Let $X$ be a normed space. If the sequence $x_n \rightarrow x$ in norm, then $\|x_n\| \rightarrow \|x\|$

Question

Let $(X, \|.\|)$ be a normed space. Let $(x_n)_{n \in N}$ be a sequence in $X$. Show that, if $x_n \rightarrow x$ (w.r.t $\|.\|)$ then $\|x_n\| \rightarrow \|x\|$.

I took a subsequence and showed that if $x_n \rightarrow x$ then the subsequebce does as well. I then thought applying the definition of normed space to show it applies to the same subsequence, but I realize this is not a good method. How would you do this proof?

Answer 1

It follows from triangle inequality that: $$\forall (x,y)\in X^2,\left|\|x\|-\|y\|\right|\leqslant\|x-y\|.$$ To see that write $\|x\|=\|y+x-y\|$.

Answer 2

What does it mean $x_n\to x$? It means that given $\epsilon>0$ there exists some $N$ such that if $n>N$ then $\|x_n-x\|<\epsilon$. In particular, by the "reverse" triangle inequality, for $n>N$ we have

$$\left|\|x_n\|-\|x\|\right|\le\|x_n-x\|<\epsilon$$

so $\|x_n\|\to\|x\|$ as desired.