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Consider a convex subdifferentiable function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ (meaning that at every point in the domain of $f$, the subdifferential is non-empty).

Are there some minor conditions that I can impose on $f$ so that all the sublevel sets of $f$ are bounded?

I know that if $f$ is strongly convex then all the sublevel sets of $f$ are bounded, but that eliminates many convex functions that I care about. If $f$ is closed, then all the sublevel sets of $f$ are closed, but this doesn't necessarily mean that they are bounded. I'm looking for something more general than strong convexity that provides me with the property that all sublevel sets are bounded.

Thank you all for your help!

NicNic8
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1 Answers1

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1/ Since your function $f$ has full domain, it is automatically subdifferentiable everywhere -- you don't have to assume this. It is also automatically continuous and thus closed.

2/ If you have just one sublevel set that is bounded and nonempty, then all its sublevel sets are bounded.

3/ An equivalent condition is $\lim_{\|x\|\to+\infty}f(x)=+\infty$.

4/ An equivalent condition is $0$ belongs to the interior of the domain of the Fenchel conjugate $f^*$.

All these results can be found in Rockafellar's Convex Analysis.

max_zorn
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