Answer to Question #280691 in Real Analysis for Fredo

Question #280691

Show whether the following functions are uniformly continuous on the given domain.

1. F(x)=x^3 on [-1,1]

2. F(x)= 2x/2x-1 on [1, infinity]

3. F(x)= sinx/x on (0,1)

4. F(x)= 1/x on (0,1)



1
Expert's answer
2022-01-03T06:21:05-0500

Solution:


Question 1

Proof Let "x_{1}, x_{2}" be any two points of [-1,1] so that "\\left|x_{1}\\right| \\leq 1,\\left|x_{2}\\right| \\leq 1"

Now. "\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|=\\left|x_{1}^{3}-x_{2}^{3}\\right|=\\left|\\left(x_{1}-x_{2}\\right)\\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\\right)\\right|"

 "\\begin{aligned}\n\n&=\\left|x_{1}-x_{2}\\right|\\left|x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\\right| \\\\\n\n&\\leq\\left|x_{1}-x_{2}\\right|\\left(\\left|x_{1}\\right|^{2}+\\left|x_{1}\\right|\\left|x_{2}\\right|+\\left|x_{2}\\right|^{2}\\right) \\\\\n\n&\\leq 3\\left|x_{1}-x_{2}\\right|<\\epsilon \\text { whenever }\\left|x_{1}-x_{2}\\right|<\\frac{\\epsilon}{3}\n\n\\end{aligned}"  

choose "\\delta=\\frac{\\epsilon}{3}" , then "\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|<\\epsilon whenever \\left|x_{1}-x_{2}\\right|<\\delta"

therefore f is uniformly continuous

Question 2

Proof Let "\\epsilon>0" be given

Let "x_{1}, x_{2}" be any two points of "[1, \\infty)" so that "1 \\leq x_{1}" and "1 \\leq x_{2}"

"\\Rightarrow 2 x_{1}-1 \\geq 1 \\text { and } 2 x_{2}-1"

"\\begin{aligned}\n&\\quad \\Rightarrow\\left|2 x_{1}-1\\right| \\geq 1 \\quad \\text { and }\\left|2 x_{2}-1\\right| \\geq 1 \\\\\n&\\quad \\Rightarrow \\frac{1}{\\left|2 x_{1}-1\\right|} \\leq 1 \\quad \\text { and } \\quad \\Rightarrow \\frac{1}{2 x_{2}-1} \\leq 1 \\\\\n&\\text { Now }\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|=\\left|\\frac{2 x_{1}}{2 x_{1}-1}-\\frac{2 x_{2}}{2 x_{2}-1}\\right| \\\\\n&\\quad=\\left|\\frac{2 x_{1}\\left(2 x_{2}-1\\right)-2 x_{2}\\left(2 x_{1}-1\\right)}{\\left(2 x_{1}-1\\right)\\left(2 x_{2}-1\\right)}\\right|=\\frac{2\\left|x_{1}-x_{2}\\right|}{2 x_{1}-1|| 2 x_{2}-1 \\mid} \\\\\n&\\quad \\leq 2\\left|x_{1}-x_{2}\\right|<\\epsilon \\text { whenever }\\left|x_{1}-x_{2}\\right|<\\frac{\\epsilon}{2}\n\\end{aligned}"

"\\begin{aligned}\n&\\text { choose } \\delta=\\frac{\\epsilon}{2}, \\text { then }\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|<\\epsilon \\text { whenever }\\left|x_{1}-x_{2}\\right|<\\delta \\\\\n&\\therefore f \\text { is uniformly continuous } \\\\\n&\\text { Question } 3 \\\\\n&\\text { Proof Let } \\epsilon>0 \\text { be given } \\\\\n&\\text { Let } x_{1}, x_{2} \\text { be any two points of }(0,1) \\text { so that } 0<x_{1}<1 \\text { and } 0<x_{2}<1 \\\\\n&\\text { Now }\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|=\\left|\\frac{\\sin \\left(x_{1}\\right)}{x_{1}}-\\frac{\\sin \\left(x_{2}\\right)}{x_{2}}\\right|=\\left|\\frac{x_{2} \\sin \\left(x_{1}\\right)-x_{1} \\sin \\left(x_{2}\\right)}{x_{1} x_{2}}\\right| \\\\\n&\\qquad=\\frac{\\left|x_{2} \\sin \\left(x_{1}\\right)-x_{1} \\sin \\left(x_{2}\\right)\\right|}{\\left|x_{1}\\right|\\left|x_{2}\\right|}<\\left|x_{2} x_{1}-x_{1} x_{2}\\right|<\\left|x_{1}-x_{2}\\right|<\\epsilon \\\\\n&\\text { take } \\delta=\\epsilon, \\text { then }\\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|<\\epsilon \\text { whenever }\\left|x_{1}-x_{2}\\right|<\\delta \\\\\n&\\therefore f \\text { is uniformly continuous } \\\\\n&\\text { Question } 4 \\\\\n&\\text { Proof Since } x \\text { is continuous on } I \\text { and } x \\neq 0 \\text { in } I \\\\\n&\\therefore \\frac{1}{x} \\text { is continuous. }\n\\end{aligned}"

 "\\begin{aligned}\n& \\text{Now, for any}\\; \\delta>0\\; \\exists m \\in \\mathbb{N} \\;such\\; that \\;\\frac{1}{n}<\\delta \\forall n>m\\\\\n& \\text{Let}\\; x_{1}=\\frac{1}{2 m}\\; and \\;x_{2}=\\frac{1}{m} so\\; that \\;x_{1}, x_{2} \\in I\\\\\n& \\text{and} \\left|x_{1}-x_{2}\\right|=\\left|\\frac{1}{2 m}-\\frac{1}{m}\\;\\right|=\\frac{1}{2 m}<\\delta\\\\\n& \\text{but} \\left|f\\left(x_{1}\\right)-f\\left(x_{2}\\right)\\right|=|2 m-m|=\\;m \\;\\text{which cannot be less than} \\;\\epsilon>0\\\\\n&\\therefore \\text{f is not uniformly continuous.}\\\\\n\\end{aligned}"





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