y\|.$$ Now, by and , we have $$\|x-y\|\leq \|x-T_\mu u_\lambda\|+\|T_\mu u_\lambda-y\|\leq \|u_\lambda- x\|+\|u_\lambda- y\|=\|\lambda (x-y)\|$$ $$+\|(1-\lambda)(x- y)\|=\|x-y\| \Rightarrow \|x-T_\mu u_\lambda\|+\|T_\mu u_\lambda-y\|=\|x-y\|.$$ The last equality implies the existence of some nonnegative numbers $a,b$ with $a,b\leq 1$, such that $$x-T_\mu u_\lambda=a(x-u_\lambda)$$ and $$y-T_\mu u_\lambda=b(y-u_\lambda).$$ Hence, for all $\lambda\in[0,1]$, $$\label{eq8u}
\|x-y\|=\|x-T_\mu u_\lambda\|+\|T_\mu u_\lambda-y\|=a\cdot \|x-u_\lambda\|+b\cdot \|y-u_\lambda\|.$$ Now, just take $\lambda=1$ in to get $a=1$ and then take $\lambda=0$ in to get $b=1$.

This shows that $T_\mu u_\lambda= u_\lambda$, that is, $u_\lambda\in Fix\,(T_\mu)$. So $Fix\,(T_\mu)$ is convex and hence $Fix\,(T)$, is convex, too. Thus, the first part of our theorem is proven.

In order to prove the last part of the theorem, consider the sequence $\{x_n\}_{n=0}^{\infty}$ given by $$x_{n+1}=(1-\lambda) x_n+\lambda T_\mu x_n,\,n\geq 0.$$ It is obvious that $\{x_n\}_{n=0}^{\infty}$ lies in $C$ and hence it is bounded.

Denote $U_\lambda=(1-\lambda) I+\lambda T_\mu$, where $I$ is the identity map. Then, since $T_\mu$ is nonexpansive, by Lemma \[lem1\] it follows that $U_\lambda$ is asymptotically regular, i.e., $$\|x_n-U_\lambda x_{n}\|\rightarrow 0,\, \textnormal{ as } n\rightarrow \infty.$$ We also have $$\label{eq8a}
U_\lambda x-x=\lambda (T_\mu x-x)=\lambda \mu (Tx-x),$$ and hence $$\|x_n-T_\mu x_{n}\|\rightarrow 0,\, \textnormal{ as } n\rightarrow \infty.$$ Since, by hypothesis, $T$ is demicompact, it follows by that $T_\mu$ is demicompact, too. Hence, 