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Math 521 – Fall 2008
July 28, 2009
The text references in these notes are to R.C. Buck,Advanced Calculus, 3rd edition, Waveland Press. In these notes we shall often use the abbrevia- tions =⇒ for “implies”, ⇐⇒ for “if and only if”, ∃ for “there exists”, ∀ for “for all”.
These abbreviations help clarify the logical structure of the definitions and theorems. The following standard notations are also used:
Z+ the set of positive integers (natural numbers). N the set of nonnegative integers. Z the set of integers. Q the set of rational numbers. R the set of real numbers. C the set of complex numbers.
The usual interval notation is used, e.g.
[a, b) := {x ∈ R : a ≤ x < b}, (−∞, b] := {x ∈ R : x ≤ b}, etc.
We write := to indicate that two objects are equal by definition. We also signal definitions by writing iff instead of if and only if.
Throughout Rn^ denotes the vector space of all ntuples of real numbers, so R^1 = R, R^2 = R × R, R^3 = R × R × R, etc. (Buck uses the term n space as a synonym for Rn.)
1 Sets, Functions, and Maps
- The notation x ∈ A means that the point x is an element of the set A. The notation A ⊆ B means that A is a subset of B, i.e. for all x we have x ∈ A =⇒ x ∈ B. By definition, two sets are equal if each is a subset of the other: A = B ⇐⇒ A ⊆ B and B ⊆ A.
The notation {x : P (x)} denotes the set of all x for which the property P (x) is true. The notation {x ∈ A : P (x)} denotes the set of all x ∈ A for which the property P (x) is true. If A and B are sets, then the sets
A ∪ B := {x : x ∈ A or x ∈ B}, A ∩ B := {x : x ∈ A and x ∈ B},
are called respectively the union and intersection of A and B. The empty set is denoted ∅: x /∈ ∅ for all x. Two sets are disjoint iff they have no elements in common, i.e iff A ∩ B = ∅. (See Buck page 5.) The set
B \ A := {y ∈ B : y /∈ A}
is called the complement of A in B. (See page 30.) The set
A × B := {(x, y) : x ∈ A and y ∈ B}
of all ordered pairs (x, y) with x ∈ A and y ∈ B is called the Cartesian product of A and B.
- An indexed family of sets is a function which assigns a set Ai to each element i of a set I. The set I is called the index set of the family and the family is usually denoted {Ai}i∈I. The union and intersection of the indexed family are defined by
x ∈
i∈I
Ai ⇐⇒ ∃i ∈ I x ∈ Ai.
x ∈
i∈I
Ai ⇐⇒ ∀i ∈ I x ∈ Ai.
The notation ∃i ∈ I is an abbreviation for “there exists i ∈ I”, and the notation ∀i ∈ I is an abbreviation for “for all i ∈ I”. In these definitions the
- When A ⊆ X, B ⊆ Y , and f : X → Y , the sets
f (A) := {y ∈ Y : ∃x ∈ A such that y = f (x)}, f −^1 (B) := {x ∈ X : f (x) ∈ B},
are called respectively the image of A by f and the pre-image (or inverse image) of B by f. (See Buck page 76.)
Problem 6. Let f : X → Y , {Ai}i∈I be a family of subsets of X, and {Bi}i∈I be a family of subsets of Y. Say which of the following inclusions are (always) true. Prove each true inclusion and give a counterexample for each false one.
f −^1
i∈I
Bi
i∈I
f −^1 (Bi)?
i∈I
f −^1 (Bi) ⊆ f −^1
i∈I
Bi
f −^1
i∈I
Bi
i∈I
f −^1 (Bi)?
i∈I
f −^1 (Bi) ⊆ f −^1
i∈I
Bi
f
i∈I
Ai
i∈I
f (Ai)?
i∈I
f (Ai) ⊆ f
i∈I
Ai
f
i∈I
Ai
i∈I
f (Ai)?
i∈I
f (Ai) ⊆ f
i∈I
Ai
- If f : X → Y and g : Y → Z, then the composition of f and g is the map g ◦ f : X → Z defined by
(g ◦ f )(x) = g(f (x))
for x ∈ X. For any set X the identity map of X is the map idX : X → X defined by idX (x) = x
for x ∈ X. Clearly idY ◦ f = f and f ◦ idX = f
for f : X → Y.
- A map g : Y → X is said to be a left inverse for the map f : X → Y iff g ◦ f = idX , i.e. iff g(f (x)) = x for all x ∈ X. A map g : Y → X is
said to be a right inverse for the map f : X → Y iff f ◦ g = idY , i.e. iff f (g(y)) = y for all y ∈ Y. A map g : Y → X is said to be a (two sided) inverse to the map f : X → Y iff it is both a left inverse and a right inverse to f. If g is a left inverse to f and g′^ is a right inverse to f then g = g′. (Proof: g = g ◦ idX = g ◦ f ◦ g′^ = idY ◦ g′^ = g′.) In this case there is a unique inverse and it is denoted f −^1. So if f : X → Y has an inverse f −^1 : Y → X, then y = f (x) ⇐⇒ x = f −^1 (y)
for x ∈ X and y ∈ Y.
- A map f : X → Y is said to be one-one iff
∀x 1 , x 2 ∈ X [f (x 1 ) = f (x 2 ) =⇒ x 1 = x 2 ]
and it is said to be onto iff Y = f (X), i.e.
∀y ∈ Y ∃x ∈ X y = f (x)
Thus
(1) A map is one-one if and only if it has a left inverse;
(2) A map is onto if and only if it has a right inverse;
(3) A map is one-one and onto if and only if it has an inverse.
Remark 10. The ’only if’ part of item (2) is called the Axiom of Choice. It was once controversial because one can imagine a situation where one can prove that a map f is onto but where one can not give an explicit formula for a right inverse.
Remark 11. The above assertions (1-3) are false if continuity (defined later) is required: There is a continuous one-one and onto map whose inverse is not continuous (and hence continuous one-one map which does not have a continuous left inverse, and a continuous onto map which does not have a continuous right inverse). See Example 64 below.
Example 12. Consider the four maps
f 1 : R → R, f 2 : [0, ∞) → R, f 3 : R → [0, ∞), f 4 : [0, ∞) → [0, ∞)
(Multiplication) 0 < a, b =⇒ 0 < ab.
The other order notations are defined as usual, i.e. a < b ⇐⇒ b > a and a ≤ b ⇐⇒ b ≥ a ⇐⇒ either a < b or a = b.
Remark 15. All the rules of algebra used in College Algebra (Math 112) follow from the Algebraic Axioms 13 and Order Axioms 14. For example, (a + b)^2 = a^2 + 2ab + b^2 , a^2 ≥ 0, etc.
Definition 16. The set S of real numbers is said to be bounded above iff there is a number b ∈ R such that x ≤ b for all x ∈ S; the number b is then called an upper bound for S. A number b ∈ R is called a least upper bound for S iff it is an upper bound for S and b ≤ b′^ for every other upper bound b′^ for S. Similarly the set S is said to be bounded below iff there is an number a ∈ R such that a ≤ x for all x ∈ S; the element b is then called a lower bound for S. An element a ∈ R is called a greatest lower bound iff it is an lower bound for S and a′^ ≤ a for every other lower bound a′^ for S. The words infimum and greatest lower bound are synonymous as are the words supremum and least upper bound. The least upperbound of the set S will be denoted sup(S) and the greatest lower bound of the set S will be denoted inf(S).
- Completeness Axiom. Every set S of real numbers which is bounded above has a least upper bound, i.e.
if x ≤ b for all x ∈ S, then x ≤ sup(S) ≤ b for all x ∈ S.
Because multiplication by −1 reverses the order it is the same to say that every set which is bounded below has a greatest lower bound. Thus
if a ≤ x for all x ∈ S, then a ≤ inf(S) ≤ x for all x ∈ S.
Theorem 18 (Archimedean Property). There is neither an infinite real number nor an infinitesimal real number. More precisely,
(1) There is no real number which is larger than every integer.
(2) For every positive real number ε > 0 there is a positive integer n such that 1 /n < ε.
Problem 19. Prove Theorem 18. Hint: If ω > n for every integer n what about ω − 1?
Problem 20. Let R denote the set of real valued rational functions, i.e. f ∈ R iff f (x) = p(x)/q(x) where p(x) and q(x) are polynomials with real coefficients (and q(x) is not the zero polynomial). For f, g ∈ R define an order relation by the condition that f > g iff there exists an M such that f (x) > g(x) for all x > M. Then the set R satisfies the algebraic axioms and order axioms given above. View R (and hence Z) as a subset of R by identifying the real number c with the constant function whose value is always c. Exhibit (in the lingo of Theorem 18) an infinite element f ∈ R and an infinitesimal element g ∈ R.
3 Distance
- The distance d(p, q) between two points p = (x 1 , x 2 ,... , xn) and q = (y 1 , y 2 ,... , yn) in Rn^ is defined by
d(p, q) =
(x 1 − y 1 )^2 + (x 2 − y 2 )^2 + · · · + (xn − yn)^2.
The distance d(v, 0) from a vector v ∈ Rn^ to the origin is called the norm of v denoted by |v| so d(p, q) := |p − q|.
The norm satisfies the following laws for v, w ∈ Rn:
(Zero Norm) |v| = 0 ⇐⇒ v = 0,
(Homogeneity) |av| = a |v| if a > 0,
(Symmetry) | − v| = |v|,
(Triangle Inequality) v + w ≤ |v| + |w|.
(The Zero Norm Law holds because a sum of squares vanishes only if each summand vanishes and the Triangle Inequality is proved in Corollary on page 14 of Buck.) The laws for the norm imply that the distance function satisfies the following:
(Zero Distance) d(p, q) = 0 ⇐⇒ p = q,
(Symmetry) d(p, q) = d(q, p),
(Triangle Inequality) d(p, r) ≤ d(p, q) + d(q, r).
Definition 26. A set S is disconnected iff there are disjoint open sets U and V such that S ⊆ U ∪ V and both S ∩ U and S ∩ V are nonempty. A set is connected iff it is not disconnected.
Theorem 27. A subset S ⊆ R of the real line is connected if and only if S is an interval, i.e. [a, b] ⊆ S whenever a, b ∈ S.
Proof. We prove only if. Assume that there exist a, b ∈ S with [a, b] 6 ⊆ S. Then there is a c ∈ [a, b] with c /∈ S. Let U = (−∞, c) and V = (c, ∞). The point c lies in the open interval (a, b) as a, b ∈ S so a ∈ U and b ∈ V. Hence both S ∩ U and S ∩ V are nonempty and clearly S ⊆ U ∪ V (as c /∈ S). Hence the open sets U and V separate S so S is disconnected as required. We prove if. Assume that S is disconnected, i.e. that there exist open sets U, V ⊆ R with S ⊆ U ∪ V , S ∩ U 6 = ∅, S ∩ V 6 = ∅, and U ∩ V = ∅. Choose a ∈ S ∩ U and b ∈ S ∩ V. Then a 6 = b as U ∩ V = ∅. Assume without loss of generality that a < b. (The case b < a is the same.) We must show that [a, b] 6 ⊆ S, and for this it is enough (as S ⊆ U ∪ V ) to show that [a, b] 6 ⊆ U ∪ V. The set T := {x ∈ [a, b] : [a, x) ⊆ U } is nonempty (as it contains a) and bounded above (as b is an upperbound) so it has a supremum c. Clearly [a, c) ⊆ T. Also a < c (as (a − ε, a + ε) ⊆ U for sufficiently small ε > 0) and c < b (as (b − ε, b + ε) ⊆ V ⊆ R \ U for sufficiently small ε > 0). If c ∈ U , then (c − ε, c + ε) ⊆ U for sufficiently small ε > 0 so [a, c + ε) = [a, c) ∪ (a − ε, a + ε) ⊂ U so [a, c + ε) ⊆ T contradicting the fact that c is an upperbound for T. If c ∈ V , then (c−ε, c+ε) ⊆ V for sufficiently small ε > 0 so [a, c − ε) ∩ V = ∅ contradicting the fact that c is the least upperbound of T. Hence c /∈ U ∪ V as required.
Definition 28. A set S is bounded iff it is contained in some large ball, i.e. iff there exists M > 0 such that |p| < M for all p ∈ S. Thus a set or real numbers is bounded if and only if it is bounded above and bounded below. (See Definition 16.)
5 Limits
- Let p 0 be a cluster point of a set S and F be a function defined on S (but possibly not at p 0 ). The notation
lim p→p 0 F (p) = L
means that for every ε > 0 there exists δ > 0 such that for all p ∈ S ∩ B(p 0 , δ) \ {p 0 } we have |f (p) − L| < ε, i.e. iff
∀ε > 0 ∃δ > 0 ∀p ∈ S [0 < |p − p 0 | < δ =⇒ |f (p) − L| < ε].
When p 0 ∈ S and p 0 is a cluster point of S we have that a function f defined on S is continuous at p 0 (see Definition 48 below) if and only if
lim p→p 0 f (p) = f (p 0 )
(and the function is trivially continuous at a point p 0 ∈ S which is not a cluster point of S). However, the limit notation is usually used in situations where (p 0 is a cluster point of S but) p 0 ∈/ S. For example, the derivative of a real valued function f : I → R defined on an open interval I ⊆ R is defined by
f ′(x 0 ) := lim x→x 0
f (x) − f (x 0 ) x − x 0
The ratio in the limit is undefined when x = x 0 but is defined for nearby values of x.
- For a real valued function f defined on a subset of R we can extend the definition of the notation limx→a F (x) = L to include the cases where a = ±∞ and/or L = ±∞ as follows. Let
Rˆ := {−∞} ∪ R ∪ {∞}
consist of the set of real numbers together with two additional points which we think of as located at infinity. The set Rˆ is sometimes called the set of extended real numbers. For b ∈ Rˆ, a set U ⊆ R is called neighborhood of b iff
either b ∈ R and U contains an open interval (b − ρ, b + ρ) for some ρ > 0,
or else b = ∞ and U contains an open interval (M, ∞) for some M > 0,
or else b = −∞ and U contains an open interval (−∞, −M ) for some M > 0.
Because B(p, ρ) = (b − ρ, b + ρ) this definition agrees with the definition in paragraph 24. A point a ∈ Rˆ is called a cluster point of a subset S ⊆ R
Remark 36. Every sequence {pn}n determines a set {pn : n ∈ Z+}. Buck calls this set the trace of the sequence, but that terminology is uncommon. The trace can be finite, for example the trace of the sequence pn = (−1)n^ is the two element set {− 1 , 1 }. If the trace of a sequence is finite then there must be at least one constant subsequence and the common value is a limit point of the sequence. By definition only an infinite set has a cluster point.
Theorem 37 (Bolzano-Weierstrass). Every bounded infinite subset of Rn has an cluster point.
Corollary 38. Every bounded sequence in Rn^ has a limit point. (Buck The- orem 22 page 62.)
Definition 39. A sequence {an}n of real numbers is said to be increasing iff a 1 ≤ a 2 ≤ · · · , is said to be decreasing iff a 1 ≥ a 2 ≥ · · · , and is called monotonic iff it is either increasing or decreasing.
Theorem 40. A bounded monotonic sequence is convergent. (Buck page 47.)
Proof. Assume the sequence {an} is increasing and bounded above. Let a = sup{an}. Then a ≥ an for all n (as a is an upper bound) but for ε > 0 aN > a − ε for some N (as a is the least upper bound). Hence an > a − ε for n ≥ N (as the sequence is increasing).
- For any sequence {ak}k of real numbers we have {ak : k ≥ n} ⊆ {ak : k ≥ m} for m < n. If the sequence {ak}k is bounded above, then the sequence sn := sup{ak : k ≥ n} is decreasing. The limit of the latter sequence is denoted lim sup n→∞
an := lim n→∞ sup{ak : k ≥ n}.
Similarly for a sequence which is bounded below,
lim inf n→∞
an := lim n→∞
inf{ak : k ≥ n}.
Definition 42. A sequence {pn} is called Cauchy iff
lim m,n→∞ |pn − pm| = 0
i.e. iff for every ε > 0 there exists N > 0 such that n, m ≥ N =⇒ |pn−pm| < ε.
Theorem 43 (Cauchy Convergence Criterion). A sequence in Rn^ converges if and only if it is a Cauchy sequence. (Buck Theorem 23 and its corollary on pages 62-63.)
7 Compact Sets
Definition 44. An open cover of a set S is a collection {Ui}i∈I of open sets such that S ⊆
i∈I Ui.^ The subset^ S^ is^ compact^ iff every compact cover {Ui}i∈I of S has finite subcover, i.e. there are indices i 1 , i 2 ,... , in ∈ I such that S ⊆ Ui 1 ∪ Ui 2 ∪ · · · ∪ Uin.
Theorem 45. The closed interval [a, b] is compact. (Buck Theorem 24 page 65.)
Theorem 46 (Heine Borel). A subset of Rn^ is compact if and only if it is closed and bounded. (Buck Theorem 25 page 65.)
Remark 47. Call a set S sequentially compact iff for every sequence {pn}n of points in S there is a point p ∈ S and a subsequence {pnk }k which converges to p. Combining Theorems 37 and 46 we see that a set is sequen- tially compact if and only if it is compact if and only if it is closed and bounded. It follows
8 Continuity
Let f : D → Rm^ where D ⊆ Rn.
Definition 48. The function f is said to be continuous at a point p ∈ D iff for every ε > 0 there exists a δ > 0 such that for all q ∈ D
|q − p| < ε =⇒ |f (q) − f (p)| < ε.
Theorem 49. The function f is continuous at p ∈ D if and only if for every sequence {pn} of points in D we have
lim n→∞
pn = p =⇒ lim n→∞
f (pn) = f (p). (1)
Proof. We prove ‘only if ’. Assume f is continuous at p. Choose sequence {pn} of points in D. Assume
lim n→∞
pn = p. (2)
Choose ε > 0. Because f is assumed to be continuous at p there is a δ > 0 such that for all q ∈ D
|q − p| < δ =⇒ |f (q) − f (p)| < ε. (3)
Theorem 53. Assume that D is compact and f is continuous. The f is uniformly continuous.
Proof. Choose ε > 0. Then, because f is continuous, for every p ∈ D there is a δ = δ(p) > 0 such that
|q − p| < δ(p) =⇒ |f (q) − f (p)| <
ε 2
Let Up := B(p, δ(p)/2). The sets Up cover D (since p ∈ Up), i.e. D ⊆
p Up. As D is compact, finitely many of these sets cover D, i.e.
D ⊆ Up 1 ∪ Up 2 ∪ · · · ∪ Upn. (5)
Define δ := 12 min{δ(p 1 ), δ(p 2 ),... , δ(pn)}.
Choose p, q ∈ D. Assume |q − p| < δ. By (5) we have that p ∈ Upk for some k. Hence |p − pk| < δ(pk)/ 2. (6)
But δ ≤ δ(pk)/2 by its definition so
|q − pk| ≤ |q − p| + |p − pk| ≤ δ +
δ(pk) 2
δ(pk) 2
δ(pk) 2
= δ(pk). (7)
Hence, by the definition of δ(·) and Equations (6) and (7) we have
|f (p) − f (q)| ≤ |f (p) − f (pk)| + |f (q) − f (pk)| <
ε 2
ε 2
= ε
as required.
Remark 54. A proof using the Bolzano-Weierstrass theorem instead on the Heine-Borel theorem is given in Buck page 85. The proof given here is like the proof sketched in Exercises 11 and 12 in Buck page 89.
Theorem 55. Assume U ⊆ Rn^ is open. Then the map f : U → Rm^ is continuous if and only if the preimage of every open subset of Rm^ is an open subset of Rn, i.e. if and only if whenever V ⊆ Rm^ is open, the set f −^1 (V ) is also open. (Buck Theorem 3 page 76.)
Theorem 56. Assume that D ⊆ Rn, E ⊆ Rm, and that f : D → E and g : E → R^ are continuous. Then g ◦ f : D → R^ is continuous. (Buck Theorem 5 page 78.)
Theorem 57. The continuous image of a compact set is compact: If f : D → Rm^ is continuous and D is compact, then f (D) is compact. (Buck Theorem 13 on page 93.)
Corollary 58. The continuous image of compact set is bounded. (Buck Theorem 10 on page 90.)
Theorem 59. If f : D → R is continuous and D is compact, then f assumes its maximum on D, i.e. there exists p ∈ D such that f (q) ≤ f (p) for all q ∈ D. Similarly for the minimum. (Buck Theorem 11 page 91.)
Theorem 60. The continuous image of a connected set is connected: If f : D → Rm^ is continuous and D is connected, then f (D) is connected. (Buck Theorem 15 on page 94.)
Corollary 61 (Intermediate Value Theorem). Assume that S is connected and that f : S → R is continuous. Suppose that a, b ∈ f (S) and that a < c < b. Then c ∈ f (S). (Buck Theorem 14 on page 93.)
Remark 62. The Intermediate Value Theorem from calculus is a special case. It says that if f : [α, β] → R is a real valued continuous function on the closed interval [α, β] ⊆ R, {a, b} = {f (α), f (β)}, and a ≤ c ≤ b, then the equation f (x) = c has a solution x ∈ [α, β].
Theorem 63. Let f : D → E be continuous, one-one, and onto, and assume D (and hence by Theorem 57 also E) is compact. Then f −^1 : E → D is continuous.
Proof. Choose a convergent sequence {qn}n in E and a let
q := lim n→∞ qn
be its limit. We will show that
f −^1 (q) = lim n→∞ f −^1 (qn); (#)
the Theorem will then follow by Theorem 49. By Bolzano Weierstrass and Heine Borel there is a convergent subsequence {f −^1 (qnk )}k. Let
p := lim k→∞ f −^1 (qnk )
Problem 67. Prove Theorem 66. (This theorem is proved in Buck The- orem 18 page 96 and Theorem 25 page 114, but Buck assumes that the intervals are closed and bounded. This assumption can be removed.)
Theorem 68. Let S ⊆ Rn^ be and f : S → Rm^ be uniformly continuous.Then the function f can be continuously extended to the closure S¯ of S. i.e. there is a continuous function F : S¯ → Rm^ such that F (p) = f (p) for p ∈ S. (Buck Theorem 25 on page 109.)
Example 69. The function f : (0, 1] → R defined by f (x) = sin(1/x) cannot be extended to a continuous function on the closure [0, 1] of (0, 1].
Problem 70. Fix a positive number a ∈ R. The purpose of this problem is to define the exponential ax^ for x ∈ R. Define a^0 := 1 and for n a positive integer define an^ := a︸ · a︷︷ · · · a︸ n
, a−n^ := 1/an.
Then
(1) Prove that for every nonzero integer n there is a unique solution b > 0 to the equation bn^ = a. Define a^1 /n^ to be this unique solution, i.e. a^1 /n^ = b ⇐⇒ bn^ = a.
(2) For a rational number q define aq^ by aq^ = (am)^1 /n^ where q = m/n. Prove that this definition is independent of the choice of the integers m and n such that q = m/n.
(3) Prove that there is a unique continuous function
R → (0, ∞) : x 7 → ax
such that ax^ = aq^ when x is a rational number q.
In your proof make clear which theorems from these notes you are appealing to. Also make your proof self contained so that a person who doesn’t have access to the statement of the problem can follow it. (You needn’t provide proofs for the theorems you use, but do provide references to them.) In your proof of (3) you may use the inequality
|ap^ − aq| ≤ M |p − q|
which is true when a > 1, N is a positive integer, 1 ≤ p, q ≤ N + 1, and
M :=
( N
k=
k
aN^.
You need not prove this inequality but use calculus to show where it comes from. Hint: What is the definition of ln x, ex, and ax^ used in calculus? The natural logarithm function ln x is usually defined as an integral. How do you bound an integral by a sum? How does the Mean Value Theorem (Buck Theorem 3 page 118) give inequalities like this?
9 Derivatives
Definition 71. The function f : I → R is said to be differentiable at the point x 0 ∈ I iff the limit
f ′(x 0 ) := lim x→x 0
f (x) − f (x 0 ) x − x 0
exists; we say that f is differentiable on a set iff it is differentiable at each point x 0 in the set. The function f ′^ is called the derivative of f.
Theorem 72. A differentiable function is continuous.
Theorem 73 (Mean Value Theorem). Suppose that f is differentiable of (a, b) and continuous on [a, b]. The there exists c ∈ (a, b) such that
f ′(c) =
f (b) − f (a) b − a
Corollary 74. Assume that f is differentiable on I. Then the derivative f ′ vanishes identically on I if and only if f is constant on I.
10 The Integral
- A partition of the closed interval [a, b] is an increasing finite sequence {xk{ 0 ≤k≤n with x 0 = a and xn = b, For any bounded function f defined on [a, b] and any partition P = {xk} 0 ≤k≤n of [a, b] define the upper sum by
S¯(f, P ) :=
∑^ n
k=
f^ ¯k(xk − xk− 1 ), f¯k := sup{f (x) : xk− 1 ≤ x ≤ xk},
