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If $\Omega$ is a bounded convex subset of $\mathbb{R}^n$ and $f$ is a convex function on $\Omega$, can we say $f$ is lower bounded on $\Omega$? If it is not true, can any one provided a counter example?

Lin Xuelei
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    At the question you linked the function is assumed to be twice differentiable hence continous… then the answer to the question above is quite trivial. Consider that the questioner doesn't make more assumptions to $f$ then convexity. – Gono Jul 18 '18 at 09:47

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I assume that $f$ maps into the real numbers. Then it is continuous on the relative interior of $\Omega$. Using a separation argument on the epigraph proves that $f$ is bounded from below by an affine function. Hence it is bounded on the bounded set.

daw
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  • Isn't it possible that the relative interior of $\Omega$ is the empty set? – Gono Jul 18 '18 at 13:53
  • No, relative interior of convex sets in $\mathbb R^n$ is always non-empty, for example $ri({x}) = {x}$. – daw Jul 18 '18 at 14:41