convex function with global minimum but no local minimum

Question

I am trying to find a convex set and function that satisfy having a global minimum but not a local minimum. I have been told that this is achievable but I am having a hard time conceptualizing such a scenario.

Thanks for any help.

Edit:

A local minimizer is defined in terms of an open ball set:

$$B_R(x_0)=\{x\in\mathbb{R}^n|∥x-x_0∥_2<R\}$$

where $x_0\in{\mathbb{R}^n}$ and $R>0$. And a local minimizer is defined as a point $x^*\in\Omega$ such that $R>0$ and $B_R(x)\in\Omega$ with:

$$f(x^*)\leq{f(x)}$$

for $x\in{B_R(x)}$

I don't think this is possible. The global minimum is the minimum point in the open neighborhood consisting of the entire space, which makes it a local minimum as well. — Joseph Camacho, Jun 14 '22 at 05:09
@JosephCamacho I found the solution to the problem. Check my answer. — binky, Jun 14 '22 at 05:48
This does not make sense. A global $\min$ is a local $\min$. — copper.hat, Jun 14 '22 at 05:51

score 0 · Accepted Answer · answered Jun 14 '22 at 05:48

Consider the case where $f(x) = x$ on the interval $I = [-1, 1]$. It is certain that the global minimum occurs at $x=-1$. However, if we consider the definition of a open ball:

$$B_R(x_0)=\{x\in\mathbb{R}^n|∥x-x_0∥_2<R\}$$

Then a local minimizer is defined as a point in the set (in this case $I$) if there exists a radius $R>0$ such that $B_R(x)\subset{I}$. In this situation, at $x=-1$, the open ball set is not a proper subset of interval/domain. Thus, there exists a global minimum but not a local minimum.

That's not the usual definition of a local minimum. If that's what you are using, then that definition belongs into the question, not the answer. — dxiv, Jun 14 '22 at 06:21

convex function with global minimum but no local minimum

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