# Proving uniform continuity on an interval

If I know that $f:(0,\infty)\rightarrow \Bbb R$ is uniformly continuous on the intervals $[a,\infty)$ and $(0,a]$, where $a$ is in $(0,\infty)$, how can I prove that it is uniformly continuous on $(0,\infty)$? I know the general definition of uniform continuity using epsilon-delta, but I am not sure how to apply it to the above.

Thanks

Edit: I meant Uniformly continuous on the first two intervals

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That $f$ is continuous on $(0,a]$ and on $[a,+\infty)$ is equivalent to the fact that $f$ is continuous on $(0,+\infty)$, which does not imply that $f$ is uniformly continuous on $(0,+\infty)$. – Did Nov 10 '11 at 18:11
Do you want uniform continuity on those intervals? If so, select $\delta_1$ and $\delta_2$ separately for each, then take the minimum of the two to serve as your $\delta$ for the entire interval. – David Mitra Nov 10 '11 at 18:14
$f(x)=x^2$ satisfies your hypotheses, but is not uniformly continuous on $(0,\infty)$. If you know that $f$ is uniformly continuous on $(0,a]$ and on $[a,\infty)$, though, that would be a different matter... – Arturo Magidin Nov 10 '11 at 18:14
Sorry I meant uniformly continuous on the first two intervals. – rioneye Nov 10 '11 at 18:55

Set $I=(0,a]$ and $J=[a,\infty)$. Let $\epsilon>0$.

Choose $\delta_1>0$ such that $|f(x)-f(y)|<\epsilon/2$ for all $x\in I$ and $y\in I$ with $|x-y|<\delta_1$.

Choose $\delta_2>0$ such that $|f(x)-f(y)|<\epsilon/2$ for all $x\in J$ and $y\in J$ with $|x-y|<\delta_2$.

Let $\delta=\min\{\delta_1,\delta_2\}$.

Then if $|x-y|<\delta$:

If $x$ and $y$ are both in $I$ or both in $J$, $|f(x)-f(y)|<\epsilon/2<\epsilon$.

If $x\in I$ and $y\in J$, or $x\in J$ and $y\in I$

$$|f(x)-f(y)|\le |f(x)-f(a)|+|f(a)-f(y)|<\epsilon/2 +\epsilon/2=\epsilon.$$

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Thanks. On a side note, I am having trouble with my analysis course, and was wondering if you could suggest a site that offered practice problems on the subject? – rioneye Nov 10 '11 at 21:19

if a function F(x) continuou on any compact subset of (0, T], also it continuou at the zero. Then can we get F(x) continuou on [0, T]? Why?

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