Continuous Functions on an Interval
Intervals are convex (4.1.4) and closed intervals are compact (4.5.4); these special properties of intervals are reflected in special properties of continuous functions defined on them. Applications include the construction of n’th roots, a characterizati
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Continuous Functions on an Interval §6.1. §6.2. §6.3. §6.4. §6.5. §6.6.
Intermediate value theorem n'th roots Continuous functions on a closed interval Monotonic continuous functions Inverse function theorem Uniform continuity
Intervals are convex (4.1.4) and closed intervals are compact (4.5.4); these special properties of intervals are reflected in special properties of continuous functions defined on them. Applications include the construction of n'th roots, a characterization of injectivity (§6.4) and automatic continuity of inverse functions (§6.5).
6.1. Intermediate Value Theorem If a continuous real-valued function on a closed interval has opposite signs at the endpoints, then it must be zero somewhere in between:
6.1.1. Lemma. If f : [a, b] ---> lR is a continuous function such that f(a)f(b) < 0, then there exists a point c E (a,b) such that f(c) = O.
Proof We can suppose f(a) > 0 and feb) < 0 (otherwise consider - f ). The idea behind the following proof: there are points x in [a, b] (for example, x = a) for which f(x) ?: 0, and b isn't one of them; the 'last' such point x is a likely candidate for c. Theset A={xE[a, b]: f(x)?:O} isnonempty(because aEA)and bounded. It is also closed: for, if Xn E A and Xn ---> x, then x E [a, b] (4.2.7) and f(xn) ---> f(x) by the continuity of f; since f(xn)?: 0 for all n, f(x)?: 0 by 3.4.8, (8) , thus x EA . Let c be the largest element of A (proof of 4.5.7). In particular, f (c) ?: 0, therefore c -I- b, thus c < b. If c < x < b then x ~ A (all elements of A are ~ c), so f(x) < O. Choose a sequence (x n ) with c < Xn < band Xn ---> c; then f(c) = limf(xn) ~ 0, consequently f(c) = O. ¢ 98
S. K. Berberian, A First Course in Real Analysis © Springer Science+Business Media New York 1994
§6.1. Intermediate Value Theorem
99
6.1.2. Theorem. (Intermediate Value Theorem) If I is an interval in lR. and f: I -+ lR. is continuous, then f(I) is also an interval.
Proof Assuming r , s E f(I), r < s , it will suffice to show that [r, s] C f(I) (4.1.4). Let r < k < s; we seek eEl such that f(c) = k. {The message of the theorem: If rand s are values of f, then so is every number between rand s.} By assumption r = f(a) and s = f(b) for suitable points a, b of I; since r -:j:. s, also a -:j:. b. Let J be the closed interval with endpoints a and b (whether a < b or b < a does not interest us); since I is an interval, J c I. Define 9 : J -+ lR. by the formula g(x)=f(x)-k (xEJ). Since
f is continuous, so is
g; moreover,
g(a) = f(a) - k = r - k < 0, g(b) = f(b) - k = s - k > O. By the lemma, there exists a point c E J such that 9 (c) = 0; thus eEl and f(c) - k = 0 , so k = f(c) E f(I). 0 (\I x E I) or f(x) < 0 (\Ix E I).
Proof The alternative is that f(a) < 0 and f(b) > 0 for suitable points a, b of I; then 0 E f(I) by the theorem, contrary to the hypothesis on f. 0 with yn = c. Let
A = {x E lR: x::; 0 } U {x E lR: x B
= {x
E lR:
x> 0 and xn
~
> 0 and xn < c }
c}.
Argue that A has no largest element as follows. Suppose x > 0 , xn < c. The sequence
(x+l/k)n-xn
(k=I,2,3, .
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