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But by hypothesis g(a) = 0, and by assertion 20of the theorem there exists a neighborhood ŨE (a) at every point of which g(x) = 0,(x)that is, fg(x)is defined in ŨE (a).1584Continuous FunctionsAssertion 40 of Theorem 1 is a consequence of the theorem on the limit of acomposite function, by virtue of whichlim(g ◦ f )(x) = lim g(y) = g(b) = g f (a) = (g ◦ f )(a),BaBbwhich is equivalent to the continuity of (g ◦ f ) at a.However, to apply the theorem on the limit of a composite function, we must verify that for any element UY (b) of the base Bb there exists an element UE (a) of thebase Ba such that f (UE (a)) ⊂ UY (b). But in fact, if UY (b) = Y ∩U (b), then by definition of the continuity of f : E → Y at the point a, given a neighborhood U (b) =U (f (a)), there is a neighborhood UE (a) of a in E such that f (UE (a)) ⊂ U (f (a)).Since the range of f is contained in Y , we have f (UE (a)) ⊂ Y ∩ U (f (a)) = UY (b),and we have justified the application of the theorem on the limit of a composite function.Example 1 An algebraic polynomial P (x) = a0 x n + a1 x n−1 + · · · + an is a continuous function on R.Indeed, it follows by induction from 30 of Theorem 1 that the sum and product of any finite number of functions that are continuous at a point are themselvescontinuous at that point.
We have verified in Examples 1 and 2 of Sect. 4.1 thatthe constant function and the function f (x) = x are continuous on R. It then follows that the functions ax m = a · x · .. . · x are continuous, and consequently them factorspolynomial P (x) is also.P (x)– a quotient of polynomials – is continExample 2 A rational function R(x) = Q(x)uous wherever it is defined, that is, where Q(x) = 0. This follows from Example 1and assertion 30 of Theorem 1.Example 3 The composition of a finite number of continuous functions is continuous at each point of its domain of definition. This follows by induction from asser2tion 40 of Theorem 1.
For example, the function esin (ln | cos x|) is continuous on allof R, except at the points π2 (2k + 1), k ∈ Z, where it is not defined.4.2.2 Global Properties of Continuous FunctionsA global property of a function, intuitively speaking, is a property involving theentire domain of definition of the function.Theorem 2 (The Bolzano–Cauchy intermediate-value theorem) If a function that iscontinuous on a closed interval assumes values with different signs at the endpointsof the interval, then there is a point in the interval where it assumes the value 0.4.2 Properties of Continuous Functions159In logical symbols, this theorem has the following expression.5f ∈ C[a, b] ∧ f (a) · f (b) < 0 ⇒ ∃c ∈ [a, b] f (c) = 0 .Proof Let us divide the interval [a, b] in half.
If the function does not assume thevalue 0 at the point of division, then it must assume opposite values at the endpointsof one of the two subintervals. In that interval we proceed as we did with the originalinterval, that is, we bisect it and continue the process.Then either at some step we hit a point c ∈ [a, b] where f (c) = 0, or we obtaina sequence {In } of nested closed intervals whose lengths tend to zero and at whoseendpoints f assumes values with opposite signs. In the second case, by the nestedinterval lemma, there exists a unique point c ∈ [a, b] common to all the intervals.By construction there are two sequences of endpoints {xn } and {xn } of the intervals In such that f (xn ) < 0 and f (xn ) > 0, while limn→∞ xn = limn→∞ xn = c.By the properties of a limit and the definition of continuity, we then find thatlimn→∞ f (xn ) = f (c) ≤ 0 and limn→∞ f (xn ) = f (c) ≥ 0.
Thus f (c) = 0.Remarks to Theorem 2 10 The proof of the theorem provides a very simple algorithm for finding a root of the equation f (x) = 0 on an interval at whose endpointsa continuous function f (x) has values with opposite signs.20 Theorem 2 thus asserts that it is impossible to pass continuously from positiveto negative values without assuming the value zero along the way.30 One should be wary of intuitive remarks like Remark 20 , since they usuallyassume more than they state. Consider, for example, the function equal to −1 onthe closed interval [0, 1] and equal to 1 on the closed interval [2, 3]. It is clearthat this function is continuous on its domain of definition and assumes values withopposite signs, yet never assumes the value 0.
This remark shows that the propertyof a continuous function expressed by Theorem 2 is actually the result of a certainproperty of the domain of definition (which, as will be made clear below, is theproperty of being connected).Corollary to Theorem 2 If the function ϕ is continuous on an open interval andassumes values ϕ(a) = A and ϕ(b) = B at points a and b, then for any number Cbetween A and B, there is a point c between a and b at which ϕ(c) = C.Proof The closed interval I with endpoints a and b lies inside the open intervalon which ϕ is defined.
Therefore the function f (x) = ϕ(x) − C is defined andcontinuous on I . Since f (a) · f (b) = (A − C)(B − C) < 0, Theorem 2 impliesthat there is a point c between a and b at which f (c) = ϕ(c) − C = 0.Theorem 3 (The Weierstrass maximum-value theorem) A function that is continuous on a closed interval is bounded on that interval. Moreover there is a point in5 We recall that C(E) denotes the set of all continuous functions on the set E. In the case E = [a, b]we often write, more briefly, C[a, b] instead of C([a, b]).1604Continuous Functionsthe interval where the function assumes its maximum value and a point where itassumes its minimal value.Proof Let f : E → R be a continuous function on the closed interval E = [a, b].By the local properties of a continuous function (see Theorem 1) for any pointx ∈ E there exists a neighborhood U (x) such that the function is bounded on the setUE (x) = E ∩ U (x).
The set of such neighborhoods U (x) constructed for all x ∈ Eforms a covering of the closed interval [a, b] by open intervals. By the finite covering lemma, one can extract a finite system U (x1 ), . . . , U (xn ) of open intervals thattogether cover the closed interval [a, b]. Since the function is bounded on each setE ∩ U (xk ) = UE (xk ), that is, mk ≤ f (x) ≤ Mk , where mk and Mk are real numbersand x ∈ UE (xk ), we havemin{m1 , . . . , mn } ≤ f (x) ≤ max{M1 , . .
. , MN }at any point x ∈ E = [a, b]. It is now established that f (x) is bounded on [a, b].Now let M = supx∈E f (x). Assume that f (x) < M at every point x ∈ E. Thenthe continuous function M − f (x) on E is nowhere zero, although (by the definition of M) it assumes values arbitrarily close to 0. It then follows that the function1M−f (x) is, on the one hand, continuous on E because of the local properties of continuous functions, but on the other hand not bounded on E, which contradicts whathas just been proved about a function continuous on a closed interval.Thus there must be a point xM ∈ [a, b] at which f (xM ) = M.1Similarly, by considering m = infx∈E f (x) and the auxiliary function f (x)−m,we prove that there exists a point xm ∈ [a, b] at which f (xm ) = m.We remark that, for example, the functions f1 (x) = x and f2 (x) = x1 are continuous on the open interval E = ]0, 1[, but f1 has neither a maximal nor a minimalvalue on E, and f2 is unbounded on E.
Thus, the properties of a continuous functionexpressed in Theorem 3 involve some property of the domain of definition, namelythe property that from every covering of E by open intervals one can extract a finitesubcovering. From now on we shall call such sets compact.Before passing to the next theorem, we give a definition.Definition 1 A function f : E → R is uniformly continuous on a set E ⊂ R if forevery ε > 0 there exists δ > 0 such that |f (x1 ) − f (x2 )| < ε for all points x1 , x2 ∈ Esuch that |x1 − x2 | < δ.More briefly,(f : E → R is uniformly continuous) :== ∀ε > 0 ∃δ > 0 ∀x1 ∈ E ∀x2 ∈ E |x1 − x2 | < δ ⇒ f (x1 ) − f (x2 ) < ε .Let us now discuss the concept of uniform continuity.10 If a function is uniformly continuous on a set, it is continuous at each point ofthat set.
Indeed, in the definition just given it suffices to set x1 = x and x2 = a, and4.2 Properties of Continuous Functions161we see that the definition of continuity of a function f : E → R at a point a ∈ E issatisfied.20 Generally speaking, the continuity of a function does not imply its uniformcontinuity.Example 4 The function f (x) = sin x1 , which we have encountered many times, iscontinuous on the open interval ]0, 1[ = E. However, in every neighborhood of 0in the set E the function assumes both values −1 and 1. Therefore, for ε < 2, thecondition |f (x1 ) − f (x2 )| < ε does not hold.In this connection it is useful to write out explicitly the negation of the propertyof uniform continuity for a function:(f : E → R is not uniformly continuous) :== ∃ε > 0 ∀δ > 0 ∃x1 ∈ E ∃x2 ∈ E |x1 − x2 | < δ ∧ f (x1 ) − f (x2 ) ≥ ε .This example makes the difference between continuity and uniform continuityof a function on a set intuitive.
To point out the place in the definition of uniformcontinuity from which this difference proceeds, we give a detailed expression ofwhat it means for a function f : E → R to be continuous on E:(f : E → R is continuous on E) :== ∀a ∈ E ∀ε > 0 ∃δ > 0 ∀x ∈ E |x − a| < δ ⇒ f (x) − f (a) < ε .Thus the number δ is chosen knowing the point a ∈ E and the number ε, and sofor a fixed ε the number δ may vary from one point to another, as happens in thecase of the function sin x1 considered in Example 1, or in the case of the functionloga x or a x studied over their full domain of definition.In the case of uniform continuity we are guaranteed the possibility of choosing δknowing only ε > 0 so that |x − a| < δ implies |f (x) − f (a)| < ε for all x ∈ E anda ∈ E.Example 5 If the function f : E → R is unbounded in every neighborhood of afixed point x0 ∈ E, then it is not uniformly continuous.Indeed, in that case for any δ > 0 there are points x1 and x2 in every 2δ neighborhood of x0 such that |f (x1 ) − f (x2 )| > 1 although |x1 − x2 | < δ.Such is the situation with the function f (x) = x1 on the set R\0.
In this casex0 = 0. The same situation holds in regard to loga x, which is defined on the set ofpositive numbers and unbounded in a neighborhood of x0 = 0.Example 6 The function f (x) = x 2 , which is continuous on R, is not uniformlycontinuous on R.1624Continuous Functions√√In fact, at the points xn = n + 1 and xn = n, where n ∈ N, we have f (xn ) =n + 1 and f (xn ) = n, so that f (xn ) − f (xn ) = 1. But√√lim ( n + 1 − n) = lim √n→∞n→∞1n+1+√ = 0,nso that for any δ > 0 there are points xn and xn such that |xn − xn | < δ, yet f (xn ) −f (xn ) = 1.Example 7 The function f (x) = sin(x 2 ), which is continuousand bounded on R,is not uniformly continuous on R.
Indeed, at the points xn = π2 (n + 1) and xn =π2 n, where n ∈ N, we have |f (xn ) − f (xn )| = 1, while limn→∞ |xn − xn | = 0.After this discussion of the concept of uniform continuity of a function and comparison of continuity and uniform continuity, we can now appreciate the followingtheorem.Theorem 4 (The Cantor–Heine theorem on uniform continuity) A function that iscontinuous on a closed interval is uniformly continuous on that interval.We note that this theorem is usually called Cantor’s theorem in the literature. Toavoid unconventional terminology we shall preserve this common name in subsequent references.Proof Let f : E → R be a given function, E = [a, b], and f ∈ C(E).
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