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If $A\subseteq \mathbb{R}$ is an interval, and $f:A\to \mathbb{R}$ is convex on the interior $A^\circ$ of $A$. Can $f$ then be convex on whole $A$? If not, which condition does $f$ need to make this work?

Mr.MathDoctor
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    Upper semicontinuous on $A$ ? – Gribouillis Jan 07 '22 at 21:41
  • @Gribouillis I was thinking about the continuity at first. Intuitively, it makes sense, but I do not know how to verify mathematically if $f$ truly is convex on the whole set due to continuity. – Mr.MathDoctor Jan 07 '22 at 21:43
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    @Mr.MathDoctor Continuity should certainly be enough. Convexity of $f$ on a set $B$ can be defined as $f(x_1t+(1-t)x_2) \le tf(x_1) + (1-t)f(x_2)$ for all $t\in [0,1], x_1,x_2 B$ so since inequalities are preserved when taking limits, you still have this intequality if you (with $A=(a,b)$) let $x_1\to a$ and $x_2 \to b$ – Snildt Jan 07 '22 at 21:49
  • @Snildt Nice, I forgot about the preservation of limits! l use a slightly different definition of convextiy. A function $f:A\to \mathbb{R}$ is said to be convex if $$ \frac{(x_3-x_2)f(x_1)+(x_1-x_3)f(x_2)+(x_2-x_1)f(x_3)}{(x_1-x_2)(x_2-x_3)(x_3-x_1)}\geq 0 $$ for all distinct $x_1,x_2,x_3\in A$. If $f$ is convex on $A^\circ$, I can do your way, but there are three points; what should I do? Letting $x_1\to a$, $x_2\to b$ and $x_3$ be the midpoint of $a,b$ (where $A=[a,b]$, $a<b$)? – Mr.MathDoctor Jan 07 '22 at 22:02
  • @Mr.MathDoctor Yeah, I think that should work – Snildt Jan 08 '22 at 15:50

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