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If $f(x)$ is continuous on $\mathbb R$ and it has a local maximum at $x_0$ and no other maximum or minimum points, then prove that $x_0$ is a global maximum.

I was hinted to use the fact that a continuous function at a closed area $[a,b]$ is bounded, but I don't see how this helps.

Thank you for the help

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    The first sentence tells you the *minimum* cannot be at $x_0$. You also know the minimum is not at $y$ (why?). So the minimum must be somewhere *inside* $(x_0,y)$. Call that point $z$. So $f(z)$ is the smallest value that $f$ takes in all of $[x_0,y]$, and hence in $(x_0,y)$. What does that mean? It means that $f$ has a *local* minimum at $z$ (because $f(z)$ is the smallest value that $f$ takes on the *open* interval $(x_0,y)$).2011-12-08

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Suppose that there is some point $a$ such that $f(a)>f(x_0)$. There’s no harm in assuming that $a; the argument when $a>x_0$ is entirely similar. Consider how $f$ behaves on the interval $[a,x_0]$.

  1. Must $f$ have a minimum on $[a,x_0]$? (This is where the hint is relevant.)
  2. Can its minimum on $[a,x_0]$ occur at $x_0$?
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    @Brian: I agree generally; the possible point of confusion is that if one says "minimum on $\mathbb{R}$", that does suggest globality...2011-12-08
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Suppose that the function attained (at $x=b$) some value greater or equal to your local max (at $x=a$.) Then show that there is a local min between $a$ and $b$.

By the way, this isn't true for functions of more than one variable.