Let $X\subset\mathbb{R}^n$ be a nonempty convex set such that $\text{int}(X)$ is nonempty and $f:X\rightarrow \mathbb{R}$ be a continuous function. Suppose that $f$ is convex on the interior of $X$. Is $f$ convex everywhere on $X$?
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$\begingroup$ By "interior" do you mean the relative interior of $X$? (Otherwise if $X$ is a nonempty convex set with empty topological interior, any continuous function will be allowed.) $\endgroup$– RigelJun 17, 2017 at 15:03
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$\begingroup$ @Rigel: I'm not sure what the relative interior is exactly (I'm looking it up), but you can assume that the interior of $X$ is nonempty. $\endgroup$– user_lambdaJun 17, 2017 at 15:06
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$\begingroup$ Ok. If $\text{int}(X) \neq \emptyset$, then the relative interior of $X$ coincides with $\text{int}(X)$. $\endgroup$– RigelJun 17, 2017 at 15:11
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1$\begingroup$ Hint: it is enough to consider the one-dimensional case. $\endgroup$– RigelJun 17, 2017 at 15:15
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Yes. If $X$ is a convex set and $\operatorname{int}X\ne \emptyset$, then $\operatorname{cl}\operatorname{int}X=\operatorname{cl}X$. Therefore, for all $x\in X$ there is a sequence $x_n\to x$ with $x_n\in\operatorname{int}X$.
Now, consider $x,y\in X$ and $x_n\to x$, $y_n\to y$ as above. For $t\in[0,1]$, $$f(tx_n+(1-t)y_n)\le tf(x_n)+(1-t)f(y_n)$$ Thus, by continuity, \begin{align}f(tx+(1-t)y)&=\lim_{n\to\infty}f(tx_n+(1-t)y_n)\le \lim_{n\to\infty}tf(x_n)+(1-t)f(y_n)=\\&=tf(x)+(1-t)f(y)\end{align}