User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 04:49, 29 July 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 $f (x) = M$ . The idea now is to locate the leftmost point where $f$ attains $M$ by taking the supremum of $X$ . But we have a small problem, which is that $X$ might be empty (it is however always bounded, so we don't need to worry about that part). This can happen if $f (a) = M$ , in which case $sup V_{a} = M$ . But if that's the case, we have already found a point where $f$ equals $M$ , so we're actually done!

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 for sake of contradiction that $f (c) < M$ . We want to find $M^{'} < M$ such that $f (t) \leq M^{'}$ 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 2]} 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 $f (t) \leq M^{'}$ .^{[note 3]}

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 $f (c) = M$ .

Takeaways

"less than" vs "bounded away from"

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 $ϵ$ 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 $sup V_{c} \leq M$ .
↑ This part of the proof uses quite a bit of "low-level" argumentation, so it can be easy to miss the broader point. The reason we split the interval $[a, c]$ into two parts is that we know two facts about $f$ : (1) near $c$ , continuity shows that $f$ must be close to the value of $f (c)$ ; since we assumed $f (c) < M$ , this means we can find a neighborhood around $c$ where $f$ is bounded away from $M$ . (2) up to $c$ , our choice of $c$ means the value of $f$ is bounded away from $M$ . Then we pick $c - δ / 2$ as a "handing off point" to pass from one side to the other.

[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 $ϵ$ 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 $sup V_{c} \leq M$ .

[3] This part of the proof uses quite a bit of "low-level" argumentation, so it can be easy to miss the broader point. The reason we split the interval $[a, c]$ into two parts is that we know two facts about $f$ : (1) near $c$ , continuity shows that $f$ must be close to the value of $f (c)$ ; since we assumed $f (c) < M$ , this means we can find a neighborhood around $c$ where $f$ is bounded away from $M$ . (2) up to $c$ , our choice of $c$ means the value of $f$ is bounded away from $M$ . Then we pick $c - δ / 2$ as a "handing off point" to pass from one side to the other.

[note 1]

[note 2]

[note 3]

@@ 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) \leq 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>.
+So suppose for sake of contradiction that <math>f(c) < M</math>. We want to find <math>M' < M</math> such that <math>f(t) \leq 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>.
 * Since <math>c-\delta/2 < c</math>, there exists <math>x \in X</math> such that <math>c-\delta/2 < x</math> (otherwise <math>c-\delta/2</math> would be a smaller upper bound for <math>X</math>). So <math>\sup V_{c-\delta/2} \leq \sup V_x < M</math>. This means that <math>f(t) \leq \sup V_{c-\delta/2} < M</math> for all <math>t \in [a,c-\delta/2]</math>.