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

$\text{Question 1} \\ \text{\textbf{Proof} } \text{Let } x_1, x_2 \text{ be any two points of } [-1,1] \text{ so that } \left| {{x_1}} \right| \le 1,\left| {{x_2}} \right| \le 1 \\ \text{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| \\ \hspace{95pt} = \left| {{x_1} - {x_2}} \right|\left| {x_1^2 + {x_1}{x_2} + x_2^2} \right| \\ \hspace{95pt} \le \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) \\ \hspace{95pt} \le 3\left| {{x_1} - {x_2}} \right| < \epsilon \text{ whenever } \left| {{x_1} - {x_2}} \right| < \frac{\epsilon }{3} \\ \text{choose } \delta = \frac{\epsilon }{3}, \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 2} \\ \text{\textbf{Proof} } \text{Let } \epsilon > 0 \text{ be given} \\ \text{Let } x_1, x_2 \text{ be any two points of } [1,\infty) \text{ so that } 1 \le {x_1} \text{ and } 1 \le {x_2} \\ \hspace{30pt} \Rightarrow 2{x_1} - 1 \ge 1 \quad \text{and } 2{x_2} - 1 \\ \hspace{30pt} \Rightarrow \left| {2{x_1} - 1} \right| \ge 1 \quad \text{and } \left| {2{x_2} - 1} \right| \ge 1 \\ \hspace{30pt} \Rightarrow \frac{1}{{\left| {2{x_1} - 1} \right|}} \le 1 \quad \text{and } \Rightarrow \frac{1}{{\left| {2{x_2} - 1} \right|}} \le 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| \\ \hspace{95pt} = \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|}}{{\left| {2{x_1} - 1} \right|\left| {2{x_2} - 1} \right|}} \\ \hspace{95pt} \le 2\left| {{x_1} - {x_2}} \right| < \epsilon \text{ whenever } \left| {{x_1} - {x_2}} \right| < \frac{\epsilon }{2} \\ \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{\textbf{Proof} } \text{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| \\ \hspace{95pt} = \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{\textbf{Proof} } \text{Since } x \text{ is continuous on } I \text{and } x\ne0 \text{ in } I \\ \therefore \frac{1}{x} \text{ is continuous}. \\ \text{Now, for any } \delta>0 ~\exist m\in \mathbb{N} \text{ such that } \frac{1}{n}<\delta ~\forall n>m \\ \hspace{30pt} \text{Let } {x_1} = \frac{1}{{2m}} \text{ and } {x_2} = \frac{1}{m} \text{ so that} x_1,x_2\in I \\ \hspace{30pt} \text{and } \left| {{x_1} - {x_2}} \right| = \left| {\frac{1}{{2m}} - \frac{1}{m}} \right| = \frac{1}{{2m}} < \delta \\ \text{ but } \left| {f\left( {{x_1}} \right) - f\left( {{x_2}} \right)} \right| = \left| {2m - m} \right| = m \text{ which cannot be less than } \epsilon > 0 \\ \therefore f \text{ is \textbf{not} uniformly continuous.}$