Let $H$ be a Hilbert space, $C\subseteq H$ a nonempty closed convex set and $T:C\to C$ a nonlinear contraction, that is $$ (*)\qquad|Tu - Tv| \leq |u-v|. $$ Let $(u_n)$ be a sequence in $C$ such that $u_n\rightharpoonup u$ weakly in $H$ and $u_n - Tu_n\to f$ strongly in $H$. I want to prove that $f = u - Tu$.
My attempt: From inequality $(*)$ it is easy to prove that $$ ((w-Tw)-(v-Tv),w-v)\geq 0, \qquad \forall w,v\in C. $$ Put $w = u_n$, then as $u_n - v\rightharpoonup u-v$ weakly and $u_n - Tu_n - (v - T)\to f-(v-T)$ strongly, we obtain that $$ (**)\qquad (f-(v-Tv),u-v)\geq 0, \qquad \forall v\in C. $$ Because $C$ is convex we have that $u\in C$. I believe that from inequality $(**)$ we can prove that $(u,f)$ belongs to the graph of $I-T$, but I'm stuck here.
Maybe my attempt is not the right way to prove the result, so any help willbe appreciated.
The book then asks to use this result to prove that if $C$ is bounded, then $T$ has a fixed point (this is the reason I'm adding the "fixed-point-theorems" tag), but this is a simple consequence of the Banach fixed point theorem and the above result.