Given function $f :[0, \infty] \rightarrow \mathbb{R}$. We know that $f$ is uniformly continuous on interval $(0, \infty]$ and continuous on point $0$. How to prove that $f$ is uniformly continuous on $[0, \infty]$?
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What do you mean by "constantly continuous"? – kobe Mar 27 '15 at 04:33
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I was thinking about "uniformly continous" – alex Mar 27 '15 at 04:36
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by chance, do you mean "uniformly continuous" ? – Alexander Mar 27 '15 at 04:36
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If you can prove $f$ is continuous on closed interval, then it will be uniformly continuous. On the other hand, for we don't have closed interval for $\infty$, but I think you should write your interval $[0,\infty)$ – Alexander Mar 27 '15 at 04:39
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The problem is that f'(0) is undefined. – alex Mar 27 '15 at 04:45
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$f$ is uniformly continuous on $(0,\infty)$, hence it's uniformly continuous on any subset such as $[1,\infty)$.
$f$ is continuous on the compact set $[0,2]$, so it is uniformly continuous on $[0,2]$.
Now choose $\epsilon > 0$. We need a $\delta > 0$ such that if $|x-y| < \delta$ then $|f(x)-f(y)| < \epsilon$.
Uniform continuity of $f$ on $[0,2]$ gives us a $\delta_1$ which works if $x,y \in [0,2]$.
Uniform continuity of $f$ on $[1,\infty)$ gives us a $\delta_2$ which works if $x,y \in [1,\infty)$.
If we let $\delta = \min(\delta_1, \delta_2, 1)$, then the condition $|x-y| < \delta$ ensures that at least one of the intervals $[0,2]$ and $[1,\infty)$ contains both $x$ and $y$, and hence $|f(x) - f(y)| < \epsilon$.
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Why $\delta = \min(\delta_1, \delta_2, 1)$? $\delta = \min(\delta_1, \delta_2)$ isn't enough? – alex Mar 27 '15 at 05:00
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@alex Because $\delta_1$ and $\delta_2$ might both be $> 1$, in which case $|x-y| < \min(\delta_1, \delta_2)$ would not ensure that both $x$ and $y$ are contained in the same interval $[0,2]$ or $[1,\infty)$. Of course you can fix this by choosing smaller $\delta_1$ and $\delta_2$, but setting $\delta = \min(\delta_1, \delta_2, 1)$ achieves the same thing. – Mar 27 '15 at 05:05