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But, since f is monotonic, that interval is contained in the intervalwith endpoints f (a) and f (b). Hence if a monotonic function has a point of discontinuity on the closed interval [a, b], then the closed interval with endpoints f (a)and f (b) cannot be contained in the range of values of the function.Theorem 5 (The inverse function theorem) A function f : X → R that is strictlymonotonic on a set X ⊂ R has an inverse f −1 : Y → R defined on the set Y = f (X)of values of f .
The function f −1 : Y → R is monotonic and has the same type ofmonotonicity on Y that f has on X.If in addition X is a closed interval [a, b] and f is continuous on X, then theset Y = f (X) is the closed interval with endpoints f (a) and f (b) and the functionf −1 : Y → R is continuous on it.Proof The assertion that the set Y = f (X) is the closed interval with endpointsf (a) and f (b) when X = [a, b] and f is continuous follows from Proposition 4proved above.
It remains to be verified that f −1 : Y → R is continuous. But f −1 ismonotonic on Y, Y is a closed interval, and f −1 (Y ) = X = [a, b] is also a closedinterval. We conclude by Proposition 4 that f −1 is continuous on the interval Y withendpoints f (a) and f (b).Example 8 The function y = f (x) = sin x is increasing and continuous on theclosed interval [− π2 , π2 ]. Hence the restriction of the function to the closed interval[− π2 , π2 ] has an inverse x = f −1 (y), which we denote x = arcsin y; this function isdefined on the closed interval [sin(− π2 ), sin( π2 )] = [−1, 1], increases from − π2 toπ2 , and is continuous on this closed interval.Example 9 Similarly, the restriction of the function y = cos x to the closed interval[0, π] is a decreasing continuous function, which by Theorem 5 has an inverse denoted x = arccos y, defined on the closed interval [−1, 1] and decreasing from π to0 on that interval.Example 10 The restriction of the function y = tan x to the open interval X =]− π2 , π2 [ is a continuous function that increases from −∞ to +∞.
By the first partof Theorem 5 it has an inverse denoted x = arctan y, defined for all y ∈ R, andincreasing within the open interval ]− π2 , π2 [ of its values. To prove that the function x = arctan y is continuous at each point y0 of its domain of definition, we takethe point x0 = arctan y0 and a closed interval [x0 − ε, x0 + ε] containing x0 andcontained in the open interval ]− π2 , π2 [. If x0 − ε = arctan(y0 − δ1 ) and x0 + ε =arctan(y0 + δ2 ), then for every y ∈ R such that y0 − δ1 < y < y0 + δ2 we shall havex0 − ε < arctan y < x0 + ε. Hence | arctan y − arctan y0 | < ε for −δ1 < y − y0 < δ2 .The former inequality holds in particular if |y − y0 | < δ = min{δ1 , δ2 }, which verifies that the function x = arctan y is continuous at the point y0 ∈ R.Example 11 By reasoning analogous to that of the preceding example, we establishthat since the restriction of the function y = cot x to the open interval ]0, π[ is a4.2 Properties of Continuous Functions167continuous function that decreases from +∞ to −∞, it has an inverse denotedx = arccot y, defined, continuous, and decreasing on the entire real line R from πto 0 and assuming values in the range ]0, π[.Remark In constructing the graphs of mutually inverse functions y = f (x) and x =f −1 (y) it is useful to keep in mind that in a given coordinate system the points withcoordinates (x, f (x)) = (x, y) and (y, f −1 (y)) = (y, x) are symmetric with respectto the bisector of the angle in the first quadrant.Thus the graphs of mutually inverse functions, when drawn in the same coordinate system, are symmetric with respect to this angle bisector.4.2.3 Problems and Exercises1.
Show thata) if f ∈ C(A) and B ⊂ A, then f |B ∈ C(B);b) if a function f : E1 ∪ E2 → R is such that f |Ei ∈ C(Ei ), i = 1, 2, it is notalways the case that f ∈ C(E1 ∪ E2 ).c) the Riemann function R, and its restriction R|Q to the set of rational numbersare both discontinuous at each point of Q except 0, and all the points of discontinuityare removable (see Example 12 of Sect. 4.1).2.
Show that for a function f ∈ C[a, b] the functionsm(x) = min f (t)a≤t≤xand M(x) = max f (t)a≤t≤xare also continuous on the closed interval [a, b].3. a) Prove that the function inverse to a function that is monotonic on an openinterval is continuous on its domain of definition.b) Construct a monotonic function with a countable set of discontinuities.c) Show that if functions f : X → Y and f −1 : Y → X are mutually inverse(here X and Y are subsets of R), and f is continuous at a point x0 ∈ X, the functionf −1 need not be continuous at y0 = f (x0 ) in Y .4.
Show thata) if f ∈ C[a, b] and g ∈ C[a, b], and, in addition, f (a) < g(a) and f (b) >g(b), then there exists a point c ∈ [a, b] at which f (c) = g(c);b) any continuous mapping f : [0, 1] → [0, 1] of a closed interval into itself hasa fixed point, that is, a point x ∈ [0, 1] such that f (x) = x;c) if two continuous mappings f and g of an interval into itself commute, thatis, f ◦ g = g ◦ f , then they do not always have a common fixed point;d) a continuous mapping f : R → R may fail to have a fixed point;e) a continuous mapping f : ]0, 1[ → ]0, 1[ may fail to have a fixed point;f) if a mapping f : [0, 1] → [0, 1] is continuous, f (0) = 0, f (1) = 1, and (f ◦f )(x) ≡ x on [0, 1], then f (x) ≡ x.1684Continuous Functions5.
Show that the set of values of any function that is continuous on a closed intervalis a closed interval.6. Prove the following statements.a) If a mapping f : [0, 1] → [0, 1] is continuous, f (0) = 0, f (1) = 1, andf n (x) := f ◦ · · · ◦ f (x) ≡ x on [0, 1], then f (x) ≡ x. n factorsb) If a function f : [0, 1] → [0, 1] is continuous and nondecreasing, then forany point x ∈ [0, 1] at least one of the following situations must occur: either x is afixed point, or f n (x) tends to a fixed point. (Here f n (x) = f ◦ · · · ◦ f (x) is the nthiteration of f .)7.
Let f : [0, 1] → R be a continuous function such that f (0) = f (1). Show thata) for any n ∈ N there exists a horizontal closed interval of length n1 with endpoints on the graph of this function;b) if the number l is not of the form n1 there exists a function of this form onwhose graph one cannot inscribe a horizontal chord of length l.8. The modulus of continuity of a function f : E → R is the function ω(δ) definedfor δ > 0 as follows:ω(δ) = sup f (x1 ) − f (x2 ).|x1 −x2 |<δx1 ,x2 ∈EThus, the least upper bound is taken over all pairs of points x1 , x2 of E whosedistance apart is less than δ.Show thata) the modulus of continuity is a nondecreasing nonnegative function having thelimit7 ω(+0) = limδ→+0 ω(δ);b) for every ε > 0 there exists δ > 0 such that for any points x1 , x2 ∈ E therelation |x1 − x2 | < δ implies |f (x1 ) − f (x2 )| < ω(+0) + ε;c) if E is a closed interval, an open interval, or a half-open interval, the relationω(δ1 + δ2 ) ≤ ω(δ1 ) + ω(δ2 )holds for the modulus of continuity of a function f : E → R;d) the moduli of continuity of the functions x and sin(x 2 ) on the whole real axisare respectively ω(δ) = δ and the constant ω(δ) = 2 in the domain δ > 0;e) a function f is uniformly continuous on E if and only if ω(+0) = 0.9.
Let f and g be bounded functions defined on the same set X. The quantityΔ = supx∈X |f (x)−g(x)| is called the distance between f and g. It shows how wellone function approximates the other on the given set X. Let X be a closed interval[a, b]. Show that if f, g ∈ C[a, b], then ∃x0 ∈ [a, b], where Δ = |f (x0 ) − g(x0 )|,and that such is not the case in general for arbitrary bounded functions.7 Forthis reason the modulus of continuity is usually considered for δ ≥ 0, setting ω(0) = ω(+0).4.2 Properties of Continuous Functions16910.
Let Pn (x) be a polynomial of degree n. We are going to approximate a boundedfunction f : [a, b] → R by polynomials. LetΔ(Pn ) = sup f (x) − Pn (x) and En (f ) = inf Δ(Pn ),Pnx∈[a,b]where the infimum is taken over all polynomials of degree n. A polynomial Pn iscalled a polynomial of best approximation of f if Δ(Pn ) = En (f ).Show thata) there exists a polynomial P0 (x) ≡ a0 of best approximation of degree zero;b) among the polynomials Qλ (x) of the form λPn (x), where Pn is a fixed polynomial, there is a polynomial Qλ0 such thatΔ(Qλ0 ) = min Δ(Qλ );λ∈Rc) if there exists a polynomial of best approximation of degree n, there alsoexists a polynomial of best approximation of degree n + 1;d) for any bounded function on a closed interval and any n = 0, 1, 2, .
. . thereexists a polynomial of best approximation of degree n.11. Prove the following statements.a) A polynomial of odd degree with real coefficients has at least one real root.b) If Pn is a polynomial of degree n, the function sgn Pn (x) has at most n pointsof discontinuity.c) If there are n + 2 points x0 < x1 < · · · < xn+1 in the closed interval [a, b]such that the quantity)*sgn f (xi ) − Pn (xi ) (−1)iassumes the same value for i = 0, . . . , n + 1, then En (f ) ≥ min0≤i≤n+1 |f (xi ) −Pn (xi )|. (This result is known as Vallée Poussin’s theorem.8 For the definition ofEn (f ) see Problem 10.)12.
a) Show that for any n ∈ N the function Tn (x) = cos(n arccos x) defined onthe closed interval [−1, 1] is an algebraic polynomial of degree n. (These are theChebyshev polynomials.)b) Find an explicit algebraic expression for the polynomials T1 , T2 , T3 , and T4and draw their graphs.c) Find the roots of the polynomial Tn (x) on the closed interval [−1, 1] andthe points of the interval where |Tn (x)| assumes its maximum value.d) Show that among all polynomials Pn (x) of degree n whose leading coefficient is 1 the polynomial Tn (x) is the unique polynomial closest to zero, that is,En (0) = max|x|≤1 |Tn (x)|. (For the definition of En (f ) see Problem 10.)8 Ch.J.
de la Vallée Poussin (1866–1962) – Belgian mathematician and specialist in theoreticalmechanics.1704Continuous Functions13. Let f ∈ C[a, b].a) Show that if the polynomial Pn (x) of degree n is such that there are n + 2points x0 < · · · < xn+1 (called Chebyshev alternant points) for which f (xi ) −Pn (xi ) = (−1)i Δ(Pn ) · α, where Δ(Pn ) = maxx∈[a,b] |f (x) − Pn (x)| and α is aconstant equal to 1 or −1, then Pn (x) is the unique polynomial of best approximation of degree n to f (see Problem 10).b) Prove Chebyshev’s theorem: A polynomial Pn (x) of degree n is a polynomialof best approximation to the function f ∈ C[a, b] if and only if there are at leastn + 2 Chebyshev alternant points on the closed interval [a, b].c) Show that for discontinuous functions the preceding statement is in generalnot true.d) Find the polynomials of best approximation of degrees zero and one for thefunction |x| on the interval [−1, 2].14.
In Sect. 4.2 we discussed the local properties of continuous functions. Thepresent problem makes the concept of a local property more precise.Two functions f and g are considered equivalent if there is a neighborhood U (a)of a given point a ∈ R such that f (x) = g(x) for all x ∈ U (a). This relation betweenfunctions is obviously reflexive, symmetric, and transitive, that is, it really is anequivalence relation.A class of functions that are all equivalent to one another at a point a is calleda germ of functions at a. If we consider only continuous functions, we speak of agerm of continuous functions at a.The local properties of functions are properties of the germs of functions.a) Define the arithmetic operations on germs of numerical-valued functions defined at a given point.b) Show that the arithmetic operations on germs of continuous functions do notlead outside this class of germs.c) Taking account of a) and b), show that the germs of continuous functionsform a ring – the ring of germs of continuous functions.d) A subring I of a ring K is called an ideal of K if the product of every elementof the ring K with an element of the subring I belongs to I .
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