Extreme Value Theorem

If f is continuous on [a,b] then f has a maximum and a minimum on the interval.

By the Bounded Value Theorem, f(x) is bounded on [a,b].  That means f has an upper bound, and by completeness, f has a least upper bound.

Let M be the least upper bound of f on the interval [a,b].

Assume f has no maximum on the interval, which means that f(x) < M for all x in [a,b].

Then g(x) = 1 / (M-f(x)) is continuous on [a,b], which by the Bounded Value Theorem, means g(x) has an upper bound, say, N.

N ³ g(x) = 1 / (M-f(x)) for all x in [a,b], so

0 < 1/N £ M-f(x), so

f(x) < f(x)+1/(2N) < f(x)+1/N £ M, which means there is a number strictly between f(x) and M, contradicting the choice of M as the least upper bound of f.

How is this theorem used?

Rolle's Theorem -- that if f(a)=f(b) then f'(c)=0 for some c in (a,b) -- depends on two key facts: that any function has a maximum on a closed interval (the Extreme Value Theorem), and that Relative Extrema Occur Only at Critical Numbers.

Internet References

Source of this proof: Cut-the-knot

Related Pages in this Website

Go back to Calculus Home

Upper Bound -- definition of "upper bound" and "least upper bound" of either sets or functions

Relative Extrema Occur Only at Critical Numbers -- if c is an extremum then f'(c)=0 or f'(c) is undefined

Rolle's Theorem -- that if f(a)=f(b) then f'(c)=0 for some c in (a,b)

The webmaster and author of the Math Help site is Graeme McRae.
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