Show that the norm is continuous in the following sense:

Question

Let $(E,||.||)$ be a normed space. Suppose $(x_n) \subseteq E$ converges and $x = \lim_{n \rightarrow \infty} x_n$. Then $||x|| = \lim_{n \rightarrow \infty} ||x_n||$. I tried to use the triangle inequality because I think it can applied in a certain form here, but I'm still stuck.

Answer 1

HINT: $\bigl| \|x_n\|-\|x\| \bigr| \le \|x_n - x\| < \varepsilon$ for big enough $n$.

Answer 2

Using the triangle inequality, you can prove the so-called reverse triangle inequality: $$\forall (x,y)\in E^2,\ |\Vert x\Vert - \Vert y\Vert|\leqslant \Vert x+y\Vert$$ Replacing $y$ with $-y$, you get that the norm is Lipschitz continuous, hence continuous on $E$.

Answer 3

You have for all $x,y$:

$$\big\vert \vert\vert x\vert\vert-\vert\vert y\vert\vert\big\vert\leqslant \vert\vert x-y\vert\vert.$$

So $\vert\vert.\vert\vert$ is $1$-Lipschitz continuous.

So $\vert\vert.\vert\vert$ is continuous.

So for all sequence $(x_n)$ such that $\lim x_n=x$:

$$\lim \vert\vert x_n\vert\vert=\vert\vert x\vert\vert.$$