Let $C\subset H$ be a nonempty closed convex subset of a Hilbert space $H$ and let $T:C\rightarrow C$ be a nonlinear contraction; i.e. $$|Tu-Tv|\leq|u-v|\quad\forall u,v\in C.$$ Let $(u_n)$ be a sequence in $C$ s.t. $u_n\rightharpoonup u$ weakly and $(u_n-T u_n)\rightarrow f$ strongly. Then I am asked to prove that $u-Tu=f$ and furthermore if $C$ is bounded and $T(C)\subset C$, then $T$ has a fixed point. The hint for the first part is the inequality $$((u-Tu)-(v-Tv),u-v)\geq0\forall u,v$$ and the hint for the second is to consider $T_\varepsilon u=(1-\varepsilon)Tu+\varepsilon a$ with $a\in C$ fixed and $\varepsilon>0,\varepsilon\rightarrow0$.
My idea for the first part was to use the inequality to obtain $$(Tu-Tv,u-v)\leq(u-v,u-v),$$ and then I need to show $$((u-Tu)-f,(u-Tu)-f)=0.$$ I may replace $f$ by $u_n-Tu_n$ by the strong convergence and then use weak convergence $u_n\rightharpoonup u$ to strip the terms not indexed by $n$ from the equation. I may then use 'the inequality' above to arrive at $$((u-Tu)-(u-u_n),(u-Tu)-(u-u_n))\leq(u-u_n,Tu_n-u_n)$$ but I have no reason to conclude that this goes to $0$. :(
If your last inequality is correct (haven't checked that), then the RHS goes to $0$: $$(u-u_n,Tu_n-u_n)=(u-u_n,Tu_n-u_n+f-f)=(u-u_n,-f)+(u-u_n,Tu_n-u_n+f)=(u-u_n,-f)+(u,Tu_n-u_n+f)-(u_n,Tu_n-u_n+f)$$ Here all terms tend to $0$: the first one, because $u-u_n\rightharpoonup 0$, the second because $Tu_n-u_n+f\to 0$, and the third, because $\|u_n\|$ is bounded by, say $M$, as a weakly convergent sequence and therefore $|(u_n,Tu_n-u_n+f)|\leq M\|Tu_n-u_n+f\|\to 0$