Answer to Question #113258 in Functional Analysis for Shivani saini

Question

Show that norm is continuous function.

Answer 1

EXPLANATION.

Denote by $\left( X,\left\| \right\| \right)$ the normalized space. $f(x)=\left\| x \right\| , \ f:X\rightarrow { R }_{ 1 }$

By the properties of the norm for any $x,a\in X$ the inequalities$\left\| x \right\| =||x-a+a||\le \left\| x-a \right\| +\left\| a \right\|$ , $\left\| a \right\| =||a-x+x||\le \left\| x-a \right\| +\left\| x \right\|$ are true. Hence, $-\left( \left\| x \right\| -\left\| a \right\| \right) \le \left\| x-a \right\| \le \left( \left\| x \right\| -\left\| a \right\| \right)$ , or

$\left| \left\| x \right\| -\left\| a \right\| \right| \le \left\| x-a \right\| \quad \quad$. For the function $f$ the last inequality means

$\left| f(x)-f(a) \right| \le \left\| x-a \right\|$ (1)

Let $ε>0, δ=ε$ . Then from inequality (1) we obtain ,that for all $x,a\in X$

$\left\| x-a \right\| <\delta$ implies $\left| f(x)-f(a) \right| \le \left\| x-a \right\| <\delta =ε$ (2)

By the definition of continuous function ,(2) means, that $f$ is continuous at any point $a\in X$ .