Prove that Norm is a Continuous Mapping
Theorem
Let’s call $(X, \left\| \cdot \right\|)$ a norm space. Then for a sequence $\left\{ x_{k} \right\}$ of $X$ which is $\lim \limits_{k\to\infty} x_{k} = x$, the following equation holds.
$$ \lim \limits_{k \to\infty} \left\| x_{k} \right\| = \left\| x \right\| $$
Explanation
$\left\| \cdot \right\|$ means that it is a continuous function. The limit symbol can freely enter and exit the continuous function, which is a very good property.
Proof
Assuming $\lim \limits_{k\to\infty} x_{k}=x$, the following equation holds.
$$ \lim \limits_{k\to\infty} \left\| x-x_{k} \right\| = 0 $$
Also, by the reverse triangle inequality, the following holds.
$$ \left\| x \right\| - \left\| x_{k} \right\| \le \left\| x - x_{k} \right\| $$
Taking limits on both sides,
$$ \lim \limits_{k\to\infty} \left( \left\| x \right\| - \left\| x_{k} \right\| \right) \le \lim \limits_{k\to\infty} \left\| x - x_{k} \right\| = 0 $$
Therefore,
$$ \lim \limits_{k \to\infty} \left\| x_{k} \right\| = \left\| x \right\| $$
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