Normed convergence Vs. Convergence in norm

Question

Given $X$ a normed vector space (be it finite or infinite dimensional), a vector $u\in X$ and a sequence $(u_n)\subset X$ such that $\| u_n -u \| \to 0$. I want to show that $$\|u_n\| \to \|u\|$$

I know that:

1. The converse is not true. Think if $X=L^2$, $u_n = \delta(x-n)$ and $u=\delta (x)$, both are dirac delta distribution.
2. Given an inner product $(\cdot,\cdot)$ where $\|v\| = (v,v)$ and an orthonormal basis where $u_n$ is the $n$th order expansion of $u$ with respect to this basis, the statement hold.

However, I've not been able to prove it otherwise, so, maybe its wrond. Any help?

A version of this is stated as a fact in this answer, but I'm not quite sure as of why is it true.

Thanks!

Answer 1

Just note that in general $\left|\|x\|-\|y\|\right|\leq\|x-y\|$.

Indeed, we have \begin{align} \|x\|&=\|(x-y)+y\|\\ &\leq\|x-y\|+\|y\|&&\text{(triangle inequality)} \end{align} so $\|x\|-\|y\|\leq\|x-y\|$ and, similarly, \begin{align} \|y\|-\|x\|&\leq\|y-x\|\\ &=\|x-y\|&&(\text{since } \|\alpha x\|=|\alpha|\|x\|). \end{align}