# Existence of minimum in bounded but open set

According to the Extreme Value Theorem, a continuous function achieves at least one minimum and one maximum whenever the set is bounded and closed (i.e. compact).

In my case, I have a bounded and open set, thus, although an infimum exists, a minimum is not assured.

But consider the extra information:

• my continuous function is like $z=f(x,y)$, with $f: A \rightarrow \Bbb R$, and $A \subset \Bbb R^2$
• from the plot I see that at the boundary the function is far from the minimum, and the minimum is at the interior. In other words, the infimum is not at the boundary.

The question is:

How can I formalise this graphical insight in order to apply the Extreme Value Theorem to prove that my function necessarily has a minimum? I am thinking on the following logic:

• continuous function +
• Bounded and open set +
• Infimum not at the boundary +
• Extreme Value Theorem

=> Minimum necessarily exist.

• I just realised that the issue here is not directly related to convex optimisation but to the Extreme Value Theorem. I see that if the set is open, existence of a minimum is not assured (Question 2). If it is closed and has a lower bound, then it does have a minimum (Question 1). Now, this goes back to my original question (not the edited one), where I know that the infimum is not boundary. Although the boundary cannot be part of the solution of the optimisation (for reasons related to the nature of the problem that are not interesting here) could I treat the set as if closed? – luchonacho Aug 13 '15 at 9:46
• In consequence, edited the question. – luchonacho Aug 13 '15 at 10:11

## 1 Answer

I would say yes.

From what you said, it seems that you can continuously extend your function $f$ on $A$ to a function $\tilde{f}$ on $\bar{A}$ (where $\bar{A}$ is the minimum closed set containing $A$) and apply the Extreme Value Theorem to $\tilde{f}$.

Similarly if you know that the infimum is not at the boundary than you know that the infimum is in $B \subset A$ with $B$ closed (and thus compact) and apply the Extreme Value Theorem to $f$ restricted to $B$.

• That makes sense. Is there a way to prove that the infimum is not in the boundary? For example, comparing the Hessian of $f$ on the boundary and elsewhere? I think that the key to my proof is on this step so I need a formalization of this. – luchonacho Aug 13 '15 at 11:24
• @luchonacho: Well for that we would need more information about the function $f$ and the set $A$. If you can show that $f$ is bounded in a neighbourhood of $\partial A$ (where $\partial A$ is the boundary of $A$), it's done! – Tantto Aug 13 '15 at 16:10
• The function is a mix of logarithms, exponentials, inverted error functions, all mixed with powers. Very complex. The set is $0 \leq x < y \leq 1$. The open boundary is $x=y$. What do you mean by a neighbour of $A$? How could I show that $f$ is bounded there? Can you refer me to a theorem or proof or the like that support your suggestion? :) – luchonacho Aug 13 '15 at 16:19
• Note: for a continuous extension to $\overline{A}$ to exist, $f$ must be uniformly continuous. This is not assumed in the question. – Niels J. Diepeveen Aug 13 '15 at 22:51