mapping $T:C\rightarrow C$ is called [*generalized pseudocontractive*]{} with constant $r$, see [@Ver], [@Ber02] and [@Ber07], if $$\label{vv}
\|Tx-Ty\|^2\leq r \|x-y\|^2+\|Tx-Ty-r(x-y)\|^2,\,x,y\in C.$$ It easy to prove that, in a Hilbert space, condition is equivalent to $$\label{vu}
\langle Tx-Ty, x-y\rangle\leq r \|x-y\|^2,\,x,y\in C.$$

Note also that the function $T$ in Example \[ex1\] is generalized pseudocontractive with constant $r>0$ arbitrary and Lipschitzian with constant $L=4$.

Theorem \[th1\] is also an extension of the main result in [@Ver], see also [@Ber02] and [@Ber07].

([@Ver]) \[cor2\] Let $C$ be a nonempty closed convex subset of a Hilbert space $H$ and $T:C\rightarrow C$ be a Lipschitzian and generalized pseudocontractive operator (with constants $s$ and $r$, respectively, $r<1$). Then for any given $x_0\in C$ and any fixed number $\lambda$, $0<\lambda<\frac{2(1-r)}{1-2 r+s^2}$, the Krasnoselskij iteration $\{x_n\}_{n=0}^{\infty}$ given by $$\label{w}
x_{n+1}=(1-\lambda) x_n+\lambda Tx_n,\,n\geq 0,$$ converges strongly to the unique fixed point of $T$.

Note that, in a Hilbert space, the $b$-enriched nonexpansive condition is equivalent to $$\label{uv}
2b\cdot \langle Tx-Ty, x-y\rangle+1\cdot \|Tx-Ty\|^2\leq (2b+1)\cdot \|x-y\|^2.$$ Since $T$ is a Lipschitzian and generalized pseudocontractive operator we have $$2b\cdot \langle Tx-Ty, x-y\rangle+1\cdot \|Tx-Ty\|^2\leq (2rb+s^2) \|x-y\|^2$$ and so, in order to have satisfied, it suffices to take $b\geq \dfrac{s^2-1}{2(1-r)}$, which is always possible.

