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I vaguely remember hearing that all global extrema are local extrema long ago. However, now that I look carefully at the definition, it appears that if the domain of a function is closed, then a value can be a global minimum without being a local minimum.

For example, imagine a $f(x)=x^2, 0\leq x \leq 1$. Then, 0 and 1 are a global minimum and maximum, respectively, but since $f(x)$ is not equal to $0$ or $1$ on for $x \in I$ where $I$ is an open interval and contained within the domain of the function, $0$ and $1$ cannot be local extrema.

Is my logic correct or am I missing something? If I'm wrong, what am I missing?

(Sorry for the somewhat "yes" or "no" nature of the question. I just can't find the answer elsewhere.)

Thanks.

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    You are missing the fact that to be a local extremum the required inequality has to hold in a *relative neighborhood* of the point in the domain of the function, that is, in the intersection of an open interval with the domain of the function.2012-01-04

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The interval $(0.9,1]$ is indeed an open subset of the space $[0,1]$, and the maximum at $1$ is a local maximum.

What you say you "vaguely remember" is correct.

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    $(0.9,1]$ is not an open subset of $\mathbb{R}$, but it is an open subset of $[0,1]$.2012-01-05