I had an idea for a proof of the Extreme Value Theorem, and I was wondering if it was valid. Any advice you have would be greatly appreciated. Thank you!
Prove: If a function f(x) is continuous on a closed interval [a,b], then f(x) has both a maximum and a minimum on [a,b]
Proof: Case 1: f is constant on [a,b]
If f is constant, then for any $x_1$ < $x_2$, f($x_1$) = f($x_2$). Therefore, f($x_1$) ≥ f($x_2$) and f($x_1$) ≤ f($x_2$). So, by definition of maxima and minima, f has at least one of both on [a,b]. This satisfies the theorem.
Case 2: f is strictly increasing or strictly decreasing on [a,b]:
Let's start with strictly increasing. By definition, for each $x_1$ < $x_2$, f($x_1$) ≤ f($x_2$). Because x < b for any x in (a,b), f(x) ≤ f(b). so the endpoint f(b) is a maximum on f on [a,b]. We also know that, because a < x for any x in (a,b), f(a) ≤ f(x). So f(a) is a minimum on [a,b]. This satisfies the theorem. The case for a strictly decreasing function f on [a,b] is similar.
Case 3: f is both strictly increasing and strictly decreasing on [a,b]
Divide the open interval (a,b) into smaller subintervals $U_i$ and $D_i$. On $U_i$, f is strictly increasing and possibly constant, and on $D_i$, f is strictly decreasing and possibly constant. Construct the subintervals so that, if there is a constant section of the graph in a subinterval, it is near the right endpoint of the respective subinterval. Each $U_i$ is followed by a $D_i$, and vice versa.
Assume the function has at least one strictly increasing interval. By the argument for case 2, at the right endpoint of all subintervals $U_i$, the value will be a maximum $M_i$. If f is constant near the right endpoint, the right endpoint is still a max for the total interval $U_i$ by the definition of a maximum. Because each interval $U_i$ is followed by an interval $D_i$, and because f is strictly decreasing on the neighboring $D_i$, $M_i$ will be a maximum for the combined interval $D_i$ + $M_i$. Adding all combined intervals $D_i$ + $M_i$ gives (a,b), so the largest $M_i$ found will be the total maximum/maxima of f on (a,b). Let S = {$M_i$ | $i∈ℕ$, $M∈f$ }. Because every continuous function is bounded, the maxima of f must also be bounded, so S is bounded above. We know S is non-empty because we assume there is at least one strictly increasing interval $U_i$, and therefore at least one $M_i$ in (a,b). Therefore by the Supremum Axiom, there must be a $M_{max}$ ≥ $M_i$ for all maxima in S. The endpoints f(a) and f(b) may be greater than the $M_{max}$, so the endpoints must be compared with $M_{max}$, but no matter the outcome of the comparison, there will still be at least one maximum for f on [a,b].
So, f has at least one maximum on [a,b] if it is strictly increasing on at least one subinterval of (a,b). The proof for f having at least one minimum on [a,b], assuming it is strictly decreasing on at least one subinterval of (a,b), is similar.
This concludes the proof.
