of Lemma \[lem1\], $Fix\,(T_\mu) =Fix\,(T)=\{p\}$.

To prove the theorem, it suffices to show that if $\{x_{n_j}\}_{j=0}^{\infty}$ given by $$x_{n_j+1}=(1-\lambda) x_{n_j}+\lambda T_\mu x_{n_j},\,j\geq 0,$$ converges weakly to a certain $p_0$, then $p_0$ is a fixed point of $T_\mu$ (and of $U_\lambda=(1-\lambda) I+\lambda T_\mu$) and hence of $T$ and therefore $p_0=p$.

Suppose that $\{x_{n_{j}}\}_{j=0}^{\infty}$ does not converge weakly to $p$.

Using the same arguments like in the proof of Theorem \[th1\], we obtain that $U_\lambda$ is nonexpansive and asymptotically regular, that is, $$\|x_{n_j}-T_\mu x_{n_j}\|\rightarrow 0,\, \textnormal{ as } j\rightarrow \infty.$$ On the other hand $$\|x_{n_j}-U_\lambda p_0\|\leq \|U_\lambda x_{n_j}-U_\lambda p_0\|+\|x_{n_j}-U_\lambda x_{n_j}\|$$ $$\leq \|x_{n_j}-p_0\|+\|x_{n_j}-U_\lambda x_{n_j}\|,$$ which implies that $$\label{eq7}
\limsup \left(\|x_{n_j}-U_\lambda p_0\|-\|x_{n_j}-p_0\| \right)\leq 0.$$ Similarly to the proof of Theorem \[th1\], we have $$\|x_{n_j}-U_\lambda p_0\|^2=\|(x_{n_j}-p_0)+(p_0-U_\lambda p_0\|^2$$ $$=\|x_{n_j}-p_0\|^2+\|p_0-U_\lambda p_0\|^2+2\langle x_{n_j}-p_0,p_0-U_\lambda p_0\rangle,$$ which, together with the fact that $\{x_{n_j}\}$ converges weakly to $p_0$, implies $$\label{eq8}
\lim_{j\rightarrow \infty} \left[\|x_{n_j}-U_\lambda p_0\|^2-\|x_{n_j}-p_0\|^2\right]=\|p_0-U_\lambda p_0\|^2.$$ We also have $$\|x_{n_j}-U_\lambda p_0\|^2-\|x_{n_j}-p_0\|^2=\left(\|x_{n_j}-U_\lambda p_0\|-\|x_{n_j}-p_0\|\right)\cdot$$ $$\label{eq9}
\left(\|x_{n_j}-U_\lambda p_0\|+\|x_{n_j}-p_0\|\right).$$ Since $C$ is bounded, the sequence $\{\|x_{n_j}-U_\lambda p_0\|+\|x_{n_j}-p_0\|\}$ is bounded, too, and therefore by combining , and , we get $$\|p_0-U_\lambda p_0\|=0,$$ that is, $$U_\lambda p_0=p_0,$$ which implies $$p_0\in Fix\,(U_\lambda)=Fix\,(T_\mu)=Fix\,(T)=\{p\}.$$

\[rem3\] 1) Theorem \[th2\] is an extension of Theorem 7 in Browder and Petryshyn [@BroP67], see also 