 Solution

Subtracting $T_{\alpha }{u}_n=z_n$ and $T_{\alpha }{u}_m=z_m$ and applying the coercive estimate from part i yieldsThus $(z_n)$ Cauchy implies $(u_n)$ Cauchy. Completeness of the Hilbert space $X$ gives $u_n\to u\in X$, and boundedness of $T_{\alpha }$ gives $z_n=T_{\alpha }{u}_n\to T_{\alpha }u$.