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If $f$ is uniformly continuous on every finite subinterval of $[a,\infty)$, then is it necessary that $f$ is uniformly continuous on $[a,\infty)$?

I thought a counter example $f(x)=x^2$.

Let $[a,b]$ be any finite interval.

Then for a given $\epsilon$ $>$ $0$ there exist $\delta= \frac{\epsilon}{2|c|} >0$ such that where $c=\max\{|a|,|b|,|b-a|\}$.

$|x-y|<\delta$ $\implies |f(x)-f(y)|\implies |x^2-y^2|=|x-y||x+y|< \delta(|x|+|y|)<\delta\cdot2|c|<\epsilon$

But $x^2$ is not uniformly continuous on $[a,\infty)$.

I don't know whether I miss something or my proof is incomplete.

Help me and give me example of a function which follow this property.

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    $\begingroup$ What you have done is fine. There is no better example. $\endgroup$ Jun 23 at 10:16

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