Showing that if the limit of norms converges, then the sequence converges Let $H$ be a hilbert space, and $C \subset H$ a convex set. Let $(x_n)_{n\in \mathbb{N}}$ be a sequence in $C$ with $\lim_{n\to \infty}||x_n|| = \inf_{x\in C}||x||$. Show $x_n$ converges in $H$. 
So far I have: 
Let $P_C(0) = \{x \in C: ||x|| = \inf_{x \in C}||x||\}$ be the projection of $0$ onto $C$. 
If we consider $\bar{C}$ (the closure of $C$), then there is a theorem that tells us since $\bar{C}$ is closed and convex, there is exactly one $y \in \bar{C}$ with $y=P_{\bar{C}}(0)$, i.e., $||y|| = \inf_{x\in \bar{C}}||x||$. Since the infimum of a set's closure equals the infimum of the set, we also have $||y|| = \inf_{x\in C} ||x||$. 
Now, I need to show that since the norms converge to the norm of a unique element ($y$), the sequence itself must converge to this element. This makes sense intuitively, but I can't make it rigorous! Any hints would be appreciated.  
 A: This is actually true in the more general setting of uniformly convex Banach spaces.
As $\|x_n\|$ converges to $\|y\| = \inf_{x\in C} \|x\|$ (which is unique for convex sets in any strictly convex reflexive normed space), we can take some radius $r > \|y\|$ and note that $(x_n)_{n\in \mathbb{N}}$ is eventually in $\overline{C}\cap B_r(0)$, where $B_r(0)$ is the closed ball of radius $r$ centered at $0$. The set $\overline{C}\cap B_r(0)$ is the intersection of a weakly closed set and a weakly compact set and is therefore weakly compact. For a subsequence $(x_{n_k})_{k\in \mathbb{N}}$, we let its weakly convergent subsequence be $(x_{n_{k_j}})_{j\in \mathbb{N}}$ (which exists by the Eberlein-Šmulian theorem), and we let $x_{n_{k_j}}\rightharpoonup x^*\in \overline{C}\cap B_r(0)$. By the weak lower semicontinuity of the norm, we have $$\|x^*\|\leq \liminf_{j\to \infty} \|x_{n_{k_j}}\| = \|y\|$$ which is of course impossible unless $x^* = y$ by our definition of $y$ as the unique element of $\overline{C}$ with minimal norm. Therefore, as every subsequence has a further subsequence that converges weakly to $y$, we have $x_n\rightharpoonup y$. Then, weak convergence and norm convergence together imply strong convergence in uniformly convex Banach spaces, so we have $x_n\to y$.
A: $||x_n-x_m||^{2}+||x_n+x_m||^{2}=2||x_n||^{2}+2||x_m||^{2}$. Denoting $\inf_{x\in C} ||x||$ by $A$ we get $\limsup ||x_n-x_m||^{2} \leq 2A+2A-\liminf 4||\frac {x_n+x_m} 2||^{2}$. Note that $\frac {x_n+x_m} 2 \in C$ so $\limsup ||x_n-x_m||^{2} \leq 2A^{2}+2A^{2}-4A^{2}=0$.  $\{x_n\}$ is Cauchy, hence convergent. 
