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Consider a function $f: X\rightarrow \bar{\mathbb{R}}$ where $X$ is a convex set and is not a singleton. Suppose $f$ is strictly convex. Does it imply that $f$ is properly convex?

My tentative proof is as follows.

Suppose there exists $x^*\in X$ and $x^{**} \in X$ such that $f\left(x^*\right)=-\infty$ and $f\left(x^{**}\right)=+\infty$. Then it violates the notion that $f$ is strictly convex because the sum of $+\infty$ and $-\infty$ is not well-defined. So the assumption is invalid.

Then suppose there exists $x^*\in X$ and $x^{**}\in X$ such that $f\left(x^*\right)=-\infty$ and $f\left(x^{**}\right)<+\infty$. Then for any $\lambda \in \left(0,1\right)$, it is impossible that $f\left(\alpha x^*+\left(1-\alpha\right)x^{**}\right)<\alpha f\left(x^*\right)+\left(1-\alpha\right)f\left(x^{**}\right)=-\infty$. So the assumption is invalid.

Combining the above two results, there does not exist $x^*$ such that $f\left(x^*\right)=-\infty$.

Then suppose $f\left(x\right)=+\infty$ for all $x \in X$. Then take $x^*, x^{**} \in X$ and $\alpha \in \left(0,1\right)$. $f\left(\alpha x^*+\left(1-\alpha\right)x^{**}\right)=+\infty$ contradicts the strict convexity. So the assumption is invalid.

In sum, $f$ is properly convex.

Ypbor
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  • What is your definition of strictly convex for a function with values in $\mathbb R$? Bauschke& Combettes define this notion only for proper convex functions. Your first part of the proof is not correct, as in the definition of convexity $+\infty - \infty$ is either avoided by requiring the inequality only for points in dom $f$ or by setting $+\infty -\infty:=+\infty$. – daw May 09 '23 at 06:37
  • What about the following function: $f(0)= -\infty$, $f(x)=+\infty$ for all $x\ne 0$. Is it strictly convex? – daw May 09 '23 at 06:38
  • @daw By " Bauschke& Combettes", do you mean "Convex Analysis and Monotone Operator Theory in Hilbert Spaces"? Could you please tell me which page you are referring to? – Ypbor May 09 '23 at 09:22
  • yes, Definition 8.6, page 114. – daw May 09 '23 at 09:43
  • @daw According to Definition 8.6, strict convexity is defined only for proper function (you wrote "proper convex function"). I did not find a direct definition of "proper convex function" in Bauschke& Combettes, but I guess that means a function that is both convex and proper? The definition of convex function is Definition 8.1 on page 113, which is based on the definition of convex set (Definition 3.1, page 43). But Definition 3.1 does not explicitly specify how to deal with the case of $+\infty$ and $-\infty$. I was wondering where do you get the setting $+\infty-\infty=+\infty$. – Ypbor May 09 '23 at 14:53
  • See Prop 8.4: the values $+\infty$ are omitted for the right-hand side, which is equivalent to defining $+\infty + a = +\infty$ for all $a\in \bar \R$. – daw May 09 '23 at 15:56
  • @daw Which part of Prop 8.4 are you referring to? Equation (8.1) or (8.2)? – Ypbor May 21 '23 at 13:34
  • On the other hand, the definition of strict convexity (Definition 8.6) presumes proper function. Proposition 8.4 indicates that strict convexity implies convexity. As long as "proper convex" means "proper and convex", then strict convexity implies proper convexity. (my proof may be wrong, but the conclusion is right?) – Ypbor May 21 '23 at 13:55

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