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Suppose $f : [a, b] \rightarrow R$ is continuous on $[a, b]$ and convex on the open interval $(a, b).$ Show that $f$ is convex on the closed interval $[a, b].$

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closed as off-topic by uniquesolution, Claude Leibovici, mrp, Namaste, Shailesh Mar 13 '17 at 14:19

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    $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. If you receive useful answers, consider accepting one. $\endgroup$ – Shailesh Mar 13 '17 at 7:42
  • $\begingroup$ To give you an idea, you will have to use the following very basic property of continuous functions: If $h(x) \geq 0$ on $(a,b)$, then $h(x) \geq 0$ on $[a,b]$. You have to - of course - define $h$ in such a way, that $h(x) \geq 0$ tests convexity of $f$. $\endgroup$ – MooS Mar 13 '17 at 7:50
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    $\begingroup$ I'm voting to close this question as off-topic because it is another "Do-my-homework" question showing zero effort. $\endgroup$ – uniquesolution Mar 13 '17 at 8:07
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It is enough to check that $f\bigl(tx+(1-t)y\bigr)\le tf(x)+(1-t)f(y)$ for any $t\in[0,1]$ and $x\in[a,b]$. If $x,y\in(a,b)$, then we have it by convexity. If, for instance, $x=a,y\in(a,b)$, then take $x_n=a+\frac{1}{n}$ and pass to a limit with $n\to\infty$. The remaining cases are handled in a similar way.

The statement remains true, if $f(a^+)\le f(a)$ and $f(b^-)\le f(b)$. These limits do exist, because $f$ is either monotone, or unimodal in the sense that $f$ decreases (possibly weakly) on $[a,c]$ and increases (possibly weakly) on $[c,b]$ for some $c\in(a,b)$. The monotonic functions admit one-sided limits.

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