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Check if $\ln(x), x > 0$ is uniformly continuous

My only idea on solving this was to use the definition of uniform continuity. Namely, I need to show that for all $\epsilon >0$ there exists a $\delta >0$ such that for all $x_1, x_2$ in the domain of the function $|x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

I have never before proved anything like this and I have no idea whether or not this function is uniformly continuous. Anyway, looking at the tremendous slope close to zero I am inclined to believe that this function is not uniformly continuous.
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I want to show that $$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$

And so let's pick $\epsilon = 1$ I want to show that we can pick such $x_1, x_2$ that $|\ln(x_1)- \ln(x_2)| > 1 \Rightarrow |x_1 - x_2| > \delta$ for all $\delta$. We know that $x_1 \ge ex_2$. And now, let's take an arbitrary $\delta$, such that $$|x_1 - x_2| \ge \delta$$ Now, we can simply pick $$x_1 = ex_2 + \delta$$ And $$x_2 = 1$$ And it works. And so the function is not uniformly continuous.

I know that my solution is very chaotic and confusing,is there an easier way to tackle this?

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  • $\begingroup$ Related: math.stackexchange.com/questions/2540090/… $\endgroup$ – Ethan Bolker Dec 6 '17 at 19:37
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    $\begingroup$ the statement $$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$ is not the negation of being uniformly continuous. We have that $\lnot (A\implies B)\equiv \lnot(\lnot A\lor B)\equiv A\land \lnot B$ $\endgroup$ – Masacroso Dec 6 '17 at 19:38
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    $\begingroup$ @Masacroso: In fact, as a side note to the OP, the irony is that $A \implies B \equiv \lnot (A \land \lnot B)$ is the correct intuitive way of thinking about the meaning of $A \implies B$. $\endgroup$ – stressed out Dec 6 '17 at 20:09
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A good criterion (theorem) to check whether a function is uniformly continuous or not is the following theorem, which is specially useful and powerful when you want to disprove something is uniformly continuous:

Theorem: A function is uniformly continuous on $A$ if and only if for any two sequences $\{a_n\}$ and $\{b_n\}$ in $A$:

$$ \lim_{n\to\infty}\left(a_n-b_n\right)=0 \implies \lim_{n\to\infty}\left(f(a_n)-f(b_n)\right)=0. $$

Now to disprove that $\ln(x)$ is uniformly continuous on $(0,+\infty)$, take $a_n=\{\frac{1}{n}\}$ and $b_n=\{\frac{1}{n^2}\}$

Then $$\lim_{n\to\infty}(a_n-b_n)=\lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n^2})=\lim_{n\to\infty}\frac{n-1}{n^2}=0$$ while $\lim_{n\to\infty} \left(\ln(\frac{1}{n})-\ln(\frac{1}{n^2})\right)=\lim_{n\to\infty}\ln(\frac{n^2}{n})=\ln(+\infty)=+\infty$

Q.E.D.

Remarks:

Note that uniform continuity very much depends on the given set and therefore, it's generally very hard to establish just by going through the $\epsilon-\delta$ definition. For example, if we wanted to study whether $f$ is uniformly continuous or not on $[1,+\infty)$ the answer would've been positive. Indeed, $$|\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)|=|\frac{1}{x}|\leq 1 \text{ for every x}\in [1,+\infty)$$ Therefore, $\ln(x)$ is uniformly continuous on $[1,+\infty)$.

Also, remember that every continuous function is uniformly continuous on a compact set.

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  • $\begingroup$ Idk the above theorem but (I assume) it must be true. I will be glad to know (if possible) where you see it. $\endgroup$ – Masacroso Dec 6 '17 at 20:08
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    $\begingroup$ Lipschitz$\implies$ uniform continuity and bounded derivative $\implies$ lipschitz $\endgroup$ – qbert Dec 6 '17 at 20:16
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    $\begingroup$ @Masacroso: I can't really remember right now where I saw it the first time. But it is a standard fact and I have used it many times for solving exercises and even in exams. Also, another good thing to remember about uniform continuity is that it preserves Cauchy sequences. I'm looking through my analysis books now to cite a good reference for this, I will edit my answer if I find any. Otherwise, I might include a proof later. The proof probably works on the fact that for the points $x$ in the set, the graph of $f$ is squeezed inside a deformed "cylinder" of a fixed radius centered at $f(x)$. $\endgroup$ – stressed out Dec 6 '17 at 20:21
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    $\begingroup$ @Masacroso: I found a related question on MSE that discusses the proof of the theorem: math.stackexchange.com/questions/1090756/… $\endgroup$ – stressed out Dec 6 '17 at 20:26
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I want to show that $$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$

This is not the correct negation of the definition of uniform continuity. It should instead read as follows.

I want to show that there exists $\epsilon > 0$ such that for all $\delta > 0$ there exist $x_1, x_2$ in the domain of $f$ satisfying $|x_1 - x_2| < \delta$ and $|f(x_1) - f(x_2)| \geq \epsilon$.

So, to show that $\ln(x)$ is not uniformly continuous, we can try $\epsilon = 1$ just as you have written. Now, $$ \begin{align*} &\ |\ln(x_1) - \ln(x_2)| \geq 1\\ \Leftrightarrow &\ |\ln(x_1/x_2)| \geq 1\\ \Leftrightarrow &\ \ln(x_1/x_2) \leq -1 \text{ or } \ln(x_1/x_2) \geq 1\\ \Leftrightarrow &\ x_1/x_2 \leq e^{-1} \text{ or } x_1/ x_2 \geq e \end{align*} $$ So, for each $\delta > 0$, we will search for points $x_1, x_2 > 0$ satisfying $x_2 = ex_1$, such that $|x_1 - x_2| < \delta$. But such $x_1, x_2$ can always be found, because we only need to choose $x_1 = \frac{\delta}{2(e-1)}$. Then, $$|x_1 - x_2| = |x_1 - e x_1| = |x_1| \cdot |1-e| < \frac{\delta}{2(e-1)}(e-1) = \delta.$$

Hence, $f(x) = \ln(x)$ is not uniformly continuous.

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