I have a homework question I am attempting to no avail.
Let $\mathcal{H}$ be a Hilbert space. Let $X\subseteq\mathcal{H}$ be a convex set. Suppose that $(x_{n})_{n\geq 1}$ is a sequence in $X$ such that, \begin{align} \lim_{n\rightarrow\infty}\|x_{n}\|=\inf_{x\in X}\|x\|. \end{align} Show that $x_{n}$ converges in $\mathcal{H}$. Is the limit of $x_{n}$ necessarily in $X$?
I have tried to show that \begin{align} (x_{n}-x|x_{n}-x)\rightarrow 0, \end{align} however I run into issues due to the limit of $\|x_{n}\|$ being defined as an infimum. In every case when the problem is reduced to $\|x_{n}-x\|$ I cannot make this go to zero because of the infimum. Also I have not identified the importance of $X$ being convex, this could also be the point I am missing. If this is the case could you please leave a small hint and allow me to return with more work.
EDIT: Is it true that $\inf_{x\in X}\|x\|\implies P_{X}(0)$?
Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator. Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge. Note that by the parallelogram equality we must have:
$$ \Vert x_m+x_n \Vert ^2 + \Vert x_m - x_n \Vert^2 = 2(\Vert x_n \Vert^2 + \Vert x_m \Vert^2)$$ So we have: $$ \Vert x_m - x_n\Vert^2 = 2\left(\Vert x_n \Vert^2 + \Vert x_m \Vert^2\right) - \Vert x_m +x_n \Vert ^2 = 2\left(\Vert x_n \Vert^2 + \Vert x_m \Vert^2 - 2\Vert \frac{x_m+x_n}{2}\Vert ^2\right)$$ Since $X$ is convex, we know $\frac{x_m+x_n}{2} \in X$, now if $M = \inf\{\Vert x \Vert \vert x \in X\}$ we must therfore have:
$$ \Vert x_m - x_n \Vert ^2 \leq 2(\Vert x_n \Vert^2 + \Vert x_m \Vert^2 - 2M^2) \underset{n,m\rightarrow \infty}{\rightarrow} 2(2M^2 - 2M^2) = 0$$ Thus $x_n$ is Cauchy.
The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.