# If $f : (a,b) \rightarrow \Bbb R$ is uniformly continuous then it is bounded

I have an exercise in my textbook:

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?)

• 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. Apr 1, 2020 at 16:16
• 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. Apr 1, 2020 at 16:18
• Note the difference between $f:(a,b)\to\Bbb R$ and $g:\Bbb R\to\Bbb R$. Apr 1, 2020 at 16:25
• @PantelisSopasakis: $(a,b)$ is not compact. Apr 1, 2020 at 16:49
• @JonathanZsupportsMonicaC * bounded Apr 1, 2020 at 19:14

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.

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