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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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Intuition: In general, the reason that a local maximum may fail to be a global maximum is that somewhere else there is a higher value. But if the function "slopes down" from $x_0$ on both sides, and achieves a larger value somewhere else, then at some point it has to stop "sloping down" and start moving up. That will mean that there has to be a local minimum somewhere else (where the function stops going down and starts going up). To make this precise, assume there is a point $y$ where $f(y)\gt f(x_0)$. Think about what happens on $f$ on the interval $[x_0,y]$ (or $[y,x_0]$ if $y\lt x_0$). –  Arturo Magidin Dec 8 '11 at 17:01
    
Yea I was thinking that for each f(x) f(x+-1) has to be smaller then f(x). but the question specifically says to use the concept I wrote in the question and I don't see how that fits in... –  Jhon Dec 8 '11 at 17:04
    
Which is why I said to consider the finite closed interval $[x_0,y]$ (if $x_0\lt y$) or $[y,x_0]$ if $y\lt x_0$. The restriction of the function to such an interval is bounded, and so must have a maximum and a minimum. You are assuming that $f(y)\gt f(x_0)$, and you know that $f$ has a local maximum at $x_0$. Where is the minimum of $f$ on this interval, then? Is it a local minimum? (You are aiming for a contradiction, because you are assuming that there is a point where $f(y)\gt f(x_0)$, and you are trying to show there is no such point). –  Arturo Magidin Dec 8 '11 at 17:08
    
So if there was such a $y$ value then $[x_0,y]$ would contain a minimum and a maximum because it's bounded - but the question says that there is no other min/max point other then $x_0$? is that it? –  Jhon Dec 8 '11 at 17:13
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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 $(x_0,y)$). –  Arturo Magidin Dec 8 '11 at 17:22

2 Answers 2

up vote 3 down vote accepted

Suppose that there is some point $a$ such that $f(a)>f(x_0)$. There’s no harm in assuming that $a<x_0$; 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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Hi, thanks for the answer but could you please elaborate? as you can see above I am kinda struggling :P –  Jhon Dec 8 '11 at 17:28
    
@Jhon: For (1): A continuous function on a closed interval is bounded, so it has a minimum somewhere on the interval. For (2): by the definition of local maximum there is an $\epsilon>0$ such that $f(x_0)>f(x)$ for every $x\in(x_0-\epsilon,x_0+\epsilon)$ except $x_0$ itself. Now pick $u\in(x_0-\epsilon,x_0$; $f(u)<f(x_0)$, so the minimum of $f$ on $[a,x_0]$ isn’t at $x_0$, and it certainly isn’t at $a$ (why?), so it must be in $(a,x_0)$. But then it’s a minimum of $f$ not just on $[a,x_0]$ but on $\mathbb{R}$. –  Brian M. Scott Dec 8 '11 at 17:40
    
@Brian: Well, a local minimum of $f$... –  Arturo Magidin Dec 8 '11 at 17:55
    
@Arturo: I generally assume local unless global is specified, but I should indeed have specified it here. –  Brian M. Scott Dec 8 '11 at 18:14
    
@Brian: I agree generally; the possible point of confusion is that if one says "minimum on $\mathbb{R}$", that does suggest globality... –  Arturo Magidin Dec 8 '11 at 18:15

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.

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