Unizor - Derivatives - Function Limit - Extreme Value Theorem

Notes to a video lecture on http://www.unizor.com

Extreme Value Theorem

In this lecture we will consider real functions f(x) of real argument x defined on a closed segment.

We will prove the following theorem.

Extreme Value Theorem

A continuous function defined on a segment (finite interval with endpoints) attains its extreme values.
In other words, there is at least one point within its domain where it reaches its maximum and there is at least one point where it reaches its minimum.

Proof

We will prove this theorem for a case of attaining the maximum value. The corresponding proof for minimum is completely analogous.

The proof is based on the boundedness theorem that states that any continuous function f(x) defined on segment [a,b] is bounded from above and below. It was proven in the previous lecture.

That means that a set of real values of our function {f(x)} is bounded from above. According to the axiom of completeness, there must be the least upper bound for this set of values.
Assume, M is this least upper bound for function f(x) on segment S₁=[a,b]:
∀x∈[a,b] f(x) ≤ M

Consider now a set S₂ of all points x in the domain of our function f(x), where the function value is greater than M−1/2.
This set cannot be empty because in this case M−1/2 would be an upper bound that is smaller than the least upper bound M.

Next, consider set S₃ of all points x in the domain of our function f(x), where the function value is greater than M−1/3.
This set cannot be empty because in this case M−1/3 would be an upper bound that is smaller than the least upper bound M.
In addition, set S₃ is a subset of set S₂ since the restrictions on the values of function f(x) are more strict within this set.

Continue this process with set S₄ of all points x in the domain of our function f(x), where the function value is greater than  M−1/4, set S₅ of all points x in the domain of our function f(x), where the function value is greater than M−1/5 etc.
Generally, set S_n contains all points x in the domain of our function f(x), where the function value is greater than M−1/n.
Each of these sets cannot be empty because in this case M−1/n would be an upper bound that is smaller than the least upper bound M.
Every subsequent set S_n+1 is a subset of the previous set S_n since the restrictions on the values of function f(x) are more strict within this set:
S₁⊃S₂⊃S₃⊃...⊃S_n⊃S_n+1⊃...

Now we can pick a single point from each of these sets to obtain a bounded sequence of points {x_n} within segment[a,b], which, according to Bolzano - Weierstrass theorem should have a convergent subsequence {y_k} with the limit also lying within our closed segment [a,b]. Assume this limit is L.
This limit point L must be the point of maximum for function f(x), that is f(L)=M.

To prove this, consider the following:
Since y_k→L as k→∞ and f(x) is continuous, f(y_k)→f(L).
The distance between f(y_k) and the least upper bound M is an infinitesimal as index k increases to infinity because, if y_k∈S_k, |f(y_k)−M| ≤ 1/k.
Therefore, M must be the limit of f(y_k), that is
M = lim_k→∞f(y_k) = f(L)

As we have proven, the maximum M of function f(x) is attained at point L∈[a,b].

End of proof.