📂Banach Space

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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