AP Calculus AB and BC

When the Extreme Value Theorem Fails

The Extreme Value Theorem needs continuity on an interval that is both closed and bounded. Break continuity and the function can run off to infinity inside the interval. Unbound the interval and it can run off at the far end. Either way there is no maximum to find.

Extreme Value Theorem

∃ c,d∈[a,b]:f(c)≤f(x)≤f(d)  ∀x∈[a,b]\exists\, c, d \in [a,b] : f(c) \le f(x) \le f(d) \ \ \forall x \in [a,b]

If f is continuous on a closed, bounded interval, then f actually attains an absolute maximum and an absolute minimum somewhere on that interval.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

  1. 1

    f is continuous on [a, b]

    f∈C[a,b]f \in C[a, b]

    Continuity is what stops the function from having a vertical asymptote inside the interval. Without it there is nothing to keep the values finite, and a set of values with no upper bound has no largest element to attain.

    Drop it and the theorem fails

    f(x)=1x  (x≠0),f(0)=0f(x) = \frac{1}{x} \ \ (x \ne 0), \qquad f(0) = 0

    f(x) = 1/x on [0, 1], with f(0) = 0

    The interval is closed and bounded, exactly as required, and the function is defined at every point of it. Only continuity fails, at the single point x=0x = 0. As xx shrinks the values climb without limit, so there is no largest value anywhere on the interval. No maximum exists to be attained.

  2. 2

    The interval is bounded

    [a,b] with a,b finite[a, b] \text{ with } a, b \text{ finite}

    A continuous function on an unbounded interval has all the smoothness in the world and still nowhere to stop. Boundedness is what makes the domain compact, and compactness is what the proof actually uses.

    Drop it and the theorem fails

    f(x)=x  on [0,∞)f(x) = x \ \text{ on } [0, \infty)

    f(x) = x on [0, infinity)

    Nothing could be more continuous than the identity function, and the domain contains its left endpoint. But it climbs forever, so the set of values has no upper bound and no maximum. Boundedness of the interval was doing real work.

  3. 3

    The interval contains its endpoints

    [a,b], not (a,b)[a, b] \text{, not } (a, b)

    On the open interval (0,1)(0,1) the function f(x)=xf(x) = x is continuous and bounded, with values approaching 0 and 1 but never reaching either. The supremum is 1 and the infimum is 0, and neither is attained, because the only points where they would be attained were removed from the domain. There is no counterexample function to display here, because the failure is in the domain rather than in the function: the same formula on the closed interval satisfies the theorem perfectly.

    No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

All 2 counterexamples above are checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

  1. First show ff is bounded. If it were not, there would be points xnx_n with ∣f(xn)∣|f(x_n)| climbing past every bound.
  2. A bounded sequence in a closed bounded interval has a convergent subsequence whose limit stays in the interval. This is the step that uses both closedness and boundedness.
  3. Continuity would force ff along that subsequence to converge to ff of the limit point, a finite number, contradicting that the values were climbing without bound. So ff is bounded.
  4. Now let MM be the least upper bound of the values. Pick points where ff climbs toward MM, extract a convergent subsequence again, and continuity gives ff of the limit point equal to MM. That point is where the maximum is attained.

What it does not say

It tells you where the maximum is.

It only says one exists. Finding it is the candidates test: check the critical points and both endpoints, then compare. The theorem is the licence to compare a finite list, nothing more.

Continuity alone is enough.

It is not. f(x)=xf(x) = x is continuous on all of the real line and has neither a maximum nor a minimum. The interval has to be closed and bounded as well.

A discontinuous function cannot attain a maximum.

It very often does. The theorem gives a guarantee under its hypotheses, not a prohibition without them. A step function on [0,1][0,1] can attain its maximum happily.

Frequently asked questions

Why does the Extreme Value Theorem need a closed interval?

Because the extreme value can sit at an endpoint, and an open interval does not contain its endpoints. On (0,1)(0,1) the function f(x)=xf(x) = x gets arbitrarily close to 1 without ever reaching it, so the supremum exists but the maximum does not.

Is the Extreme Value Theorem the same as the candidates test?

No. The Extreme Value Theorem says the extremes exist. The candidates test is the procedure for finding them once you know they exist, by evaluating at the critical points and the endpoints.