Answer to Question #113258 in Functional Analysis for Shivani saini

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

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 "\u03b5>0, \u03b4=\u03b5" . 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 =\u03b5" (2)

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


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