STEP BY STEP guide to prove the Extreme Value Theorem
Definition of the Extreme Value Theorem: If the function f is continuous on [a,b], then there exists constants c and d in the interval [a,b] so that f( c ) is the absolute maximum of f and f(d) is the absolute minimum of f on [a,b].
Sounds *gross* eh? All of that just means given a continuous function f, we can pick an interval [a,b] and we can always find a local maximum and a local minimum in that interval. This is a common theorem covered in first-year calculus courses.
You might think that it’s obvious.
But how do we prove that it’s true for all x? There are an infinite amount of x that we can sub into function f. It’s rather silly to test them all out. It will take us an infinite amount of time, and idk about you but I’m too lazy for that. So, we will prove this theorem, mathematically.
We will prove this theorem by infinitely dividing the interval into halves. I have the written formal proof below, but it’s not important if you just want to know the basic idea of the proof.
We will divide up our closed interval [a,b] into 2 parts.
We then look at the local maxima of each interval.
We then pick the interval containing the maximum value of the original interval to be our new interval.
Notice that everytime we split up our interval, the new one gets smaller. And when we repeat this process an infinite amount of time, our interval gets *REALLY, REALLY, REALLY* small. we want to divide it up to the point where the interval is reallyyyyyy closed to completely *disappear*. This is a very abstract thought.
Remember that while we’re dividing up interval, we’re also choosing the one where the local maximum of the original interval is. So we’re approaching a value where f reaches its maximum. We won’t be arriving exactly where the maximum is ( because then the interval is non-existent ), but we’re getting arbitrarily closed to it. We can call this point c, and the maximum value f( c ). Because the function is continuous on the interval, that point we’re approaching exists and is the maximum. The existence of the minimum can also be proven this way.
Ok so what the fuck did i write in the proof?
The proof I wrote below is a bit more technical, but using the same idea. We divide up the interval n times. Let M be the least upper bound of f(x), meaning f( c ) is smaller or equal to M. Using the epsilon-delta definiton, I proved that f( c ) is larger or equal to M. But for these 2 statements to both be true, f( c ) must be M, the supremum of f(x) on the interval [a, b], and because f( c ) is on f(x), it’s the local maximum.
