The Extreme Value Theorem is a partner to the equally obvious Intermediate Value Theorem. It is stated without proof in almost every first year calculus text.

The Extreme Value Theorem states that:

If f(x) is continous on the closed interval between a and b (that is f(x) is continous at every point on the interval and the interval includes the endpoints), then f(x) is bounded for all x on the interval. So there is some value that f(x) never exceeds and another that f(x) never gets under. But the extreme value theorem is stronger than just that -- It also says that over all the x's on the closed interval there is a value of x that gives you a maximum f(x) and another that gives you a minimum f(x).

All this is saying is that if you fasten one end of a clothes line to one pole and the other to another pole, it doesn't make any difference how much slack you leave or how many branches you drape it through, there will be a highest point on the clothes line and a lowest point on the clothes line.
This is true because:

  1. The clothes line is continuous; and
  2. The clothes line includes its endpoints.

In other words, a function that is continuous on a closed interval cannot run off to infinity or to minus infinity on that interval. It must remain always between some minimum value and some maximum value. Not only that, there must be a point, c, on the interval such that f(c) is the maximum value, and another point, d, on the interval such that f(d) is the minimum.

It is important that the interval be closed -- that is that it include the endpoints. If the interval does not include the endpoints, it is easy to come up with a counterexample:

The following function is continuous on the open interval (that is excluding the endpoints) that runs -1 < x < 1. Yet it is not bounded.

 f(x) =   x   
        x2 - 1

...and it does run for infinity and for minus infinity as it approaches those endpoints, and hence we say it is unbounded on the open interval.

The proof of this theorem is even harder than the proof of the intermediate value theorem, so I will not get into that here :)

Let's give a quick proof. I'll be using the Bolzano-Weierstrass theorem, as well as the fact that continuous functions preserve limits.


First, boundedness. If f weren't bounded above, then for any natural number n we could always find a number xn in the interval with f(xn)>n. By the Bolzano Weierstrass theorem, a subsequence of xn converges to some number x in the closed bounded interval. By construction, f(xn) tends to infinity, and so so does the subsequence in question. But since continuous functions preserve limits, this is a contradiction (we'd have f(x)=infinity). Similarly for boundedness below.


So, f is bounded. Let M=sup f, so f(x) is at most M for all x in the interval. On the other hand, because M is the least upper bound, any number less than M will be exceeded by some value of f. So, for all n, there exists xn with M-1/n<f(xn)<=M. Again by the Bolzano Weierstrass theorem, some subsequence of xn converges to x, and by design, f(xn) converges to M, and then so does the subsequence in question. Since continuous functions preserve limits, we have f(x)=M. Again, lower bounds are similar.

Log in or register to write something here or to contact authors.