Suppose that $f:(a,b] \rightarrow \mathbb{R}$ is continuous and that the limit as $\lim\limits_{x \rightarrow a}f(x)$ exists. Show that $f$ is uniformly continuous.

I am really struggling with this one. HELP

  • 1
    $\begingroup$ You can start by remembering that $F$ is uniformly continuous on $[a+\epsilon,b]$ for any $\epsilon >0$ (because $F$ is continuous on that interval, which is closed). Now, what you have to do, is try to show that if you pick $\epsilon$ small enough, $F$ is also bounded on $(a,a+\epsilon]$ for the same bound given above. This is where you use that $\lim_{x\rightarrow a} f(x)$ exists. $\endgroup$ – James Apr 20 '15 at 17:32
  • $\begingroup$ You can easily extend $f$ to a function from $[a,b]$ to $\mathbb R$. $\endgroup$ – Akiva Weinberger May 12 '15 at 11:23

Do you already know the fact that continuous functions from a compact set are uniformly continuous? (This follows from the fact that continous functions coming from compact sets always have a maximum)

With this in mind and with the existence of the limit you get that your function can be extended to $[a,b]$, which is compact. Since the extension is uniformly continuous the restriction to $(a,b]$ is uniformly continuous as well.


"A continuous function on a closed set is always Uniform"

if it is not closed then just check whether the limit exists at the end points, if it exists then the function is uniformly continuous otherwise not.

  • $\begingroup$ Not true...a continuous function on a compact set is uniformly continuous. $\endgroup$ – Math1000 May 12 '15 at 11:30
  • $\begingroup$ Actually, he is mentioning the interval (subset of R ) it is always bounded, a closed and bounded set is compact, so what u said is right and even mine. $\endgroup$ – Sam Christopher May 12 '15 at 11:37
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    $\begingroup$ That is correct, but that is not the same as what you said in your answer. It would be correct to say "a continuous function on a closed interval is uniformly continuous". $\endgroup$ – Math1000 May 12 '15 at 17:13

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