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Supose $f$ is increasing, bounded and continuous on $[a,+\infty)$.

Is $f$ uniformly continuous ?

I think yes. how to prove that?

My idea is to show there exists $X$ , $f$ is uniformly continuous on $[X,+\infty)$.

How to fix such $X$?

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Since the function is bounded, $s=\sup\{f(x):x\geq a\}$ exists. let $\epsilon>0$. Pick $X$ such that $f(X)>s-\epsilon$. Such an $X$ exists by the definition of the supremum. Since $f$ is incresing, for all $y\geq X$, $f(X)\leq f(y)\leq s$. So if $y_1\geq X$ and $y_2\geq X$, $|f(y_1)-f(y_2)|<\epsilon$. –  Michael Greinecker Feb 3 '12 at 13:24
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2 Answers

up vote 3 down vote accepted

Since $f$ is increasing and bounded, there exists $l\in\mathbb R$ such that $\lim_{x\to+\infty}f(x)=l$. Fix $\varepsilon>0$; there is $t_0>a$ such that if $x\geq t_0$ then $|f(x)-l|\leq\varepsilon/2$. Using the continuity of $f$, we get that $f$ is uniformly continuous on $[a,t_0+1]$, so there is a $\delta\in (0,1)$ such that if $a\leq x,y \leq t_0+1$ and $|x-y|\leq \delta$ then $|f(x)-f(y)|\leq\varepsilon/2$. Now, let $x,y\geq a$ such that $|x-y|\leq \delta$. If $x,y\in [a,t_0]$ we have $|f(x)-f(y)|\leq\varepsilon$; if $x, y>t_0$ then $|f(x)-f(y)|\leq |f(x)-l|+|f(y)-l|\leq \varepsilon$ and if $x\leq t_0$ and $y>t_0$ then $y\in [a,t_0+1]$ so $|f(x)-f(y)|\leq\varepsilon$.

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Davide, if you don't mind can you clarify somethings for me. There is a lot going on here and I am confused. In your third line, why is $f$ uniformly cont. on $[a,t_0+1]$ where did the $t_0+1$ come from and why does that equate to $\delta\in (0,1)$ where did the $(0,1)$ come from? –  Q.matin Apr 22 '13 at 6:36
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Yes, it is. Let $\epsilon > 0$ Since $f$ is bounded and increasing, $\lim_{x\to\infty }f(x)$ exists; let us denote this by $M$. Choose $R\ge a$ so that $f(R) > M - \epsilon//2$. Since $f$ is continuous on $[a, R]$ is is uniformly continuous there. Chose $\delta > 0$ so $|x - y|< \delta \implies |f(x - f(y)| < \epsilon/2$.

I is an easy matter now to argue that on the entire line if $|x - y| < \delta\implies |f(x)-f(y)|< \epsilon.$

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