The Weierstrass theorem
Not every function has a maximum/minimum. For example, the function f(x) = x2 (for -∞ < x < +∞ ) has no maximum or minimum. Also, even though the function f(x) = arctan(x) has values which are arbitrarily close to π/2, this value in never achieved for any real value of x.
Definition 5.1 - A real-valued function f is called bounded if the following holds:
∃ m,M ∈ ℝ: m ≤ f(x) ≤ M, ∀x ∈ Dom f
If in the above definition we only require the existence of M then we say f is upper bounded; and if we only require the existence of m then say that f is lower bounded.
So a function can be continuous and bounded, yet if M = Sup {f(x): x ∈ Dom f} and m = Inf {f(x) :x ∈ Dom f}, there may be no x value in dom f such that f(x) = M or f(x) = m.
The next theorem gives the conditions that guarantee that a given function has both maximum and minimum value.
Theorem 5.0 (The Weierstrass theorem). If f: [a, b] → ℝ, is continuous on [a,b], then .
f is bounded on [a,b].
f has both a maximum and minimum on [a,b].
Proof. Proposition i) is proved observing that there exists a sequence {xn} ∈ [a,b] ∀n ∈ ℕ, convergent to a bound, considering the upper one:
it is enough to consider a sequence {xn} for the following two cases:
f(xn) > n if sup f(I) = +∞
f(xn) > s − 1/n if sup f(I) = s < +∞
in the case of the upper bounded interval, it follows from the definition of upper bound that:
s − 1/n < f(xn) ≤ s
For n tending to +∞, from the comparison test follows the convergence of f(xn) to s.
To prove ii), we note that {xn} is bounded because contained in [a, b] and by the Theorem of Bolzano-Weierstrass that each bounded sequence has a convergent subsequence {xnk}. Since all xnk are within [a,b] the limit, let call it xM will be also in the same interval xM ∈ [a,b]. By the convergenza delle sottosuccessioni and from the continuity of f:
This relation would be a contradiction in case sup f(I) = ∞. Moreover, given that ξ ∈ Im f, the followin is true:
