Answer to Question #160019 in Real Analysis for Luna

Question

Which of the following functions is uniformly continuous on the given set?

a) f(x)= x³ on [1,infinity)

b) f(x)= x³ on [1,5]

c) f(x)=1/x on (0,1)

d) f(x)=1/x on [0,1]

e) All of the above

Answer 1

Which of the following functions is uniformly continuous on the given set?

a) f(x)= x³ on [1,infinity)

b) f(x)= x³ on [1,5]

c) f(x)=1/x on (0,1)

d) f(x)=1/x on [0,1]

e) All of the above

A function is said to uniformly continuous a set say, S, for every $\epsilon>0$ , $\exists$ $\delta>0$ such that $x,a\isin S$ and $\mid x-a\mid<\delta$ then $\mid f(x)-f(a)\mid <\epsilon$

Also a function is said to be uniformly continuous A function is said to uniformly continuous a set say, S, for every $\epsilon>0$ , $\exists$ $\delta>0$ such that $x,a\isin S$ and $\mid x-a\mid<\delta$ then $\mid f(x)-f(a)\mid \ \ge \epsilon$

(c) d) f(x)=1/x on (0,1)

suppose f is uniformly continuous

then pick $\epsilon=1$ then $\exists$ $\delta>0: \forall$ $x,a \isin (0,1)$ with $\mid x-a\mid<\delta$

the $\mid {1 \over x} - {1 \over a}\mid<\epsilon$

Pick x(0,1) with $x<\delta$ then set $a = {x \over 2}$

$\mid x-a\mid< \mid x- {x \over 2}\mid=\mid {x \over2}\mid< {\delta \over 2}$

$\mid {1 \over x} - {1 \over a}\mid=\mid {1 \over x} - {1 \over{x \over 2}}\mid=\mid {1 \over x} - {2 \over x}\mid=\mid -{1 \over x} \mid={1 \over x}$

Since x is in the interval (0,1) it implies that ${1 \over x}>1$

therefore the function is not uniformly continuous

d) f(x)=1/x on [0,1]

this has the same prove as (c) it is also not not uniformly continuous

b) f(x)= x³ on [1,5]

$\epsilon >0$ then Let $\delta={\epsilon \over 108}$

Supppose $\mid x-a\mid< \delta$ then $\mid f(x) -f(a)\mid =\mid x^3 - a^3\mid = \mid x-a\mid\mid x^2+xa+a^2\mid<\epsilon$

$<{ \epsilon \over 108}.108 = \epsilon$

$\therefore$ the function uniformly continuous

a) f(x)= x³ on [1,infinity)