Does $ \|x_n \| \rightarrow \|x\| $ $\Longrightarrow $ $x_n \rightarrow x$

Question

I am bit confused about whether it is true that for a given normed space $(X, \| \cdot \|)$, $\|x_n\| \rightarrow \|x\|$ $\Longrightarrow $ $x_n \rightarrow x$. I know the converse of above is true which is essentially the continuity of norm. I was thinking about this implication which looks like it is true but I am not getting how to prove neither any counterexample I can think of.

Answer 1

No: consider for example the real sequence $x_n=(-1)^n$.

Answer 2

No of course not.

Take any sequence $(x_n)_n$ of unit vectors in $X$ and any unit vector $x$ in $X$ with $x_n \not\to x$. Then we still have $\Vert x_n \Vert = 1 \to 1=\Vert x \Vert$. On any non-trivial normed space, you can find such a sequence.

For example, consider:

(1) in $\mathbb{R}$: $x_n=(-1)^n$

(2) in $l^p$: $x_n = \delta_n$

Answer 3

No, it's false even in $\Bbb{R}$. Simply let $x_n = (-1)^n$, and $x=1$. Then, $\lim_{n \to \infty}x_n$ doesn't even exist, so can't equal $x=1$, but clearly $|x_n| = |x| = 1$, for all $n$; so $|x_n| \to |x| = 1$.

Answer 4

No, consider $x_n=(-1)^n$ which converges in the norm but doesn't without it.