User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 23:30, 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 $ϵ > 0$ with $ϵ < M - f (c)$ .^{[note 1]} By continuity at $c$ , there exists a $δ > 0$ such that $| t - c | < δ$ implies $| f (t) - f (c) | < ϵ$ . So now pick a point like $c - δ / 2$ , and split the interval into $[a, c - δ / 2]$ and $[c - δ / 2, c]$ .

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

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

If $c < b$ then

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 $ϵ$ does not equal $M - f (c)$ ; choosing this $ϵ$ 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 $ϵ$ does not equal $M - f (c)$ ; choosing this $ϵ$ 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 14: / Line 14: @@
 * But now if <math>t \in [c-\delta/2, c]</math>, then <math>|t-c|<\delta</math>, so <math>|f(t)-f(c)|<\epsilon</math>. 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>.
+Now we can choose <math>M' = \max\{\sup V_{c-\delta/2}, f(c) + \epsilon\}</math>. Then whatever <math>t \in [a,c]</math> happens to be, we can say <math>f(t) \leq M'</math>.
 If <math>c < b</math> then
-If <math>t \leq c-\delta/2</math> then there exists some <math>x \in X</math> such that <math>t < x</math>. This means <math>\sup V_t \leq \sup V_x < M</math> so <math>t \in X</math>. Otherwise if <math>c-\delta/2 \leq t \leq c</math> then <math>|t-c|<\delta</math> so <math>f(t) < f(c) + \epsilon < M</math>. So now what can we say about <math>\sup V_c</math>? We want to say <math>\sup V_c < M</math>. We can do this by showing that there exists a number <math>M' < M</math> such that <math>f(t) < M'</math> for all <math>t \in [a,c]</math>. That way, <math>\sup V_c \leq M' < M</math>. But <math>M' = \sup V_c</math> works.