Question #113258
Show that norm is continuous function.
1
Expert's answer
2020-05-01T18:50:04-0400

EXPLANATION.

Denote by (X,)\left( X,\left\| \right\| \right) the normalized space. f(x)=x, f:XR1f(x)=\left\| x \right\| , \ f:X\rightarrow { R }_{ 1 }

By the properties of the norm for any x,aXx,a\in X the inequalitiesx=xa+axa+a\left\| x \right\| =||x-a+a||\le \left\| x-a \right\| +\left\| a \right\| , a=ax+xxa+x\left\| a \right\| =||a-x+x||\le \left\| x-a \right\| +\left\| x \right\| are true. Hence, (xa)xa(xa)-\left( \left\| x \right\| -\left\| a \right\| \right) \le \left\| x-a \right\| \le \left( \left\| x \right\| -\left\| a \right\| \right) , or

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

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

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

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

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


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