User:IssaRice/Extreme value theorem: Difference between revisions

Revision as of 05:09, 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 $f (c) < M$ . Choose $ϵ > 0$ with $ϵ < M - f (c)$ .^{[note 2]} Continuity of $f$ at $c$ implies that there exists $δ > 0$ such that $x \in (c - δ, c + δ) \cap [a, b]$ implies $| f (x) - f (c) | < ϵ$ . Now, $c - δ$ cannot be an upper bound of $X$ , so there exists some $x_{0} \in X$ such that $c - δ < x_{0} \leq c$ . Thus for $x \in [a, x_{0}]$ we have $f (x) \leq sup V_{x_{0}} < M$ . Also, for $x \in [x_{0}, c] \subseteq (c - δ, c + δ)$ we have $f (x) < f (c) + ϵ < M$ . So $f$ is bounded above by $max {sup V_{x_{0}}, f (c) + ϵ} < M$ on $[a, c]$ , which means $sup V_{c} < M$ .^{[note 3]} If $c < b$ , then there are points to the right of $c$ where $f$ continues to stay below $f (c) + ϵ < M$ , which contradicts the fact that $c$ is an upper bound of $X$ . So $c = b$ , which means $M = sup V_{b} = sup V_{c} < M$ , a contradiction. 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 $x_{0}$ 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 $x_{0}$ 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 for sake of contradiction that <math>f(c) < M</math>. 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> Continuity of <math>f</math> at <math>c</math> implies that there exists <math>\delta > 0</math> such that <math>x \in (c-\delta, c+\delta)\cap [a,b]</math> implies <math>|f(x)-f(c)|<\epsilon</math>. Now, <math>c-\delta</math> cannot be an upper bound of <math>X</math>, so there exists some <math>x_0 \in X</math> such that <math>c-\delta < x_0 \leq c</math>. Thus for <math>x \in [a,x_0]</math> we have <math>f(x) \leq \sup V_{x_0} < M</math>. Also, for <math>x \in [x_0, c] \subseteq (c-\delta, c+\delta)</math> we have <math>f(x) < f(c) + \epsilon < M</math>. So <math>f</math> is bounded above by <math>\max\{\sup V_{x_0}, f(c) + \epsilon\} < M</math> on <math>[a,c]</math>, which means <math>\sup V_c < M</math>.<ref group="note">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 <math>[a,c]</math> into two parts is that we know two facts about <math>f</math>: (1) near <math>c</math>, continuity shows that <math>f</math> must be close to the value of <math>f(c)</math>; since we assumed <math>f(c) < M</math>, this means we can find a neighborhood around <math>c</math> where <math>f</math> is bounded away from <math>M</math>. (2) up to <math>c</math>, our choice of <math>c</math> means the value of <math>f</math> is bounded away from <math>M</math>. Then we pick <math>x_0</math> as a "handing off point" to pass from one side to the other.</ref> If <math>c < b</math>, then there are points to the right of <math>c</math> where <math>f</math> continues to stay below <math>f(c) + \epsilon < M</math>, which contradicts the fact that <math>c</math> is an upper bound of <math>X</math>. So <math>c=b</math>, which means <math>M = \sup V_b = \sup V_c < M</math>, a contradiction.
+So suppose for sake of contradiction that <math>f(c) < M</math>. 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> Continuity of <math>f</math> at <math>c</math> implies that there exists <math>\delta > 0</math> such that <math>x \in (c-\delta, c+\delta)\cap [a,b]</math> implies <math>|f(x)-f(c)|<\epsilon</math>. Now, <math>c-\delta</math> cannot be an upper bound of <math>X</math>, so there exists some <math>x_0 \in X</math> such that <math>c-\delta < x_0 \leq c</math>. Thus for <math>x \in [a,x_0]</math> we have <math>f(x) \leq \sup V_{x_0} < M</math>. Also, for <math>x \in [x_0, c] \subseteq (c-\delta, c+\delta)</math> we have <math>f(x) < f(c) + \epsilon < M</math>. So <math>f</math> is bounded above by <math>\max\{\sup V_{x_0}, f(c) + \epsilon\} < M</math> on <math>[a,c]</math>, which means <math>\sup V_c < M</math>.<ref group="note">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 <math>[a,c]</math> into two parts is that we know two facts about <math>f</math>: (1) near <math>c</math>, continuity shows that <math>f</math> must be close to the value of <math>f(c)</math>; since we assumed <math>f(c) < M</math>, this means we can find a neighborhood around <math>c</math> where <math>f</math> is bounded away from <math>M</math>. (2) up to <math>c</math>, our choice of <math>c</math> means the value of <math>f</math> is bounded away from <math>M</math>. Then we pick <math>x_0</math> as a "handing off point" to pass from one side to the other.</ref> If <math>c < b</math>, then there are points to the right of <math>c</math> where <math>f</math> continues to stay below <math>f(c) + \epsilon < M</math>, which contradicts the fact that <math>c</math> is an upper bound of <math>X</math>. So <math>c=b</math>, which means <math>M = \sup V_b = \sup V_c < M</math>, a contradiction. The assumption that <math>f(c) < M</math> was false, and we conclude <math>f(c) = M</math>.
-If <math>c < b</math> then by continuity we can find points <math>t</math> to the right of <math>c</math> where <math>\sup V_t < M</math>, which contradicts the fact that <math>c</math> is an upper bound of such points.
-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>.
 ==Takeaways==