Showing this sequence is Cauchy

Question

Let $X$ be a normed space and let $p \in X$. Then for all $n \in \mathbb{N}$ I have an element $x_n \in L$ (where $L$ is a finite-dimensional subspace of $X$) with the property that $|| x_n - p || \leq d + 1/n$, EDIT where $$ d = \inf_{l \in L} ||x - l|| $$ (end of EDIT). What I want to prove is that $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence. (Then since $\dim L < \infty$, the sequence converges, in $L$, to an element $x$ with $||p - x|| \leq d$, which is what I want.) However, I don't know how to arrive at $$ ||x_m - x_n|| \leq \text{something} $$ where "something" tends to zero as $m, n$ grow larger. All I can do is $$ \left|\frac{1}{m} - \frac{1}{n}\right| \leq ||x_m - x_n|| \leq 2d + \frac{1}{m} + \frac{1}{n} $$ What can I do? The parallelogram rule? Any other inequality?

Answer 1

$(x_n)$ is not always a CauChy sequence, take $X=R$ the field of real numbers, $L=R$, $x_n=(-1)^n$, $p=0$, and $d=1$. But since $L$ is finite dimensional, $\|x_n-x_1\|\leq \|x_n-p\|+\|x_1-p\|\leq 2p+2$, so the sequence $(x_n)$ is a bounded sequence of a finite dimensional vector space, and you can extract a converging subsequence from it.

Answer 2

You may find a convergent subsequence $(x_{n_k})$ because $(x_n)$ is bounded and in the finite dimensional subspace $L$. The subsequence converges to $x$ which is at a distance at most $d$ from $p$. And this seems to be sufficient for your purpose?