User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 23:20, 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 image of $f$ up to and including $x$ .

Let $M=\sup\{f(x):a\leq x\leq b\}=\sup V_{b}$ and $X=\{x\in [a,b]:\sup V_{x}<M\}$ .

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

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 1]} 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 $[c-\delta /2,c]$ .

Consider $t=c-\delta /2$ . Then since $t<c$ , there exists $x\in X$ such that $t<x$ . So $\sup V_{t}\leq \sup V_{c}<M$ so $f(t)\leq \sup V_{t}<M$ . This also means that for all $x\in [a,c-\delta /2]$ we have $f(x)\leq \sup V_{t}<M$ .
But now if $c-\delta /2\leq t\leq c$ , then This means $f(t)<f(c)+\epsilon <M$ .

Now we can choose $M'=\max\{\sup V_{c-\delta /2},f(c)+\epsilon \}$ .

If $c<b$ then

If $t\leq c-\delta /2$ then there exists some $x\in X$ such that $t<x$ . This means $\sup V_{t}\leq \sup V_{x}<M$ so $t\in X$ . Otherwise if $c-\delta /2\leq t\leq c$ then $|t-c|<\delta$ so $f(t)<f(c)+\epsilon <M$ . So now what can we say about $\sup V_{c}$ ? We want to say $\sup V_{c}<M$ . We can do this by showing that there exists a number $M'<M$ such that $f(t)<M'$ for all $t\in [a,c]$ . That way, $\sup V_{c}\leq M'<M$ . But $M'=\sup V_{c}$ works.

Therefore, $c=b$ , which implies that . So the assumption that $f(c)<M$ was false, and we conclude $f(c)=M$ .

If $c=b$ then $M=\sup V_{b}=\sup V_{c}<M$ , a contradiction.

Notes

↑ 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] 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]

@@ Line 9: / Line 9: @@
 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.
-So suppose <math>f(c) < M</math>. We want to find <math>M' < M</math> such that <math>f(t) < M'</math> for all <math>t \in [a,c]</math>. That would mean that <math>\sup V_c \leq M' < M</math>. To do this, we split the interval into two parts. By continuity of <math>f</math> at <math>c</math>, we can choose <math>\epsilon > 0</math> with <math>\epsilon < M - f(c)</math>.<ref group="note">it is important here that <math>\epsilon</math> does not equal <math>M - f(c)</math>; choosing this <math>\epsilon</math> would be too weak and we would not be able to conclude <math>\sup V_c < M</math>, rather only that <math>\sup V_c \leq M</math>.</ref> So now pick a point like <math>c - \delta/2</math>, and split the interval into <math>[a,c-\delta/2]</math> and <math>[c-\delta/2,c]</math>.
+So suppose <math>f(c) < M</math>. We want to find <math>M' < M</math> such that <math>f(t) < M'</math> for all <math>t \in [a,c]</math>. That would mean that <math>\sup V_c \leq M' < M</math>. To do this, we split the interval into two parts. Choose <math>\epsilon > 0</math> with <math>\epsilon < M - f(c)</math>.<ref group="note">It is important here that <math>\epsilon</math> does not equal <math>M - f(c)</math>; choosing this <math>\epsilon</math> would be too weak and we would not be able to conclude <math>\sup V_c < M</math>, rather only that <math>\sup V_c \leq M</math>.</ref> By continuity at <math>c</math>, there exists a <math>\delta > 0</math> such that <math>|t-c|<\delta</math> implies <math>|f(t)-f(c)|<\epsilon</math>. So now pick a point like <math>c - \delta/2</math>, and split the interval into <math>[a,c-\delta/2]</math> and <math>[c-\delta/2,c]</math>.
 * Consider <math>t = c - \delta/2</math>. Then since <math>t < c</math>, there exists <math>x \in X</math> such that <math>t < x</math>. So <math>\sup V_t \leq \sup V_c < M</math> so <math>f(t) \leq \sup V_t < M</math>. This also means that for all <math>x \in [a,c-\delta/2]</math> we have <math>f(x) \leq \sup V_t < M</math>.
-* But now if <math>c-\delta/2 \leq t \leq c</math>, then  By continuity at <math>c</math>, there exists a <math>\delta > 0</math> such that <math>|t-c|<\delta</math> implies <math>|f(t)-f(c)|<\epsilon</math>. This means <math>f(t) < f(c) + \epsilon < M</math>.
+* But now if <math>c-\delta/2 \leq t \leq c</math>, then   This means <math>f(t) < f(c) + \epsilon < M</math>.
 Now we can choose <math>M' = \max\{\sup V_{c-\delta/2}, f(c) + \epsilon\}</math>.