Extreme Value Theorem Proof

1. f is continuous on a closed interval

• $\therefore$ there is an absolute max or min on that interval

Proof

1. Suppose $f$ is continuous on closed interval $[a,b]$
2. by Boundedness Theorem, $f$ is bounded on $[a,b]$
3. by Completeness Axiom, $f$ has a supremum and infimum
4. Let $\sup (range(f))=M$
5. Then, $\forall f(x)\in range(f),f(x)\le M$
6. Let $\inf (range(f))m$
7. Then, $\forall f(x)\in range(f),f(x)\ge m$
8. Suppose for sake of contradiction that $\forall x\in [a,b],f(x)\ne M$
   1. Then, this means that $\forall f(x)\in range(f),f(x)<M$
   2. Then $M-f(x)>0$
   3. This means $g(x)=\frac{1}{M-f(x)}>0$
   4. Note that $g(x)$ is continuous as $f(x)$ is cont by Continuity Theorem
   5. Then $g(x)$ is bounded by Boundedness Theorem
   6. Then, there exists a upper bound $K>0$ and $K\ge g(x)$
   7. This means $K\ge \frac{1}{M-f(x)}$
   8. This means $\frac{1}{K}\ge M-f(x)$
   9. This means $f(x)\le M-\frac{1}{k}$
   10. This contradicts that $M$ is the least upper bound
   11. Thus, $\exists x\in [a,b],f(x)=M$
9. Suppose for sake of contradiciton that $\forall x\in [a,b],f(x)\ne m$
   1. Then, this means that $\forall f(x)\in range(f),f(x)>M$
   2. Then $M-f(x)<0$
   3. This means $g(x)=\frac{1}{m-f(x)}<0$
   4. Note that $g(x)$ is continuous as $f(x)$ is cont by Continuity Theorem
   5. Then $g(x)$ is bounded by Boundedness Theorem
   6. Then, there exists a upper bound $K>0$ and $K\le g(x)$
   7. This means $K\le \frac{1}{m-f(x)}$
   8. This means $\frac{1}{K}\le m-f(x)$
   9. This means $f(x)\ge m-\frac{1}{k}$
   10. This contradicts that $m$ is the greatest lower bound
   11. Thus, $\exists x\in [a,b],f(x)=m$
10. Thus, there exists $M$ and $m$ that are attainable by $x\in [a,b]$ so $f(x)$ is guaranteed a local minima/maxima