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I'm trying to prove that a function $f:[0,\infty) \to \mathbb R$ that is continuous, monotone increasing and bounded is uniformly continuous.

Here's a skech of what I've got so far: $f(x) \to L$ for some $L \in \mathbb R$. Fix $\gamma>0$ then $[f^{-1}(0),L-\gamma]:=[a,b]$ is a compact interval and so $f$ is uniformly continuous on $[a,b]$. This is where I'm stuck, If I let $x,y \in [a,b]$ then $\forall \epsilon>0, \exists \delta>0$ such that $0<|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$. If $x,y \notin[a,b]$ then $x,y>b$ and so I know $|f(x)-f(y)| \leq |f(x)-L|+|f(y)-L|$ which are each less than $L-\gamma$. What $\delta$ can I use to bound this expression? And what about the case where $x \in [a,b]$ and $y>b$?

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1 Answer 1

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Since $f$ is bounded and monotone increasing $\lim_{x\rightarrow \infty} f(x)=L $ exist. Now suppose $\varepsilon >0 $ and you want to show the uniform continuity requirement for this $\varepsilon$. Choose $R$ such that for $x>R$, $|f(x)-L|< \varepsilon$. Then for $x, y> R$, by monotonicity, $|f( x)-f(y)| < \varepsilon$.

On the other hand, $[0, R+1]$ is compact, so here you get $\delta >0$ such that $|x-y|<\delta$ implies $|f( x)-f(y)| < \varepsilon$ by standard reasoning. Wlog $\delta < 1$, so all of $[0,\infty) $ is covered.

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