Let $E$ be a uniformly convex Banach space(so E is reflexive), and $C\subset E$ a non-empty closed convex set. Let $P_Cx$ denote the point s.t. $||x-P_Cx||=\inf_{y\in C}||x-y||$. I have proved the existence and uniqueness of $P_Cx$, $\forall x$. Want to show that the minimizing sequence $y_n\to P_Cx$ strongly.
We know that $(y_n)_n$ is bounded sequence, so $\exists(y_{n_k})_k$ s.t. $y_{n_k}\rightharpoonup z$. And we know $z\in C$ as $C$ is weakly closed(as $C$ is closed and convex).
I want to show that $y_{n_k}\to z$, by the fact that $E$ is uniformly convex, so it suffices to show that $\limsup_{k}||y_{n_k}||\le ||z||$. But now I'm stuck with the proof of this inequality, can someone give me some ideas?
Best Answer
If $(y_n)$ is the minimizing sequence for the projection problem, then $z=P_C(x)$. By definition $\|y_n -x \| \to \|P_Cx-x\|= \|z-x$. Now, since $y_n -x \rightharpoonup z-x$, this implies $y_n -x \to z-x$.