User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 23:04, 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 $X = {x \in [a, b] : sup V_{x} < M}$ .

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 $sup V_{c} \leq M^{'} < M$ .

Consider $t = c - δ / 2$ . 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 $x \in [a, c - δ / 2]$ we have $f (x) \leq sup V_{t} < M$ .
But now if $c - δ / 2 \leq t \leq c$ , then by continuity at $c$ , We can 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 $| f (t) - f (c) | < ϵ$ . This means $f (t) < f (c) + ϵ < M$ .

Now we can choose $M^{'} = max {sup V_{c - δ / 2}, f (c) + ϵ}$ .

If $c < b$ then

If $t \leq c - δ / 2$ 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 $c - δ / 2 \leq t \leq c$ then $| t - c | < δ$ so $f (t) < f (c) + ϵ < M$ . 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 $ϵ$ 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] t 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 3: / Line 3: @@
 For <math>x \in [a,b]</math>, define <math>V_x = f([a,x]) = \{f(t) : a \leq t \leq x\}</math> to be the image of <math>f</math> up to and including <math>x</math>.
-Let <math>M = \sup\{f(x) : x \in [a,b]\}</math> and <math>X = \{x \in [a,b] : \sup V_x < M\}</math>.
+Let <math>M = \sup\{f(x) : a \leq x \leq b\} = \sup V_b</math> and <math>X = \{x \in [a,b] : \sup V_x < M\}</math>.
 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.
@@ Line 14: / Line 14: @@
 Now we can choose <math>M' = \max\{\sup V_{c-\delta/2}, f(c) + \epsilon\}</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.
-<math>\sup V_{c-\delta/2} < M</math> so let <math>M' = (M+\sup V_{c-\delta/2})/2</math>. Then <math>\sup V_{c-\delta/2} < M' < M</math> and if <math>v \in V_{c-\delta/2}</math> we have <math>v \leq \sup V_{c-\delta/2} < M'</math>.
+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.
-Therefore, <math>c=b</math>, which implies that <math>M = \sup V_b = \sup V_c < M</math>, a contradiction. So the assumption that <math>f(c) < M</math> was false, and we conclude <math>f(c) = M</math>.
+Therefore, <math>c=b</math>, which implies that . So the assumption that <math>f(c) < M</math> was false, and we conclude <math>f(c) = M</math>.
+If <math>c=b</math> then <math>M = \sup V_b = \sup V_c < M</math>, a contradiction.
 ==Notes==
 <references group="note" />