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So my question is actually this.

Say I have a function $F:\mathbb R^2\to\mathbb R$. If I find all the potential local extremes by finding the roots of the partial derivatives and I find that only one of them is an actual minimum while the rest are saddles, is it a good enough argument to just find a point that's larger than the local minimum I found and a point that's lower than it to prove that there are no global extremes?

Actually, since there are no local maximums, just finding a point that's lower than the local minimum should be enough, right?

If it isn't, can you provide an example where that isn't the case?

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The short answer to your question is "yes" if your problem is unconstrained (though more work would be required in the constrained case).

I'm going to guess that you classified the stationary points you found (from the roots of the first partial derivatives) by using the second derivative test. If this is the case, then you are correct in saying that there cannot be a global maximum since if no point is even locally maximal, then no point could possibly be globally maximal.

You're also completely correct about the local minimum. If you have a local minimum, $x^*$, then by definition, it is minimal within some neighborhood. If you find a point which produces a lower value of $F$ (which would need to come from outside this neighborhood), then the neighborhood within which $x^*$ is a minimizer must be bounded. Then $x^*$ is not minimal over all of $\mathbb{R}^2$ and $x^*$ is not a global minimizer.

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