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How can I prove that a convex function ƒ defined on some open interval C is continuous on C?

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marked as duplicate by Thursday, amWhy 15 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers 3

up vote 5 down vote accepted

In any closed interval subset of the open interval, the function satisfies $$\displaystyle \frac{|f(x) - f(y)|}{|x-y|} \leq C$$

($\displaystyle C$ depending on the interval).

Intuitively, the slope of the line joining $(x, f(x))$ to $(y,f(y))$ increases if you move $x$ or $y$ to the right.

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elaborating on another answer: for $x\leq y\leq z$ we have $$ \frac{f(y)-f(x)}{y-x}\leq\frac{f(z)-f(x)}{z-x} $$ employing convex combination (i.e. $y=tx+(1-t)z, t\in[0,1]$) $$ y=\frac{z-y}{z-x}x+\frac{y-x}{z-x}z $$ and the definition of convexity $$ f(tx+(1-t)y)\leq tf(x)+(1-t)f(y). $$

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For each $x_0 \in C$, $f$ is both left- and right-differentiable at $x_0$, that is, $$ \mathop {\lim }\limits_{x \uparrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R} \;\; {\rm and} \;\; \mathop {\lim }\limits_{x \downarrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R}, $$ respectively. For a proof, see Theorem 14.5(a) on p. 248 of this book or Theorem 1(2) here. It follows that $f$ is both left- and right-continuous at $x_0$, hence continuous there.

Remark: A convex function on a closed interval need not be continuous at the end points (for example, the function $f$ on $[0,1]$ defined by $f(x)=x^2$ for $|x|<1$, $f(x)=2$ for $x \in \lbrace -1,1 \rbrace)$.

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