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Let $x\in[1,\infty)$. Is $\ln x$ uniformly continuous? I took this function to be continuous and wrote the following proof which I'm not entirely sure of.

Let $\varepsilon>0 $, $x,y\in[1, ∞)$ and $x>y$. Then, $\ln x< x$ and $\ln y< y$ and this follows that $0<|\ln x-\ln y|<|x-y|$ since $x> y$. Choose $δ=ϵ$. Now suppose $|x-y|< δ$. Then, $|\ln x-\ln y|<|x-y|<\varepsilon$

It would be much appreciated if someone could validate my proof

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    $\begingroup$ Dear Rajinda, As Jonas Meyer notes in his answer below, your argument is not correct. Just because $\ln x < x $and $\ln y < y$ you can't conclude anything about $\ln x - \ln y$ vs. $x - y$; for that, you have to know something about the disance between $\ln x$ and $x$ and the distance between $\ln y$ and $y$. (Just to give an example with numbers, $3 < 10$ and $1 < 9$, but it is not true that $3 - 1$ is less than $10 - 9$.) Regards, $\endgroup$ – Matt E Jun 19 '13 at 19:16
  • $\begingroup$ A rookie mistake I suppose. However $lnx$ increases at lower rate than $x$ increases. Hence doesn't lnx<xand lny<y imply the inequality based on rate on increase i mentioned? In your example the lower side of the inequality decreases by a higher proportion than the higher side $\endgroup$ – Heisenberg Jun 19 '13 at 19:27
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    $\begingroup$ Dear Rajinda, Yes, your discussion of rates of increase is exactly the point, and this is what you have to incorporate into your proof! My suggestion is that, if you really want to get your head round this, you try writing out a more careful proof on your own, trying to incorporate this idea of rate of increase in a careful way. (Maybe you will just end up coming back to one of the arguments given below, but maybe you will find a different argument!) Best wishes, $\endgroup$ – Matt E Jun 19 '13 at 19:33
  • $\begingroup$ Thanks for helping me clear things up. Realized the flaw in the proof. $\endgroup$ – Heisenberg Jun 19 '13 at 19:35
  • $\begingroup$ Qualitatively, the natural log function does not seem uniformly continous, because as we approach zero, it rapidly becomes large and negative. There is no upper bound on how big of a vertical delta corresponds to any given horizontal delta. $\endgroup$ – Kaz Jun 19 '13 at 20:38
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You can prove something more general:

PROP Suppose $f:[a,\infty)\to\Bbb R$ has bounded derivative. Then $f$ is uniformly continuous on its domain.

P Pick $x,y\in[a,\infty)$ arbitrarily. By the mean value theorem, we can write $$|f(x)-f(y)|=|f'(\xi)||x-y|$$

Let $M=\sup\limits_{x\in[a,\infty)}|f'(x)|$. Then $$|f(x)-f(y)|\leq M|x-y|$$

Thus, for any $\epsilon$ we may take $\delta=\frac{\epsilon}{2M}$. Note that in your case $M=1$. I only divide by $2$ to turn $\leq$ into $<$.

ADD This means, for example, that $\log x$ (over $[a,\infty)$, $a>0$), $\sin x$, $\cos x$, $x$, and similar functions are all uniformly continuous. Note, for example, that $\sin(x^2)$ is not uniformly continuous. Note that we actually prove $f$ is $1$-Lipschitz with constant $M$, so this might be of interest.

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  • $\begingroup$ A very useful answer! Thanks alot. $\endgroup$ – Heisenberg Jun 19 '13 at 18:44
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An easier argument is to note that the derivative of $\ln x$ is bounded by 1 on the interval $[1,\infty)$. Therefore $\ln x$ is Lipschitz and in particular uniformly continuous.

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  • $\begingroup$ OK but I was hoping to validate my proof as well. Is it correct? $\endgroup$ – Heisenberg Jun 19 '13 at 18:14
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    $\begingroup$ I am not sure how you deduce that $|\ln x-\ln y|<|x-y|$. I don't think it is enough to write that $\ln x < x$. $\endgroup$ – Mikhail Katz Jun 19 '13 at 18:19
  • $\begingroup$ If $x∈[1,∞)$ then $e^x> x$ so $ln x < x$. Why is it wrong ? $\endgroup$ – Heisenberg Jun 19 '13 at 18:27
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    $\begingroup$ $x<a$ and $y<b$ does not imply $|x-y|<|a-b|$, for example let $x=1$ $y=5$ and $a=5$ and $b=6$ $|1-5|$ is not less than $|5-6|$ $\endgroup$ – user10444 Jun 19 '13 at 18:41
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Alternatively, you can prove a function is uniformly continuous based off the following idea:

$f$ is uniformly continuous if and only if for any sequence $\{a_n\},\{b_n\}$

$$ \lim\left(a_n-b_n\right)=0 \Rightarrow \lim\left(f\circ a_n-f\circ b_n\right)=0. $$

Let $\{a_n\}, \{b_n\}$ satisfy our hypothesis ($\lim\left(a_n-b_n\right)=0$), then we have $\lim a_n = \lim b_n$ and so

$$ \lim\left(f\circ a_n-f\circ b_n\right) =\lim\left(\ln(a_n)-\ln(b_n)\right)=\lim\ln\left(\frac{a_n}{b_n}\right) = \ln(1) = 0. $$

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  • $\begingroup$ What you used, is it a theorem ? Haven't come across it $\endgroup$ – Heisenberg Jun 19 '13 at 18:40
  • $\begingroup$ It is, Uniform continuity via sequences. It is worthwhile to prove this result, I was given a homework assignment to prove this and it was very useful for many uniform continuity proofs. $\endgroup$ – Kenny Hegeland Jun 19 '13 at 18:46
  • $\begingroup$ True, at times using sequences does make life much easier. $\endgroup$ – Heisenberg Jun 19 '13 at 18:49
  • $\begingroup$ As a tip, there are a lot of things which can be proven to be not uniformly continuous from this theorem. Many times you can take $a_n=n+\frac{1}{n}, b_n = n$ and contradict this characterization. $\endgroup$ – Kenny Hegeland Jun 19 '13 at 18:53
  • $\begingroup$ I think your proof is not correct, since for any two sequences ${a_n},{b_n}$ such that $\lim(a_n-b_n)=0$ we do not know if $\lim{a_n}$ and $\lim{b_n}$ exist. And even if the limits $\lim{a_n}$ and $\lim{b_n}$ exist, we do not have $\lim \frac{a_n}{b_n}=1$ (unless the limits are not equal to zero): $a_n=\frac{1}{n}$ and $b_n=\frac{2}{n}$, we have $\lim(a_n-b_n)=0$ but $\lim \frac{a_n}{b_n}\neq 0$. In fact this can be used to prove that $x\mapsto ln(x)$ is not uniformly continuous on $(0,\infty)$ $\endgroup$ – user50618 Dec 21 '14 at 22:01
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No, your proof has a problem. If $f(x)<x$ for all $x\in[1,\infty)$, it does not follow that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.

You have $x>y$, and using the fact that $\ln$ is increasing, $|\ln x -\ln y|=\ln x - \ln y$. But how do you conclude that this is less than $x-y=|x-y|$? We know that $\ln x<x$, which gives $\ln x - \ln y <x-\ln y$. But $\ln y<y$ applied to the last expression gives $x-\ln y>x-y$, which doesn't help. Replacing $\ln x$ with $x$ makes the expression bigger, while replacing $\ln y$ with $y$ makes the expression smaller. To ensure that the net result is bigger, you need to know that $x-\ln x > y- \ln y$. But this is just a rearrangement of the inequality that you want to prove.

In summary: The conclusion that $|\ln x -\ln y|\leq |x-y|$ for all $x,y\geq 1$ is true, but more is needed to show it. Some methods to complete the proof are given in the other answers.

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  • $\begingroup$ So basically it lacks detail but I haven't stated anything incorrect? $\endgroup$ – Heisenberg Jun 19 '13 at 19:09
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    $\begingroup$ @Rajinda: It is incorrect to say that from $\ln x<x$ and $\ln y<y$ it follows that $|\ln x-\ln y|<|x-y|$. The inequality is correct, but the stated justification is not. $\endgroup$ – Jonas Meyer Jun 19 '13 at 19:10
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    $\begingroup$ @Rajinda: I don't know what you mean by "putting forward the idea". You can state the inequality, and that puts forward the idea. But in your proof, I guess you want to give valid justification for why the inequality is true. Their are several methods of doing so. For example, the mean value theorem can be used as in Peter's answer. Here is another: If $x>y\geq 1$, then $\ln x-\ln y=\ln(x/y)=\ln(1+((x/y)-1))\leq (x/y)-1\leq x-y$. Here I used the inequality $\ln(1+t)\leq t$ for all $t\geq 0$, which can be proved using positivity of the derivative of $t-\ln(1+t)$ for all $t>0$. $\endgroup$ – Jonas Meyer Jun 19 '13 at 19:23
  • $\begingroup$ Sorry, I meant how can I justify in another way. Thanks for the help! $\endgroup$ – Heisenberg Jun 19 '13 at 19:32
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Assume $x>y>1$, Then by triangle inequality and the fact that $y>1$:

$$\frac{x}{y} < \frac{|x-y|}{|y|} + 1 < |x-y|+1$$

Let $\epsilon > 0$, choose $\delta = e^{\epsilon}-1$, then we have

$$|\ln(x)- \ln(y)| = \ln(\frac{x}{y}) < \ln(|x-y|+1) < \ln(e^{\epsilon} - 1+1) = \epsilon$$

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