrac{1}{2}$.

We need some definitions and results for proving our main results.

[@Pet] \[def1\] Let $H$ be a Hilbert space and $C$ a subset of $H$. A mapping $T:C\rightarrow H$ is called *demicompact* if it 