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all. I have two questions, one a slightly weaker version than the other.

  1. Suppose $f:(0,1] \to \Bbb R$ is continuous, and that its derivative $f'$ exists and is continuous throughout $(0,1)$. Suppose further that the left-sided limit of $f'$ as $x$ approaches $1$ exists. Is it then true that $f$ is left-differentiable at $1$, with $f'(1)$ being the limit of $f'(x)$ as $x$ approaches $1$?

  2. If (1) is not true in general, is it true when $f$ is strictly increasing on $(0,1]$?

Thanks in advance for any help.

Eric Stucky
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Jeff
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  • The mean value theorem will help. Assume for simplcity that the function is already defined in $x=0$ and is continuous there. Then application of the MVT gives the result realtively direct. However, you still need to show that the function can be extended continuously. Can you show that the function is uniformly continuous? – Quickbeam2k1 Jul 11 '14 at 20:02
  • ^ Sorry...I'm not sure I follow. The function is certainly uniformly continuous on [a,1] for any a>0, and since I'm just concerned with showing that f' exists at 1, this is good enough. – Jeff Jul 11 '14 at 20:10
  • ^ continuation..didn't mean to post yet. But how does the mean value theorem help? It requires differentiability in an open interval. I'm trying to show differentiability on an endpoint. Sorry if I'm missing something obvious. – Jeff Jul 11 '14 at 20:11
  • has a positive answer. The derivative from the left at $1$ has to be the value of the limit. See this slightly different problem. (Note you don't need to assume $f'$ is continuous.)
    • – David Mitra Jul 11 '14 at 20:17
  • Wonderful! That's exactly what I needed...this result is going into my thesis right now. Thank you so much for hunting down that answer. :) – Jeff Jul 11 '14 at 20:21
  • sorry, I mixed up the points zero and one. So the uniform continuity is indeed not an issue there. – Quickbeam2k1 Jul 11 '14 at 20:29