Mathematics 242 – Principles of Analysis Discussion 5 – Between the Max and the Min October 20, 2000

Background

Last time, we saw the proof of the Extreme Value Theorem. If f : [a, b] → R is a function that is continuous at every point x in the closed interval [a, b], then f attains its maximum and and minimum values on [a, b] – that is:

a) The set S = {f(x) : x ∈ [a, b]} is bounded above and below in R,

b) By completeness of R, S has a least upper bound, M, and a greatest lower bound, m, AND

c) There exists some x_{1} ∈ [a, b], such that f(x_{1}) = M. Similarly, there is some x_{2} ∈ [a, b] such that f(x_{2}) = m.

Today, we will ask: What happens for y ∈ R between m and M – for y in the range m ≤ y ≤ M?

Discussion Questions

A) Use our intuitive notion of continuity (the graph y = f(x) for x ∈ [a, b] has no “holes, breaks, or jumps”) to formulate a response to the following question: Is it necessarily that case that for each y ∈ [m, M], there exists some (at least one) x ∈ [a, b] such that f(x) = y?

B) Can you prove your conjecture? One suggestion: Suppose you try something like our proof of the Bolzano-Weierstrass Theorem, attempting to “zero in” on a solution of f(x) = y. Consider the midpoint x_{1} = \frac{a+b}{2} of [a, b]. If f(x_{1}) = y, then of course, you are done! But if not, then f has a max M_{1} and a min m_{1} on [a, x_{1}], and a max M′_{1} and a min m′_{1} on [x_{1}, b]. Can you discard one half in such a way that y is still between the max and the min on the half you keep? Now repeat the process on the remaining interval (forever) to generate a nested sequence of intervals:

[a, b] = I_{2} ⊃ I_{2} ⊃ I_{3} ⊃ I_{4} ⊃ ⋯ ⊃ I_{n} ⊃ ⋯

What happens in the limit as n → ∞?

C) Suppose f(x) is a polynomial of odd degree:

f(x) = c_{2k+1}x^{2k+1} + c_{2k}x^{2k} + ⋯ + c_{1}x + x_{0}

(c_{2k+1} ≠ 0). We know that f(x) is continuous at all x ∈ R. Show that the equation f(x) = 0 must have at least one solution x ∈ R.

Assignment

Group write-ups due Wednesday, October 25.