# Let $f: I \to \mathbb{R}$ be a continuous monotonic and bounded function on an interval $I$, then $f$ is uniformly continuous

The question is as the title says. Here, the interval $$I$$ is not necessary bounded, for example, $$f(x)=1/{(1+e^{x})}$$ with $$f: \mathbb{R} \to \mathbb{R}$$ is an exemple of such function.

My attempts:

a) I have tried to construct $$\phi$$ such that $$\phi(x)=\lim_{y \rightarrow x} f(y)$$ , and then look the range (the range is bounded and "almost" closed), but without success.

b) I have tried "to bind functions" in such sense: if $$f$$ is continuous at b and $$f|{[c,b]}$$ and $$f|[b,d]$$ is uniformly continuous, then $$f:[c,d] \to \mathbb{R}$$ is uniformly continuous, but without success, cos I cant "bind" indefinitely.

c) I have tried by contradiction, but again without success.

Hint:

Assume $f$ is nondecreasing, and write $I=(q,p)$ where $p$ and $q$ may be infinite. Show that $\lim_{x \to p} f(x) = u$, where $u:=\sup_{x \in I} f(x)$. Similarly show that $\lim_{x \to q} f(x) = v$, where $v := \inf_{x \in I} f(x)$.

Next fix $\epsilon>0$. Choose $M, N \in I$ with $M \geq N$, such that $x\geq M$ implies that $u-f(x)<\epsilon/2$, and such that $x \leq N$ implies that $f(x)-v < \epsilon/2$. Conclude that $|f(x)-f(y)|<\epsilon/2$ whenever $x,y \in (q, N]$ or $x,y \in [M, p)$.

Since $[N,M]$ is a compact interval, $f$ is uniformly continuous on $[M,N]$, so there exists $\delta>0$ such that for all $x,y\in [M,N]$, if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon/2$.

Conclude that for all $x,y \in I$, if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.

• In case the O.P. doesn't know, any continuous $f:J\to R$ is uniformly continuous when $J$ is a closed bounded interval of $R$. Oct 28, 2015 at 2:24

Without loss of generality, one may assume that $$f$$ is nondecreasing. The continuity and boundedness of $$f$$ shows that $$J:=f(I)$$ is a bounded interval. Let the closure of $$J$$ be $$[a,b]$$. To prove that $$f$$ is uniformly continuous, let $$\epsilon>0$$ be given. Make a partition of $$[a,b]$$ such that $$a=y_0 and $$|y_i-y_{i-1}|<\frac {\epsilon} 2$$ for $$1\leq i\leq n$$. For each $$y_i$$ with $$1\leq i\leq n-1$$, let $$f^{-1}(y_i)=x_i$$ (if $$f$$ is not strictly monotone, $$x_i$$ may not be unique, so just choose one in this case). These $$x_i$$'s together with the (possibly infinite) boundary points of $$I$$ form a partition $$P$$ of $$I$$. Note that $$I$$ is the union of the subintervals $$I_1,I_2,\cdots,I_n$$, where $$I_1=I\cap \{x~|~x\leq x_1\},$$ $$I_n=I\cap \{x~|~x\geq x_{n-1}\}$$, and $$I_k=[x_{k-1},x_k]$$ for $$2\leq k\leq n-1.$$ For convenience, one may use less precise notations $$I_1=[x_0,x_1]$$ and $$I_n=[x_{n-1},x_n]$$ with the understanding that $$x_0,x_n$$ may not be in $$I$$, and one sets $$f(x_0):=\lim_{x\rightarrow x_0+}f(x)$$ and $$f(x_n):=\lim_{x\rightarrow x_n^-}f(x),$$ where $$x_0=-\infty$$ or $$x_n=\infty$$ is allowed. Let $$\delta=\|P\|$$ be the norm of the partition, i.e. the minimum length among the subintervals determined by the partition. Now to prove uniform continuity, take $$z_1 with $$z_2-z_1<\delta$$, one needs to show that $$|f(z_2)-f(z_1)|=f(z_2)-f(z_1)<\epsilon.$$ To see this, look at the interval $$[z_1,z_2]$$. By construction, $$(z_1,z_2)$$ can contain at most one of $$x_i$$'s with $$1\leq i\leq n-1,$$ so one has two cases to consider. First if $$(z_1,z_2)$$ contains none of the $$x_i$$'s, then $$[z_1,z_2]$$ is contained in exactly one of the subintervals, say $$I_i=[x_{i-1},x_i],$$ so $$f(z_2)-f(z_1)\leq f(x_i)-f(x_{i-1})<\frac{\epsilon}2<\epsilon.$$ Secondly, if $$(z_1,z_2)$$ contains $$x_i$$ with $$1\leq i\leq n-1$$, then $$[z_1,x_i]\subset I_i=[x_{i-1},x_i]~{\rm and~}[x_i,z_2]\subset I_{i+1}=[x_i,x_{i+1}].$$ By construction, $$f(z_2)-f(z_1)=(f(z_2)-f(x_i))+(f(x_i)-f(z_1))$$$$\leq (f(x_{i+1})-f(x_i))+(f(x_i)-f(x_{i-1})<\frac {\epsilon}2+\frac {\epsilon}2=\epsilon,$$ which combined with the first case finishes the proof.

• Wow, you developed perfectly my first idea. Good answear! =) Jul 12, 2021 at 15:09