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I'm having trouble on Part (c). I read the solution, which assumes that $f$ has a minimum on the interval $[a, b]$, but we have not proved that convexity has any implication of continuity or anything. How does one show that $f$ does indeed have a minimum on this interval?

minimario
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  • are you sure that you aren't meant to assume $f$ is continuous? – operatorerror Jan 07 '18 at 16:31
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    Aren't you still assuming $f$ continuous? Because otherwise $f(0)=1$ and $f(x)=e^{x}-1$ on $x\in (0,1]$ seems to be convex in $[0,1]$ but doesn't have a minimum, right? – Pedro Jan 07 '18 at 16:31
  • @AndrésE.Caicedo I think that only works on open intervals. If I stand correct, convexity means the set above the graph is convex. This seems true to me for the example I gave in the previous comment and in the answer below, isn't it? – Pedro Jan 07 '18 at 16:37
  • @AndrésE.Caicedo in fact if you read the comments on the question you mentioned you have another example of Michael Grant showing that the interval being open is necessary here – Pedro Jan 07 '18 at 16:41
  • @Pedro You are absolutely right. I didn't read carefully your example. :-) Sorry. – Andrés E. Caicedo Jan 07 '18 at 16:41
  • @AndrésE.Caicedo no problem! It is always good to keep a skeptic attitude (especially with me, as I make mistakes very often) :) – Pedro Jan 07 '18 at 16:44
  • Where's your efforts @James Li. You have created a pattern of copying and pasting images, instead of asking and clarifying your question, and adding only "I'm stuck, I don't know anything, I'm clueless, help me, poor me." Please start acting responsibly, or you'll be found out it no time. – amWhy Jan 08 '18 at 00:42

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The exercise says without assuming $f$ differentiable, but we still assume $f$ to be continuous. Otherwise the solution would be wrong:

Consider the interval $[0,1]$ and define $f(0)=1$ and $f(x)=e^{x}-1$ for all $x\in (0,1]$. This is a non continuous convex function which doesn't have a minimum on $[0,1]$.

Pedro
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  • Thanks, I see now. The problem didn't explicitly say f is continuous, so that's why I was trying to prove it without this condition! – minimario Jan 07 '18 at 16:51
  • Not convinced. For $c=0$, the conditions of (a) are still satisfied: the function is increasing on the right of $0$, and vacuously decreasing on the left of $0$ (because there is nothing on the left of $0$). –  Jan 07 '18 at 17:00
  • I didn't mean to say that the statement isn't true without assuming $f$ continuous (although when I read my answer I see that it can be understood like that), only the solution (unless there is more to the solution than what was mentioned by the OP). For example, in the example I gave, the statement remains true, as you pointed out. – Pedro Jan 07 '18 at 17:07