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This is a question asked in exam and I tried to use a continuous extension theoram but this theoram doesn't mentions that continuous extension must exist and necessary conditions.
I think I am not thinking in right direction.

Please help

Question is - Let f :(0, 1) -> R be continuous. It can be extended to a continuous function f' : [0, 1] ->R, iff, it is................. ................... Answer -

Uniformly Continuous

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    That's very confusing, what exactly is your question? You wrote it yourself "unformly continuous" (don't know why you hide it). Are you looking for a proof? Or what? – freakish Jul 05 '19 at 13:06
  • Probably the same origin as https://math.stackexchange.com/q/3283768? – Paul Frost Jul 05 '19 at 13:08
  • @freakish I am looking for a proof or logical reasoning as why It must be uniformly continuous –  Jul 05 '19 at 13:14
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    @user686624 For "$\Leftarrow$" see https://math.stackexchange.com/questions/245237/extension-of-a-uniformly-continuous-function-between-metric-spaces . For "$\Rightarrow$" see https://math.stackexchange.com/questions/110573/continuous-mapping-on-a-compact-metric-space-is-uniformly-continuous together with a standard argument that a restriction of uniformly continuous function is uniformly continuous. – freakish Jul 05 '19 at 13:23
  • FYI - hiding answers is something that makes sense on the puzzle forum. Here, the only purpose it serves is to mildly aggravate readers, who are forced to highlight the text in order to understand what you are asking. – Paul Sinclair Jul 06 '19 at 00:31

1 Answers1

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A continuous extension exists iff
lim$_{x->0^+}$f(x) and lim$_{x->1^-}$f(x) are finite.
As f' is a continuous function with compact domain,
it is uniformly continuous. Thus f is uniformly continuous.

William Elliot
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