Which of the following is not uniformly continuous?

Question

1.$f_1(x)=|x|$

2.$f_2(x)=\frac{1}{1+x^2}$

3.$f_3(x)=\sin x^2$

4.$f_4(x)=\ln(1+x^2)$

5.$f_5(x)=e^{-x}$

My solution:$f_1(x)=|x|$ is lipschitz so uniformly continuous.

$\lim_{ x\to\pm \infty}\frac{1}{1+x^2}=0$.Also $f_2(x)$ is continuous,so it is uniformly continuous

$f_3(x),f_4(x),f_5(x)$ are not uniformly continuous.Am i right?

@AlexR My argument is "a continuous function on an unbounded interval is uniformly continuous if either it is lipschitz or limits at infinity exists".You can check this result from some analysis book. — Flip, Sep 24 '14 at 10:02
@Math So the correct statement is "a continuous function on an unbounded interval is uniformly continuous if either it is lipschitz or limits at the interval bounds exist". Yes, your application is correct; I will delete all prior comments. Sorry for the inconvenience. — AlexR, Sep 24 '14 at 10:10
@Mesih for $f_3$ consider $x_n=\sqrt{(2n+1)\frac{\pi}{2}}$ and $y_n=\sqrt{n\pi}$ then you can see $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\to1\neq 0$ — Flip, Sep 24 '14 at 10:13

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

A function with bounded derivative is uniformly continuous (see for instance Prove that a function whose derivative is bounded is uniformly continuous.). Hence 4) is uniformly continuous.

edited Apr 13 '17 at 12:21

Community

answered Sep 24 '14 at 10:18

paw88789

ok yes you are right,and what are your thoughts about 5th? – Flip Sep 24 '14 at 10:27
1

I agree with you about 5 (not uniformly continuous) because of the rapid (unbounded) growth rate of $e^{-x}$ as $x\to-\infty$. – paw88789 Sep 24 '14 at 10:29
In the case of $log(1+x^2)$, there is a rapid growth at $\infty$. – Oct 23 '17 at 03:36
@paw88789 If a person uses this result before 'A function with bounded derivative is uniform convergence'. Why does this function show dual behaviour? Please explain. – Oct 23 '17 at 03:39

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