Showing that if the limit of norms converges, then the sequence converges

Question

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.

I've just realized that this is actually true in the more general setting of uniformly convex Banach spaces if you allow for the result that nonempty closed convex sets have unique elements of minimal norm in strictly convex, reflexive normed spaces (which implies the existence of unique minimizers in Hilbert space settings as well). Uniformly convex Banach spaces satisfy both these requirements, and uniform convexity is a stipulation of the last theorem I use in my proof anyway. — Michael L., Apr 30 '18 at 09:33

score 2 · Accepted Answer · answered Apr 30 '18 at 06:10

2

$||x_n-x_m||^{2}+||x_n+x_m||^{2}=2||x_n||^{2}+2||x_m||^{2}$. 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.

answered Apr 30 '18 at 06:10

Kavi Rama Murthy

311,013

Thats much simpler than what I was trying to do. Interesting that we don't need to show the element with smallest norm is unique. Thanks! – user2139 Apr 30 '18 at 06:18

Michael L. · Answer 2 · 2018-04-30T11:19:10.863

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 $x_n\to y$.

Showing that if the limit of norms converges, then the sequence converges

2 Answers2