Let $H$ be a Hilbert space, and let $T: H\to H$ be a nonlinear contraction, i.e. $$||Tu-Tv||\le||u-v||, \forall u,v\in C$$
Let $(u_n)$ be a sequence in $H$ s.t. $$u_n\rightharpoonup u\text{ weakly and } (u_n-Tu_n)\to f \text{ strongly}$$. Prove that $u-Tu = f$. Hint: use inequality $(u-Tu)-(v-Tv),u-v)\ge 0, \forall u,v$
My thought is to show $||u_n-Tu_n-(u-Tu)||\to 0$, but when apply the inequality in the hint, it doesn't lead to the convergence to $0$. Could anyone shed light on how to use this inequality?
From the assumptions it follows $Tu_n\rightharpoonup u-f$. Expanding the norm-square and using the hint $$ \| u_n - Tu_n -(u-Tu)\|^2 = (u_n - Tu_n -(u-Tu), u_n -u) + (u_n - Tu_n -(u-Tu), Tu-Tu_n) \ge (u_n - Tu_n -(u-Tu), Tu-Tu_n). $$ The left-hand side tends to zero, $u_n - Tu_n \to f$ strongly, $Tu_n \rightharpoonup u-f$, hence the right-hand side converges to $(f-(u-Tu),Tu-u+f)=\|f-(u-Tu)\|^2$, which implies $u-Tu=f$.