%%visits: 3 ## intuition Suppose f is unbounded. Then ∃xn ∈ [a, b], n = 1, 2, 3… : |f(xn)| → ∞ as n → ∞ (def on unbounded)
By Bolzano-Weierstrass, ∃xnk → c ∈ [a, b] as k → ∞. (the domain)
f is continous, so f(xnk) → f(c) as k → ∞. (tends to a constant, by def)
Contradiction with |f(xnk)| → ∞
Corollary of the boundedness_theorem, If f ∈ C[a, ∞) and ∃limx → ∞f(x) then f is bounded.
Proof. f(x) → A as x → ∞: ∀ϵ > 0∃a0 ≥ a : x ≥ a0 ⟹ |f(x) − A| < ϵ
Take ϵ = 1: $$ a_0 a: x a_0 | f(x) -A | < 1,
| f(x) | = | f(x) - A + A | | f(x) - A | + | A | + | A | $$ f is contious on all intervals greater than a by def, so f is continous on $$ so by the boundedness_theorem: |f(x)| ≤ M for x ∈ [a, a0]. We conclude |f(x)| ≤ max {m, 1 + |A|}
boundedness_theorem requires continuity in a closed, bounded interval
Intuitively this makes sense, however rigour is often a bit harder
than intuition. The proof shows by contradiction that it is an extension
on continuity ## rigour boundedness_theore := If f ∈ C[a, b],
then f is bounded ## exam
clinic ## examples and non-examples ## resources tags :math: