User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 23:38, 1 June 2019

Working through the proof in Pugh's book by filling in the parts he doesn't talk about.

For $x\in [a,b]$ , define $V_{x}=f([a,x])=\{f(t):a\leq t\leq x\}$ to be the values that $f$ takes on as the input ranges from $a$ to $x$ (inclusive).

Let $M=\sup\{f(x):a\leq x\leq b\}=\sup V_{b}$ (this number exists by the boundedness theorem) and $X=\{x\in [a,b]:\sup V_{x}<M\}$ .^{[note 1]}

Our goal now is to find some $x$ such that $f(x)=M$ . If $f(a)=M$ this is immediate.

So now suppose $f(a)<M$ . Then $a\in X$ . We already know that $X$ is bounded above, for instance by the number $b$ . We can thus take the least upper bound of $X$ , say $c=\sup X$ . We already know $f(c)\leq M$ , so if we can just eliminate the possibility that $f(c)<M$ , we will be done.

So suppose $f(c)<M$ . We want to find $M'<M$ such that $f(t)<M'$ for all $t\in [a,c]$ . That would mean that $\sup V_{c}\leq M'<M$ . To do this, we split the interval into two parts. Choose $\epsilon >0$ with $\epsilon <M-f(c)$ .^{[note 2]} By continuity at $c$ , there exists a $\delta >0$ such that $|t-c|<\delta$ implies $|f(t)-f(c)|<\epsilon$ . So now pick a point like $c-\delta /2$ , and split the interval into $[a,c-\delta /2]$ and $[c-\delta /2,c]$ .

Since $c-\delta /2<c$ , there exists $x\in X$ such that $c-\delta /2<x$ (otherwise $c-\delta /2$ would be a smaller upper bound for $X$ ). So $\sup V_{c-\delta /2}\leq \sup V_{x}<M$ . This means that $f(t)\leq \sup V_{c-\delta /2}<M$ for all $t\in [a,c-\delta /2]$ .
But now if $t\in [c-\delta /2,c]$ , then $|t-c|<\delta$ , so $|f(t)-f(c)|<\epsilon$ . This means $f(t)<f(c)+\epsilon <M$ .

Now we can choose $M'=\max\{\sup V_{c-\delta /2},f(c)+\epsilon \}$ . Then whatever $t\in [a,c]$ happens to be, we can say $f(t)\leq M'$ .

If $c<b$ then by continuity we can find points $t$ to the right of $c$ where $\sup V_{t}<M$ , which contradicts the fact that $c$ is an upper bound of such points.

Therefore, $c=b$ , which implies that $M=\sup V_{b}=\sup V_{c}<M$ , a contradiction. So the assumption that $f(c)<M$ was false, and we conclude $f(c)=M$ .

Notes

↑ If we had used " $\leq$ " in the definition of $X$ , then when we take the supremum we would just end up with $b$ , regardless of where $f$ achieves the maximum.
↑ It is important here that $\epsilon$ does not equal $M-f(c)$ ; choosing this $\epsilon$ would be too weak and we would not be able to conclude $\sup V_{c}<M$ , rather only that $\sup V_{c}\leq M$ .

[1] If we had used " $\leq$ " in the definition of $X$ , then when we take the supremum we would just end up with $b$ , regardless of where $f$ achieves the maximum.

[2] It is important here that $\epsilon$ does not equal $M-f(c)$ ; choosing this $\epsilon$ would be too weak and we would not be able to conclude $\sup V_{c}<M$ , rather only that $\sup V_{c}\leq M$ .

[note 1]

[note 2]

@@ Line 5: / Line 5: @@
 Let <math>M = \sup\{f(x) : a \leq x \leq b\} = \sup V_b</math> (this number exists by the boundedness theorem) and <math>X = \{x \in [a,b] : \sup V_x < M\}</math>.<ref group="note">If we had used "<math>\leq</math>" in the definition of <math>X</math>, then when we take the supremum we would just end up with <math>b</math>, regardless of where <math>f</math> achieves the maximum.</ref>
-Our goal now is to find some <math>x</math> such that <math>f(x) = M</math>. If <math>f(a)=M</math> this is easy.
+Our goal now is to find some <math>x</math> such that <math>f(x) = M</math>. If <math>f(a)=M</math> this is immediate.
 So now suppose <math>f(a) < M</math>. Then <math>a \in X</math>. We already know that <math>X</math> is bounded above, for instance by the number <math>b</math>. We can thus take the least upper bound of <math>X</math>, say <math>c = \sup X</math>. We already know <math>f(c) \leq M</math>, so if we can just eliminate the possibility that <math>f(c) < M</math>, we will be done.