Answer to Question #260304 in Real Analysis for scholar

Question

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)

Answer 1

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

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

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 $1 \leq x_{2}$

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

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

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

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