that, for any given $x_0\in C$, the Krasnoselskij iteration $\{x_n\}_{n=0}^{\infty}$ given by $$\label{eq4}
x_{n+1}=(1-\lambda) x_n+\lambda Tx_n,\,n\geq 0,$$ converges strongly to a fixed point of $T$.

Since $T$ is enriched nonexpansive, by Definition \[def0\] it follows that there exists a constant $b$, $b\in[0,ü\infty)$, such that $$\|b(x-y)+Tx-Ty\|\leq (b+1) \|x-y\|,\forall x,y \in C.$$ By putting $b=\dfrac{1}{\mu}-1$, it follows that $\mu\in (0,1]$ and the previous inequality is equivalent to $$\label{eq5}
\|(1-\mu)(x-y)+\mu Tx-\mu Ty\|\leq  \|x-y\|,\forall x,y \in C.$$ Denote $T_\mu x=(1-\mu) x+\mu T x$. Then inequality expresses the fact that $$\|T_\mu x-T_\mu y\|\leq  \|x-y\|,\forall x,y \in C,$$ i.e., that the averaged operator $T_\mu$ is nonexpansive.

By means of Browder-Goede-Kirk fixed point theorem (e.g., Theorem 4 in [@BroP67]), reproduced as Theorem 3.1 in [@Ber07], it follows that $T_\mu$ has at least one fixed point.

Note also that, in view of Lemma \[lem1\], $Fix\,(T)=Fix\,(T_\mu)\neq \emptyset$.

Since $H$ is a Hilbert space, $Fix\,(T)$ is also convex, see for example Theorem 6 in [@BroP67]. We include here a proof of the fact that $Fix\,(T_\mu)$ is convex, i.e., when $x,y\in Fix\,(T_\mu)$ and $\lambda\in[0,1]$, we have $$u_\lambda=(1-\lambda)x+\lambda y\in Fix\,(T_\mu).$$ Indeed, since $T_\mu$ is nonexpansive, we have $$\label{eq6}
\|T_\mu u_\lambda- x\|=\|T_\mu u_\lambda-T_\mu x\|\leq \|u_\lambda- x\|$$ and, similarly, $$\label{eq7a}
\|T_\mu u_\lambda- y\|\leq \|u_\lambda- 