I have an exercise in my textbook:

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The first one I try to prove by contrapositive:

Suppose $f$ is not bounded. Then $\forall M \in \Bbb R : |f(x)| \gt M \text{ },\forall x \in X$

$\Rightarrow f(x) \gt M $ or $f(x) \lt -M$

Hence $\forall \epsilon \text{ }\exists \delta \leq |b-a| : |x-y| \lt \delta$ but $|f(x)-f(y)| \gt \epsilon$ since

$$|f(x)| \gt M \Leftrightarrow f(x) \gt M \text{ or } f(x) \lt -M$$

$\Rightarrow f(x) \gt \epsilon + f(y) \text{ or } f(x) \lt -\epsilon-f(y)$

$\Leftrightarrow |f(x) -f(y)| \gt \epsilon$

Hence, by definition, $f$ is not uniformly continuous.

I think I'm doing something wrong since in the following task I observe that the function $g(x) =x$, $\Bbb R\rightarrow \Bbb R$ is not bounded but it is uniformly continuous.

(also, $g(x)= \text{sin}(x)$ is uniformly continuous, right?)

  • $\begingroup$ This is not correct. The definition of unbounded is for all $M>0$, there exists $x$ such that $|f(x)|>M$. Moreover, $|f(x)|>M$ means $f(x)>M$ or $f(x)<-M$. It can't be both. $\endgroup$
    – ProfOak
    Apr 1, 2020 at 16:16
  • $\begingroup$ Regarding your examples, the exercise states that $f$ is defined on a bounded interval, $(a, b)$. Uniformly continuous functions functions are bounded on compact subsets of their domains. $\endgroup$ Apr 1, 2020 at 16:18
  • $\begingroup$ Note the difference between $f:(a,b)\to\Bbb R$ and $g:\Bbb R\to\Bbb R$. $\endgroup$
    – TonyK
    Apr 1, 2020 at 16:25
  • $\begingroup$ @PantelisSopasakis: $(a,b)$ is not compact. $\endgroup$
    – JonathanZ
    Apr 1, 2020 at 16:49
  • $\begingroup$ @JonathanZsupportsMonicaC * bounded $\endgroup$ Apr 1, 2020 at 19:14

1 Answer 1


First of all, I think you need to brush up on negations.

Bounded: There exists $M>0$ such that $|f(x)|\leq M$ for all $x \in (a,b)$.

Negation: For all $M>0$, there exists $x\in(a,b)$ such that $|f(x)|>M$.

Uniformly continuous: For all $\epsilon>0$, there exists $\delta>0$ such that for all $x,y$ with $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$.

Negation: There exists $\epsilon>0$ such that for all $\delta>0$, there exists $x,y$ with $|x-y|<\delta$, but $|f(x)-f(y)|\geq \epsilon$.

In this problem, I don't see much value from using a proof by contradiction. For the direct proof, here are some hints. (Uniformly) continuous functions are bounded on closed intervals. Therefore, for any small $\delta>0$, we have $f$ is bounded on $[a+\delta,b-\delta]$.

Let $\epsilon>0$ be fixed. Then there exists $\delta>0$ such that for all $x,y\in(a,b)$ with $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$. Consider the interval $(a,a+\delta]$. Restrict $\delta$ so that $(a,a+\delta]\subset (a,b)$ and $[b-\delta,b)\subset (a,b)$. Fix $y\in (a,a+\delta]$. Then what can you say about $f(x)$ for all $x \in (a,a+\delta]$? Apply a similar argument on $[b-\delta,b)$. You will get three bounds, on the intervals $(a,a+\delta],\, [a+\delta,b-\delta],\, [b-\delta,b)$ respectively.

  • $\begingroup$ How would you bound the function for $(a,a+\delta]$? $\endgroup$
    – Lev Bahn
    Apr 1, 2020 at 16:56
  • $\begingroup$ Take any fixed $y\in (a,a+\delta]$. Then for all $x\in (a,a+\delta]$, we have $|x-y|<\delta$ so $|f(x)-f(y)|<\epsilon \implies -\epsilon+f(y)<f(x)<\epsilon+f(y)$. $\endgroup$
    – ProfOak
    Apr 1, 2020 at 17:19
  • $\begingroup$ Lev.You have for fixed $y \in (a,a+\delta] $: $|f(x)-f(y)| <\epsilon,$ or $-\epsilon +f(y) <f(x)<f(y)+\epsilon$, where $x \in (a,a+\delta].$ $\endgroup$ Apr 1, 2020 at 17:19
  • $\begingroup$ @PeterSzilas and zugzug Ah! Thanks $\endgroup$
    – Lev Bahn
    Apr 1, 2020 at 17:26
  • $\begingroup$ @Lev Bahn. :):) $\endgroup$ Apr 1, 2020 at 17:32

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