Introduction to Analysis in Several Variables (Advanced Calculus) Michael E. Taylor

Transcription

1 Introduction to Analysis in Several Variables (Advanced Calculus) Michael E. Taylor 1

2 2 Contents 0. One-variable calculus 1. The derivative 2. Inverse function and implicit function theorem 3. Systems of differential equations 4. The Riemann integral in n variables 5. Surfaces and surface integrals 6. Differential forms 7. Products and exterior derivatives of forms 8. The general Stokes formula 9. The classical Gauss, Green, and Stokes formulas 10. Holomorphic functions and harmonic functions 11. Homotopies of maps and actions on forms 12. Differential forms and degree theory 13. Fourier series 14. The Fourier transform A. Metric spaces, convergence, and compactness B. Partitions of unity C. Differential forms and the change of variable formula D. Complements on power series E. The Weierstrass theorem and the Stone-Weierstrass theorem F. Sard s theorem G. Morse functions H. Inner product spaces

3 3 Introduction This text was produced for the second semester of a two-semester sequence on advanced calculus, whose aim is to provide a firm logical foundation for analysis, for students who have had 3 semesters of calculus and a course in linear algebra. The first semester treats analysis in one variable, and the text [T4] was written to cover that material. The text at hand treats analysis in several variables. These two texts can be used as companions, but they are written so that they can be used independently, if desired. We begin with an introductory section, 0, presenting the elements of one-dimensional calculus. We first define the Riemann integral of a class of functions on an interval. We then introduce the derivative, and establish the Fundamental Theorem of Calculus, relating differentiation and integration as essentially inverse operations. Further results are dealt with in the exercises, such as the change of variable formula for integrals, and the Taylor formula with remainder. This section distills material developed in more detail in the companion text [T4]. We have included it here to facilitate the independent use of this text. In 1 we define the derivative of a function F : O R m, where O is an open subset of R n, as a linear map from R n to R m. We establish some basic properties, such as the chain rule. We use the one-dimensional integral as a tool to show that, if the matrix of first order partial derivatives of F is continuous on O, then F is differentiable on O. We also discuss two convenient multi-index notations for higher derivatives, and derive the Taylor formula with remainder for a smooth function F on O R n. In 2 we establish the Inverse Function Theorem, stating that a smooth map F : O R n with an invertible derivative DF (p) has a smooth inverse defined near q = F (p). We derive the Implicit Function Theorem as a consequence of this. As a tool in proving the Inverse Function Theorem, we use a fixed point theorem known as the Contraction Mapping Principle. In 3 we treat systems of differential equations. We establish a basic existence and uniqueness theorem and also study the smooth dependence of a solution on initial data. We interpret the solution operator as a flow generated by a vector field and introduce the concept of the Lie bracket of vector fields. In 4 we take up the multidimensional Riemann integral. The basic definition is quite parallel to the one-dimensional case, but a number of central results, while parallel in statement to the one-dimensional case, require more elaborate demonstrations in higher dimensions. This section is one of the most demanding in this text, and it is in a sense the heart of the course. One central result is the change of variable formula for multidimensional integrals. Another is the reduction of multiple integrals to iterated integrals. In 5 we define the notion of a smooth m-dimensional surface in R n and study properties of these objects. We associate to such a surface a metric tensor, and make use of this to define the integral of functions on a surface. This includes the study of surface area. Examples include the computation of areas of higher dimensional spheres. We also explore

4 4 integration on the group of rotations on R n, leading to the notion of averaging over rotations. In 6 we introduce a further class of objects that are defined on surfaces, differential forms. A k-form can be integrated over a k-dimensional surface, endowed with an extra piece of structure, an orientation. The change of variable formula established in 4 plays a central role in establishing this. Properties of differential forms are developed further in the next two sections. In 7 we define exterior products of forms, and interior products of forms with vector fields. Then we define the exterior derivative of a form. Section 8 is devoted to the general Stokes formula, an important integral identity which contains as special cases classical identities of Green, Gauss, and Stokes. These special cases are discussed in some detail in 9. In 10 we use Green s Theorem to derive fundamental properties of holomorphic functions of a complex variable. Sprinkled throughout earlier sections are some allusions to functions of complex variables, particularly in some of the exercises in 1 2. Readers with no previous exposure to complex variables might wish to return to these exercises after getting through 10. In this section, we also discuss some results on the closely related study of harmonic functions. One result is Liouville s Theorem, stating that a bounded harmonic function on all of R n must be constant. When specialized to holomorphic functions on C = R 2, this yields a proof of the Fundamental Theorem of Algebra. In 11 we define the notion of smoothly homotopic maps and consider the behavior of closed differential forms under pull back by smoothly homotopic maps. This material is then applied in 12, which introduces degree theory and derives some interesting consequences. Applications include the Brouwer fixed point theorem, the non-existence of smooth vector fields tangent to an even-dimensional sphere, and the Jordan-Brouwer separation theorem (in the smooth case). We also show how degree theory yields another proof of the fundamental theorem of algebra. Section 13 treats Fourier series on the n-dimensional torus T n, and 14 treats the Fourier transform for functions on R n. These two sections are written to be read independently, for the benefit of those who decide to cover only one of these topics. Those who read both sections will therefore notice a bit of repetition, hopefully not too annoying. For those who do cover both sections, 14 ends with a topic that ties them together, known as Poisson s summation formula. We apply this formula to establish a classical result of Riemann, his functional equation for the Riemann zeta function. Appendix A at the end of this text covers some basic notions of metric spaces and compactness, used from time to time throughout the text, particularly in the study of the Riemann integral and in the proof of the fundamental existence theorem for ODE. As is the case with 0, this appendix distills material developed at a more leisurely pace in [T4], again serving to make this text independent of the first one. Further appendices provide additional material, useful as either background or complements to the main body of the text. Appendix B discusses partitions of unity, useful particularly in the proof of the Stokes formula. Appendix C presents a proof of the change of variable formula for multiple integrals, following an idea presented by Lax in [L], which is somewhat shorter than that given in 4. Appendix D discusses the remainder term in the Taylor series of a function. Appendix E gives the Weierstrass theorem on approximating

5 a continuous function by polynomials, and an extension, known as the Stone-Weierstrass theorem, which is a useful tool in analysis. Applications arise in sections 12 and 13. Appendix F proves Sard s theorem, of use in the presentation of degree theory in 12. Appendix G gives another application of Sard s theorem, to the existence of lots of Morse functions, which in turn implies the existence of lots of vector fields with only isolated critical points, also of use in 12. Appendix H gives some basic material on inner product spaces, which arise in sections We mention that the exercises in this text play a particularly important role in developing the material. In addition to exercises providing practice in applying results established in the text, there are also exercises asking the reader to supply details for some of the arguments used in the text. There are also scattered exercises intended to give the reader a fresh perspective on topics familiar from previous study. We mention in particular exercises on determinants and on the cross product in 1, and on trigonometric functions in 3. 5

6 6 0. One variable calculus In this brief discussion of one variable calculus, we introduce the Riemann integral, and relate it to the derivative. We will define the Riemann integral of a bounded function over an interval I = [a, b] on the real line. For now, we assume f is real valued. To start, we partition I into smaller intervals. A partition P of I is a finite collection of subintervals {J k : 0 k N}, disjoint except for their endpoints, whose union is I. We can order the J k so that J k = [x k, x k+1 ], where (0.1) x 0 < x 1 < < x N < x N+1, x 0 = a, x N+1 = b. We call the points x k the endpoints of P. We set (0.2) l(j k ) = x k+1 x k, maxsize (P) = max 0 k N l(j k) We then set (0.3) I P (f) = k I P (f) = k sup J k f(x) l(j k ), inf J k f(x) l(j k ). See (A.10) (A.11) for the definition of sup and inf. We call I P (f) and I P (f) respectively the upper sum and lower sum of f, associated to the partition P. Note that I P (f) I P (f). These quantities should approximate the Riemann integral of f, if the partition P is sufficiently fine. To be more precise, if P and Q are two partitions of I, we say P refines Q, and write P Q, if P is formed by partitioning each interval in Q. Equivalently, P Q if and only if all the endpoints of Q are also endpoints of P. It is easy to see that any two partitions have a common refinement; just take the union of their endpoints, to form a new partition. Note also that refining a partition lowers the upper sum of f and raises its lower sum: (0.4) P Q = I P (f) I Q (f), and I P (f) I Q (f). Consequently, if P j are any two partitions and Q is a common refinement, we have (0.5) I P1 (f) I Q (f) I Q (f) I P2 (f). Now, whenever f : I R is bounded, the following quantities are well defined: (0.6) I(f) = inf P Π(I) I P(f), I(f) = sup P Π(I) I P (f),

7 where Π(I) is the set of all partitions of I. We call I(f) the lower integral of f and I(f) its upper integral. Clearly, by (0.5), I(f) I(f). We then say that f is Riemann integrable provided I(f) = I(f), and in such a case, we set b (0.7) f(x) dx = f(x) dx = I(f) = I(f). a We will denote the set of Riemann integrable functions on I by R(I). We derive some basic properties of the Riemann integral. Proposition 0.1. If f, g R(I), then f + g R(I), and (0.8) (f + g) dx = f dx + g dx. I 7 Proof. If J k is any subinterval of I, then I I I sup (f + g) sup f + sup g, and inf (f + g) inf J k J k J k J k J k f + inf J k g, so, for any partition P, we have I P (f + g) I P (f) + I P (g). Also, using common refinements, we can simultaneously approximate I(f) and I(g) by I P (f) and I P (g), and ditto for I(f + g). Thus the characterization (0.6) implies I(f + g) I(f) + I(g). A parallel argument implies I(f + g) I(f) + I(g), and the proposition follows. Next, there is a fair supply of Riemann integrable functions. Proposition 0.2. If f is continuous on I, then f is Riemann integrable. Proof. Any continuous function on a compact interval is bounded and uniformly continuous (see Propositions A.14 and A.15). Let ω(δ) be a modulus of continuity for f, so (0.9) x y δ = f(x) f(y) ω(δ), ω(δ) 0 as δ 0. Then (0.10) maxsize (P) δ = I P (f) I P (f) ω(δ) l(i), which yields the proposition. We denote the set of continuous functions on I by C(I). Thus Proposition 0.2 says C(I) R(I). The proof of Proposition 0.2 provides a criterion on a partition guaranteeing that I P (f) and I P (f) are close to f dx when f is continuous. We produce an extension, giving a I condition under which I P (f) and I(f) are close, and I P (f) and I(f) are close, given f bounded on I. Given a partition P 0 of I, set (0.11) minsize(p 0 ) = min{l(j k ) : J k P 0 }.

8 8 Lemma 0.3. Let P and Q be two partitions of I. Assume (0.12) maxsize(p) 1 k minsize(q). Let f M on I. Then (0.13) I P (f) I Q (f) + 2M k l(i), I P (f) I Q (f) 2M k l(i). Proof. Let P 1 denote the minimal common refinement of P and Q. Consider on the one hand those intervals in P that are contained in intervals in Q and on the other hand those intervals in P that are not contained in intervals in Q. Each interval of the first type is also an interval in P 1. Each interval of the second type gets partitioned, to yield two intervals in P 1. Denote by P b 1 the collection of such divided intervals. By (0.12), the lengths of the intervals in P b 1 sum to l(i)/k. It follows that I P (f) I P1 (f) J P b 1 and similarly I P (f) I P1 (f) 2Ml(I)/k. Therefore 2Ml(J) 2M l(i) k, I P (f) I P1 (f) + 2M k l(i), I P (f) I P 1 (f) 2M k l(i). Since also I P1 (f) I Q (f) and I P1 (f) I Q (f), we obtain (0.13). The following consequence is sometimes called Darboux s Theorem. Theorem 0.4. Let P ν be a sequence of partitions of I into ν intervals J νk, 1 k ν, such that maxsize(p ν ) 0. If f : I R is bounded, then (0.14) I Pν (f) I(f) and I Pν (f) I(f). Consequently, (0.15) f R(I) I(f) = lim ν k=1 for arbitrary ξ νk J νk, in which case the limit is I f dx. ν f(ξ νk )l(j νk ),

9 9 Proof. As before, assume f M. Pick ε = 1/k > 0. Let Q be a partition such that I(f) I Q (f) I(f) + ε, I(f) I Q (f) I(f) ε. Now pick N such that ν N = maxsize P ν ε minsize Q. Lemma 0.3 yields, for ν N, I Pν (f) I Q (f) + 2Ml(I)ε, I Pν (f) I Q (f) 2Ml(I)ε. Hence, for ν N, I(f) I Pν (f) I(f) + [2Ml(I) + 1]ε, I(f) I Pν (f) I(f) [2Ml(I) + 1]ε. This proves (0.14). Remark. The sums on the right side of (0.15) are called Riemann sums, approximating f dx (when f is Riemann integrable). I Remark. A second proof of Proposition 0.1 can readily be deduced from Theorem 0.4. One should be warned that, once such a specific choice of P ν and ξ νk has been made, the limit on the right side of (0.15) might exist for a bounded function f that is not Riemann integrable. This and other phenomena are illustrated by the following example of a function which is not Riemann integrable. For x I, set (0.16) ϑ(x) = 1 if x Q, ϑ(x) = 0 if x / Q, where Q is the set of rational numbers. Now every interval J I of positive length contains points in Q and points not in Q, so for any partition P of I we have I P (ϑ) = l(i) and I P (ϑ) = 0, hence (0.17) I(ϑ) = l(i), I(ϑ) = 0. Note that, if P ν is a partition of I into ν equal subintervals, then we could pick each ξ νk to be rational, in which case the limit on the right side of (0.15) would be l(i), or we could pick each ξ νk to be irrational, in which case this limit would be zero. Alternatively, we could pick half of them to be rational and half to be irrational, and the limit would be l(i)/2.

10 10 Associated to the Riemann integral is a notion of size of a set S, called content. If S is a subset of I, define the characteristic function (0.18) χ S (x) = 1 if x S, 0 if x / S. We define upper content cont + and lower content cont by (0.19) cont + (S) = I(χ S ), cont (S) = I(χ S ). We say S has content, or is contented if these quantities are equal, which happens if and only if χ S R(I), in which case the common value of cont + (S) and cont (S) is (0.20) m(s) = It is easy to see that I χ S (x) dx. { N (0.21) cont + (S) = inf l(j k ) : S J 1 J N }, k=1 where J k are intervals. Here, we require S to be in the union of a finite collection of intervals. See the appendix at the end of this section for a generalization of Proposition 0.2, giving a sufficient condition for a bounded function to be Riemann integrable on I, in terms of the upper content of its set of discontinuities. There is a more sophisticated notion of the size of a subset of I, called Lebesgue measure. The key to the construction of Lebesgue measure is to cover a set S by a countable (either finite or infinite) set of intervals. The outer measure of S I is defined by { (0.22) m (S) = inf k 1 l(j k ) : S k 1 J k }. Here {J k } is a finite or countably infinite collection of intervals. Clearly (0.23) m (S) cont + (S). Note that, if S = I Q, then χ S = ϑ, defined by (0.16). In this case it is easy to see that cont + (S) = l(i), but m (S) = 0. Zero is the right measure of this set. More material on the development of measure theory can be found in a number of books, including [Fol] and [T2]. It is useful to note that f dx is additive in I, in the following sense. I

11 11 Proposition 0.5. If a < b < c, f : [a, c] R, f 1 = f [a,b], f 2 = f [b,c], then (0.24) f R ( [a, c] ) f 1 R ( [a, b] ) and f 2 R ( [b, c] ), and, if this holds, (0.25) c a f dx = b a f 1 dx + c b f 2 dx. Proof. Since any partition of [a, c] has a refinement for which b is an endpoint, we may as well consider a partition P = P 1 P 2, where P 1 is a partition of [a, b] and P 2 is a partition of [b, c]. Then (0.26) I P (f) = I P1 (f 1 ) + I P2 (f 2 ), I P (f) = I P1 (f 1 ) + I P2 (f 2 ), so (0.27) I P (f) I P (f) = { I P1 (f 1 ) I P1 (f 1 ) } + { I P2 (f 2 ) I P2 (f 2 ) }. Since both terms in braces in (0.27) are 0, we have equivalence in (0.24). Then (0.25) follows from (0.26) upon taking sufficiently fine partitions. Let I = [a, b]. If f R(I), then f R([a, x]) for all x [a, b], and we can consider the function (0.28) g(x) = If a x 0 x 1 b, then x (0.29) g(x 1 ) g(x 0 ) = so, if f M, a f(t) dt. x1 x 0 f(t) dt, (0.30) g(x 1 ) g(x 0 ) M x 1 x 0. In other words, if f R(I), then g is Lipschitz continuous on I. A function g : (a, b) R is said to be differentiable at x (a, b) provided there exists the limit 1 [ ] (0.31) lim g(x + h) g(x) = g (x). h 0 h When such a limit exists, g (x), also denoted dg/dx, is called the derivative of g at x. Clearly g is continuous wherever it is differentiable. The next result is part of the Fundamental Theorem of Calculus.

12 12 Theorem 0.6. If f C([a, b]), then the function g, defined by (2.28), is differentiable at each point x (a, b), and (0.32) g (x) = f(x). Proof. Parallel to (2.29), we have, for h > 0, (0.33) 1 [ ] 1 x+h g(x + h) g(x) = f(t) dt. h h x If f is continuous at x, then, for any ε > 0, there exists δ > 0 such that f(t) f(x) ε whenever t x δ. Thus the right side of (0.33) is within ε of f(x) whenever h (0, δ]. Thus the desired limit exists as h 0. A similar argument treats h 0. The next result is the rest of the Fundamental Theorem of Calculus. Theorem 0.7. If G is differentiable and G (x) is continuous on [a, b], then (0.34) b a G (t) dt = G(b) G(a). Proof. Consider the function (0.35) g(x) = x a G (t) dt. We have g C([a, b]), g(a) = 0, and, by Theorem 0.6, g (x) = G (x), x (a, b). Thus f(x) = g(x) G(x) is continuous on [a, b], and (0.36) f (x) = 0, x (a, b). We claim that (0.36) implies f is constant on [a, b]. Granted this, since f(a) = g(a) G(a) = G(a), we have f(x) = G(a) for all x [a, b], so the integral (2.35) is equal to G(x) G(a) for all x [a, b]. Taking x = b yields (2.34). The fact that (0.36) implies f is constant on [a, b] is a consequence of the following result, known as the Mean Value Theorem. Theorem 0.8. Let f : [a, β] R be continuous, and assume f is differentiable on (a, β). Then ξ (a, β) such that (0.37) f (ξ) = f(β) f(a). β a

13 Proof. Set g(x) = f(x) κ(x a), where κ is the right side of (0.37). Note that g (ξ) = f (ξ) κ, so it suffices to show that g (ξ) = 0 for some ξ (a, β). Note also that g(a) = g(β). Since [a, β] is compact, g must assume a maximum and a minimum on [a, β]. Since g(a) = g(β), one of these must be assumed at an interior point, at which g vanishes. Now, to see that (0.36) implies f is constant on [a, b], if not, β (a, b] such that f(β) f(a). Then just apply Theorem 0.8 to f on [a, β]. This completes the proof of Theorem 0.7. We now extend Theorems to the setting of Riemann integrable functions. Proposition 0.9. Let f R([a, b]), and define g by (0.28). If x [a, b] and f is continuous at x, then g is differentiable at x, and g (x) = f(x). The proof is identical to that of Theorem 0.6. Proposition Assume G is differentiable on [a, b] and G R([a, b]). Then (0.34) holds. Proof. We have for some ξ kn satisfying n 1 [ ( G(b) G(a) = G a + (b a) k + 1 ) n k=0 = b a n 1 G (ξ kn ), n k=0 a + (b a) k n < ξ kn < a + (b a) k + 1 n, ( G a + (b a) k )] n as a consequence of the Mean Value Theorem. Given G R([a, b]), Darboux s theorem (Theorem 0.4) implies that as n one gets G(b) G(a) = b a G (t) dt. Note that the beautiful symmetry in Theorems is not preserved in Propositions The hypothesis of Proposition 0.10 requires G to be differentiable at each x [a, b], but the conclusion of Proposition 0.9 does not yield differentiability at all points. For this reason, we regard Propositions as less fundamental than Theorems There are more satisfactory extensions of the fundamental theorem of calculus, involving the Lebesgue integral, and a more subtle notion of the derivative of a nonsmooth function. For this, we can point the reader to Chapters of the text [T2], Measure Theory and Integration. So far, we have dealt with integration of real valued functions. If f : I C, we set f = f 1 + if 2 with f j : I R and say f R(I) if and only if f 1 and f 2 are in R(I). Then f dx = f 1 dx + i f 2 dx. 13 I I I

14 14 There are straightforward extensions of Propositions to complex valued functions. Similar comments apply to functions f : I R n. If a function G is differentiable on (a, b), and G is continuous on (a, b), we say G is a C 1 function, and write G C 1( (a, b) ). Inductively, we say G C k( (a, b) ) provided G C k 1( (a, b) ). An easy consequence of the definition (0.31) of the derivative is that, for any real constants a, b, and c, f(x) = ax 2 + bx + c = df dx = 2ax + b. Now, it is a simple enough step to replace a, b, c by y, z, w, in these formulas. Having done that, we can regard y, z, and w as variables, along with x : F (x, y, z, w) = yx 2 + zx + w. We can then hold y, z and w fixed (e.g., set y = a, z = b, w = c), and then differentiate with respect to x; we get F = 2yx + z, x the partial derivative of F with respect to x. Generally, if F is a function of n variables, x 1,..., x n, we set F x j (x 1,..., x n ) 1 = lim h 0 h [ F (x1,..., x j 1, x j + h, x j+1,..., x n ) F (x 1,..., x j 1, x j, x j+1,..., x n ) ], where the limit exists. Section one carries on with a further investigation of the derivative of a function of several variables. Complementary results on Riemann integrability Here we provide a condition, more general then Proposition 0.2, which guarantees Riemann integrability. Proposition Let f : I R be a bounded function, with I = [a, b]. Suppose that the set S of points of discontinuity of f has the property (0.38) cont + (S) = 0. Then f R(I). Proof. Say f(x) M. Take ε > 0. As in (0.21), take intervals J 1,..., J N such that S J 1 J N and N k=1 l(j k) < ε. In fact, fatten each J k such that S is contained

15 in the interior of this collection of intervals. Consider a partition P 0 of I, whose intervals include J 1,..., J N, amongst others, which we label I 1,..., I K. Now f is continuous on each interval I ν, so, subdividing each I ν as necessary, hence refining P 0 to a partition P 1, we arrange that sup f inf f < ε on each such subdivided interval. Denote these subdivided intervals I 1,..., I L. It readily follows that N L 0 I P1 (f) I P1 (f) < 2Ml(J k ) + εl(i k) k=1 k=1 < 2εM + εl(i). Since ε can be taken arbitrarily small, this establishes that f R(I). 15 Remark. An even better result is that such f is Riemann integrable if and only if (0.38A) m (S) = 0, where m (S) is defined by (0.22). The implication m (S) = 0 f R(I) is established in Chapter 4 of [T4], and its higher dimensional generalization is established in Proposition 4.31 of this text. For the reverse implication f R(I) m (S) = 0, one can see standard books on measure theory, such as [Fol] and [T2]. We give an example of a function to which Proposition 0.11 applies, and then an example for which Proposition 0.11 fails to apply, though the function is Riemann integrable. Example 1. Let I = [0, 1]. Define f : I R by f(0) = 0, f(x) = ( 1) j for x (2 (j+1), 2 j ], j 0. Then f 1 and the set of points of discontinuity of f is S = {0} {2 j : j 1}. It is easy to see that cont + S = 0. Hence f R(I). See Exercises below for a more elaborate example to which Proposition 0.11 applies. Example 2. Again I = [0, 1]. Define f : I R by f(x) = 0 if x / Q, 1 n if x = m, in lowest terms. n

16 16 Then f 1 and the set of points of discontinuity of f is S = I Q. As we have seen below (0.23), cont + S = 1, so Proposition 0.11 does not apply. Nevertheless, it is fairly easy to see directly that I(f) = I(f) = 0, so f R(I). In fact, given ε > 0, f ε only on a finite set, hence I(f) ε, ε > 0. As indicated below (0.23), (0.38A) does apply to this function. By contrast, the function ϑ in (0.16) is discontinuous at each point of I. We mention an alternative characterization of I(f) and I(f), which can be useful. Given I = [a, b], we say g : I R is piecewise constant on I (and write g PK(I)) provided there exists a partition P = {J k } of I such that g is constant on the interior of each interval J k. Clearly PK(I) R(I). It is easy to see that, if f : I R is bounded, { I(f) = inf } f 1 dx : f 1 PK(I), f 1 f, (0.39) I { I(f) = sup I } f 0 dx : f 0 PK(I), f 0 f. Hence, given f : I R bounded, (0.40) f R(I) for each ε > 0, f 0, f 1 PK(I) such that f 0 f f 1 and (f 1 f 0 ) dx < ε. I This can be used to prove (0.41) f, g R(I) = fg R(I), via the fact that (0.42) f j, g j PK(I) = f j g j PK(I). In fact, we have the following, which can be used to prove (0.41).

17 17 Proposition Let f R(I), and assume f M. continuous. Then ϕ f R(I). Let ϕ : [ M, M] R be Proof. We proceed in steps. Step 1. We can obtain ϕ as a uniform limit on [ M, M] of a sequence ϕ ν of continuous, piecewise linear functions. Then ϕ ν f ϕ f uniformly on I. A uniform limit g of functions g ν R(I) is in R(I) (see Exercise 12). So it suffices to prove Proposition 0.12 when ϕ is continuous and piecewise linear. Step 2. Given ϕ : [ M, M] R continuous and piecewise linear, it is an exercise to write ϕ = ϕ 1 ϕ 2, with ϕ j : [ M, M] R monotone, continuous, and piecewise linear. Now ϕ 1 f, ϕ 2 f R(I) ϕ f R(I). Step 3. We now demonstrate Proposition 0.12 when ϕ : [ M, M] R is monotone and Lipschitz. By Step 2, this will suffice. So we assume M x 1 < x 2 M = ϕ(x 1 ) ϕ(x 2 ) and ϕ(x 2 ) ϕ(x 1 ) L(x 2 x 1 ), for some L <. Given ε > 0, pick f 0, f 1 PK(I), as in (0.40). Then ϕ f 0, ϕ f 1 PK(I), ϕ f 0 ϕ f ϕ f 1, and (ϕ f 1 ϕ f 0 ) dx L (f 1 f 0 ) dx Lε. This proves ϕ f R(I). I I Exercises 1. Let c > 0 and let f : [ac, bc] R be Riemann integrable. Working directly with the definition of integral, show that (0.43) b a f(cx) dx = 1 c bc ac f(x) dx. More generally, show that (0.44) b d/c a d/c f(cx + d) dx = 1 c bc ac f(x) dx. 2. Let f : I S R be continuous, where I = [a, b] and S R n. Take ϕ(y) = f(x, y) dx. I

18 18 Show that ϕ is continuous on S. Hint. If f j : I R are continuous and f 1 (x) f 2 (x) δ on I, then (0.45) I f 1 dx I f 2 dx l(i)δ. Hint. Suppose y j S, y j y S. Let S = {y j } {y}. This is compact. Thus f : I S R is uniformly continuous. Hence where ω(δ) 0 as δ 0. f(x, y j ) f(x, y) ω( y j y ), x I, 3. With f as in Exercise 2, suppose g j : S R are continuous and a g 0 (y) < g 1 (y) b. Take ϕ(y) = g 1 (y) g 0 f(x, y) dx. Show that ϕ is continuous on S. (y) Hint. Make a change of variables, linear in x, to reduce this to Exercise Suppose f : (a, b) (c, d) and g : (c, d) R are differentiable. Show that h(x) = g(f(x)) is differentiable and h (x) = g (f(x))f (x). This is the chain rule. Hint. Peek at the proof of the chain rule in If f 1 and f 2 are differentiable on (a, b), show that f 1 (x)f 2 (x) is differentiable and d ( f1 (x)f 2 (x) ) = f dx 1(x)f 2 (x) + f 1 (x)f 2(x). If f 2 (x) 0, x (a, b), show that f 1 (x)/f 2 (x) is differentiable, and d ( f1 (x) ) = f 1(x)f 2 (x) f 1 (x)f 2(x) dx f 2 (x) f 2 (x) Let ϕ : [a, b] [A, B] be C 1 on a neighborhood J of [a, b], with ϕ (x) > 0 for all x [a, b]. Assume ϕ(a) = A, ϕ(b) = B. Show that the identity (0.46) B A f(y) dy = b a f ( ϕ(t) ) ϕ (t) dt, for any f C(I), I = [A, B], follows from the chain rule and the Fundamental Theorem of Calculus. Hint. Replace b by x, B by ϕ(x), and differentiate.

19 19 7. Show that (0.46) holds for each f PK(I). Using (0.39) (0.40), show that f R(I) f ϕ R([a, b]) and (0.46) holds. (This result contains that of Exercise 1.) 8. Show that, if f and g are C 1 on a neighborhood of [a, b], then (0.47) b a b f(s)g (s) ds = f (s)g(s) ds + [ f(b)g(b) f(a)g(a) ]. a This transformation of integrals is called integration by parts. 9. Let f : ( a, a) R be a C j+1 function. Show that, for x ( a, a), (0.48) f(x) = f(0) + f (0)x + f (0) 2 where x f (j) (0) x j + R j (x), j! (0.49) R j (x) = x 0 (x s) j f (j+1) (s) ds j! This is Taylor s formula with remainder. Hint. Use induction. If (0.48) (0.49) holds for 0 j k, show that it holds for j = k + 1, by showing that (0.50) x 0 (x s) k k! f (k+1) (s) ds = f (k+1) (0) x (k + 1)! xk (x s) k+1 f (k+2) (s) ds. (k + 1)! To establish this, use the integration by parts formula (0.47), with f(s) replaced by f (k+1) (s), and with appropriate g(s). See 1 for another approach. Note that another presentation of (0.49) is 1 (0.51) R j (x) = xj+1 f (j+1)( ( 1 t 1/(j+1) ) ) x dt. (j + 1)! Assume f : ( a, a) R is a C j function. Show that, for x ( a, a), (0.48) holds, with (0.52) R j (x) = 1 (j 1)! x 0 (x s) j 1[ f (j) (s) f (j) (0) ] ds. Hint. Apply (0.49) with j replaced by j 1. Add and subtract f (j) (0) to the factor f (j) (s) in the resulting integrand.

20 Given I = [a, b], show that (0.53) f, g R(I) = fg R(I), as advertised in (0.41). 12. Assume f k R(I) and f k f uniformly on I. Prove that f R(I) and (0.54) f k dx f dx. I I 13. Given I = [a, b], I ε = [a + ε, b ε], assume f k R(I), f k M on I for all k, and (0.55) f k f uniformly on I ε, for all ε (0, (b a)/2). Prove that f R(I) and (0.54) holds. 14. Use the fundamental theorem of calculus to compute (0.56) b a x r dx, r Q \ { 1}, where 0 a < b < if r 0 and 0 < a < b < if r < Use the change of variable result of Exercise 6 to compute 1 0 x 1 + x 2 dx. 16. We say f R(R) provided f [k,k+1] R([k, k + 1]) for each k Z, and (0.57) k= k k+1 f(x) dx <. If f R(R), we set (0.58) k f(x) dx = lim f(x) dx. k k Formulate and demonstrate basic properties of the integral over R of elements of R(R). 17. This exercise discusses the integral test for absolute convergence of an infinite series,

21 which goes as follows. Let f be a positive, monotonically decreasing, continuous function on [0, ), and suppose a k = f(k). Then a k < k=0 0 f(x) dx <. 21 Prove this. Hint. Use N a k k=1 N 0 f(x) dx N 1 k=0 a k. 18. Use the integral test to show that, if p > 0, (0.59) k=1 1 < p > 1. kp Hint. Use Exercise 14 to evaluate I N (p) = N 1 x p dx, for p 1, and let N. See if you can show x 1 dx = without knowing about log N. Subhint. Show that x 1 dx = 2N N x 1 dx. In Exercises 19 20, C R is the Cantor set, defined as follows. Take a closed, bounded interval [a, b] = C 0. Let C 1 be obtained from C 0 by deleting the open middle third interval, of length (b a)/3. At the jth stage, C j is a disjoint union of 2 j closed intervals, each of length 3 j (b a). Then C j+1 is obtained from C j by deleting the open middle third of each of these 2 j intervals. We have C 0 C 1 C j, each a closed subset of [a, b]. The Cantor set is C = j 0 C j. 19. Show that cont + C j = (2/3) j (b a), and conclude that cont + C = Define f : [a, b] R as follows. We call an interval of length 3 j (b a), omitted in passing from C j 1 to C j, a j-interval. Set f(x) = 0, if x C, ( 1) j, if x belongs to a j-interval. Show that the set of discontinuities of f is C. Hence Proposition 0.11 implies f R([a, b]). 21. Generalize Exercise 8 as follows. Assume f and g are differentiable on a neighborhood of [a, b] and f, g R([a, b]). Then show that (0.47) holds. Hint. Use the results of Exercise 11 to show that (fg) R([a, b]).

22 Let f : I R be bounded, I = [a, b]. Show that { I(f) = inf } f 1 dx : f 1 C(I), f 1 f, I { I(f) = sup f 0 dx : f 0 C(I), f 0 f }. I Compare (0.39). Then show that (0.61) f R(I) for each ε > 0, f 0, f 1 C(I) such that f 0 f f 1 and (f 1 f 0 ) dx < ε. I Compare (0.40).

23 23 1. The derivative Let O be an open subset of R n, and F : O R m a continuous function. We say F is differentiable at a point x O, with derivative L, if L : R n R m is a linear transformation such that, for y R n, small, (1.1) F (x + y) = F (x) + Ly + R(x, y) with (1.2) R(x, y) y 0 as y 0. We denote the derivative at x by DF (x) = L. With respect to the standard bases of R n and R m, DF (x) is simply the matrix of partial derivatives, ( ) F 1 / x 1 F 1 / x n Fj (1.3) DF (x) = = x.., k F m / x 1 F m / x n so that, if v = (v 1,..., v n ) t, (regarded as a column vector) then ( F 1 / x k )v k k (1.4) DF (x)v =... ( F m / x k )v k k Recall the definition of the partial derivative f j / x k from 0. It will be shown below that F is differentiable whenever all the partial derivatives exist and are continuous on O. In such a case we say F is a C 1 function on O. More generally, F is said to be C k if all its partial derivatives of order k exist and are continuous. If F is C k for all k, we say F is C. In (1.2), we can use the Euclidean norm on R n and R m. This norm is defined by (1.5) x = ( x x 2 n for x = (x 1,..., x n ) R n. Any other norm would do equally well. An application of the Fundamental Theorem of Calculus, to functions of each variable x j separately, yields the following. If we assume F : O R m is differentiable in each variable separately, and that each F/ x j is continuous on O, then n [ F (x + y) = F (x) + F (x + zj ) F (x + z j 1 ) ] n = F (x) + A j (x, y)y j, (1.6) j=1 A j (x, y) = 1 0 ) 1/2 F x j ( x + zj 1 + ty j e j ) dt, j=1

24 24 where z 0 = 0, z j = (y 1,..., y j, 0,..., 0), and {e j } is the standard basis of R n. Consequently, (1.7) F (x + y) = F (x) + R(x, y) = j=1 n j=1 0 F x j (x) y j + R(x, y), n 1 { F y j (x + z j 1 + ty j e j ) F } (x) dt. x j x j Now (1.7) implies F is differentiable on O, as we stated below (1.4). Thus we have established the following. Proposition 1.1. If O is an open subset of R n and F : O R m is of class C 1, then F is differentiable at each point x O. As is shown in many calculus texts, one can use the Mean Value Theorem instead of the Fundamental Theorem of Calculus, and obtain a slightly sharper result. Let us give some examples of derivatives. First, take n = 2, m = 1, and set (1.8) F (x) = (sin x 1 )(sin x 2 ). Then (1.9) DF (x) = ((cos x 1 )(sin x 2 ), (sin x 1 )(cos x 2 )). Next, take n = m = 2 and ( ) x1 x 2 (1.10) F (x) = x 2 1. x2 2 Then ( x2 x (1.11) DF (x) = 1 2x 1 2x 2 We can replace R n and R m by more general finite-dimensional real vector spaces, isomorphic to Euclidean space. For example, the space M(n, R) of real n n matrices is isomorphic to R n2. Consider the function (1.12) S : M(n, R) M(n, R), S(X) = X 2. We have ). (1.13) (X + Y ) 2 = X 2 + XY + Y X + Y 2 = X 2 + DS(X)Y + R(X, Y ),

25 25 with R(X, Y ) = Y 2, and hance (1.14) DS(X)Y = XY + Y X. For our next example, we take (1.15) O = Gl(n, R) = {X M(n, R) : det X 0}, which, as shown below, is open in M(n, R). We consider (1.16) Φ : Gl(n, R) M(n, R), Φ(X) = X 1, and compute DΦ(I). We use the following. If, for A M(n, R), (1.17) A = sup{ Av : v R n, v 1}, then (1.18) A, B M(n, R) A + B A + B and AB A B, hence Y M(n, R) Y k Y k. Also (1.19) S k = I Y + Y 2 + ( 1) k Y k Y S k = S k Y = Y Y 2 + Y 3 + ( 1) k Y k+1 (I + Y )S k = S k (I + Y ) = I + ( 1) k Y k+1, hence (1.20) Y < 1 = (I + Y ) 1 = so ( 1) k Y k = I Y + Y 2, k=0 (1.21) DΦ(I)Y = Y. Related calculations show that Gl(n, R) is open in M(n, R). In fact, given X Gl(n, R), Y M(n, R), (1.22) X + Y = X(I + X 1 Y ), which by (1.20) is invertible as long as (1.23) X 1 Y < 1.

26 26 One can proceed from here to compute DΦ(X). See the exercises. We return to general considerations, and derive the chain rule for the derivative. Let F : O R m be differentiable at x O, as above, let U be a neighborhood of z = F (x) in R m, and let G : U R k be differentiable at z. Consider H = G F. We have (1.24) H(x + y) = G(F (x + y)) = G ( F (x) + DF (x)y + R(x, y) ) = G(z) + DG(z) ( DF (x)y + R(x, y) ) + R 1 (x, y) = G(z) + DG(z)DF (x)y + R 2 (x, y) with R 2 (x, y) y Thus G F is differentiable at x, and 0 as y 0. (1.25) D(G F )(x) = DG(F (x)) DF (x). Another useful remark is that, by the Fundamental Theorem of Calculus, applied to ϕ(t) = F (x + ty), (1.26) F (x + y) = F (x) DF (x + ty)y dt, provided F is C 1. For a typical application, see (6.6). For the study of higher order derivatives of a function, the following result is fundamental. Proposition 1.2. Assume F : O R m is of class C 2, with O open in R n. Then, for each x O, 1 j, k n, (1.27) F (x) = x j x k F (x). x k x j Proof. It suffices to take m = 1. We label our function f : O R. For 1 j n, we set (1.28) j,h f(x) = 1 h( f(x + hej ) f(x) ), where {e 1,..., e n } is the standard basis of R n. The mean value theorem (for functions of x j alone) implies that if j f = f/ x j exists on O, then, for x O, h > 0 sufficiently small, (1.29) j,h f(x) = j f(x + α j he j ),

27 for some α j (0, 1), depending on x and h. Iterating this, if j ( k f) exists on O, then, for x O, h > 0 sufficiently small, 27 (1.30) k,h j,h f(x) = k ( j,h f)(x + α k he k ) = j,h ( k f)(x + α k he k ) = j k f(x + α k he k + α j he j ), with α j, α k (0, 1). Here we have used the elementary result (1.31) k j,h f = j,h ( k f). We deduce the following. Proposition 1.3. If k f and j k f exist on O and j k f is continuous at x 0 O, then (1.32) j k f(x 0 ) = lim h 0 k,h j,h f(x 0 ). Clearly (1.33) k,h j,h f = j,h k,h f, so we have the following, which easily implies Proposition 1.2. Corollary 1.4. In the setting of Proposition 1.3, if also j f and k j f exist on O and k j f is continuous at x 0, then (1.34) j k f(x 0 ) = k j f(x 0 ). We now describe two convenient notations to express higher order derivatives of a C k function f : Ω R, where Ω R n is open. In one, let J be a k-tuple of integers between 1 and n; J = (j 1,..., j k ). We set (1.35) f (J) (x) = jk j1 f(x), j = x j. We set J = k, the total order of differentiation. As we have seen in Proposition 1.2, i j f = j i f provided f C 2 (Ω). It follows that, if f C k (Ω), then jk j1 f = lk l1 f whenever {l 1,..., l k } is a permutation of {j 1,..., j k }. Thus, another convenient notation to use is the following. Let α be an n-tuple of non-negative integers, α = (α 1,..., α n ). Then we set (1.36) f (α) (x) = α 1 1 α n n f(x), α = α α n. Note that, if J = α = k and f C k (Ω), (1.37) f (J) (x) = f (α) (x), with α i = #{l : j l = i}.

28 28 Correspondingly, there are two expressions for monomials in x = (x 1,..., x n ): (1.38) x J = x j1 x jk, x α = x α 1 1 xα n n, and x J = x α provided J and α are related as in (1.37). Both these notations are called multi-index notations. We now derive Taylor s formula with remainder for a smooth function F : Ω R, making use of these multi-index notations. We will apply the one variable formula (0.48) (0.49), i.e., (1.39) ϕ(t) = ϕ(0) + ϕ (0)t ϕ (0)t k! ϕ(k) (0)t k + r k (t), with (1.40) r k (t) = 1 k! t 0 (t s) k ϕ (k+1) (s) ds, given ϕ C k+1 (I), I = ( a, a). (See Exercise 9 of 0, Exercise 7 of this section, and also Appendix D for further discussion.) Let us assume 0 Ω, and that the line segment from 0 to x is contained in Ω. We set ϕ(t) = F (tx), and apply (1.39) (1.40) with t = 1. Applying the chain rule, we have (1.41) ϕ (t) = Differentiating again, we have (1.42) ϕ (t) = n j F (tx)x j = F (J) (tx)x J. j=1 J =1, K =1 J =1 F (J+K) (tx)x J+K = J =2 F (J) (tx)x J, where, if J = k, K = l, we take J + K = (j 1,..., j k, k 1,..., k l ). Inductively, we have (1.43) ϕ (k) (t) = Hence, from (1.39) with t = 1, or, more briefly, F (x) = F (0) + J =1 (1.44) F (x) = J =k F (J) (tx)x J. F (J) (0)x J k! J k J =k 1 J! F (J) (0)x J + R k (x), F (J) (0)x J + R k (x),

29 29 where (1.45) R k (x) = 1 k! J =k+1 ( 1 0 ) (1 s) k F (J) (sx) ds x J. This gives Taylor s formula with remainder for F C k+1 (Ω), in the J-multi-index notation. We also want to write the formula in the α-multi-index notation. We have (1.46) where J =k F (J) (tx)x J = α =k ν(α)f (α) (tx)x α, (1.47) ν(α) = #{J : α = α(j)}, and we define the relation α = α(j) to hold provided the condition (1.37) holds, or equivalently provided x J = x α. Thus ν(α) is uniquely defined by (1.48) α =k ν(α)x α = J =k x J = (x x n ) k. One sees that, if α = k, then ν(α) is equal to the product of the number of combinations of k objects, taken α 1 at a time, times the number of combinations of k α 1 objects, taken α 2 at a time, times the number of combinations of k (α α n 1 ) objects, taken α n at a time. Thus ( )( ) ( ) k k α1 k α1 α n 1 (1.49) ν(α) = = α 1 In other words, for α = k, (1.50) ν(α) = k! α!, where α! = α 1! α n! Thus the Taylor formula (1.44) can be rewritten α 2 (1.51) F (x) = where (1.52) R k (x) = α =k+1 α k k + 1 α! α n 1 α! F (α) (0)x α + R k (x), ( 1 0 ) (1 s) k F (α) (sx) ds x α. k! α 1!α 2! α n!

30 30 The formula (1.51) (1.52) holds for F C k+1. It is significant that (1.51), with a variant of (1.52), holds for F C k. In fact, for such F, we can apply (1.52) with k replaced by k 1, to get (1.53) F (x) = with (1.54) R k 1 (x) = α k 1 α =k k α! 1 α! F (α) (0)x α + R k 1 (x), ( 1 0 ) (1 s) k 1 F (α) (sx) ds x α. We can add and subtract F (α) (0) to F (α) (sx) in the integrand above, and obtain the following. Proposition 1.5. If F C k on a ball B r (0), the formula (1.51) holds for x B r (0), with (1.55) R k (x) = α =k k α! ( 1 0 (1 s) k 1[ F (α) (sx) F (α) (0) ] ) ds x α. Remark. Note that (1.55) yields the estimate (1.56) R k (x) α =k x α α! sup F (α) (sx) F (α) (0). 0 s 1 The term corresponding to J = 2 in (1.44), or α = 2 in (1.51), is of particular interest. It is (1.57) 1 2 J =2 F (J) (0) x J = 1 2 n j,k=1 2 F x k x j (0) x j x k. We define the Hessian of a C 2 function F : O R as an n n matrix: (1.58) D 2 F (y) = ( 2 ) F (y). x k x j Then the power series expansion of second order about 0 for F takes the form (1.59) F (x) = F (0) + DF (0)x x D2 F (0)x + R 2 (x),

31 31 where, by (1.56), (1.60) R 2 (x) C n x 2 sup 0 s 1, α =2 F (α) (sx) F (α) (0). In all these formulas we can translate coordinates and expand about y O. For example, (1.59) extends to (1.61) F (x) = F (y) + DF (y)(x y) (x y) D2 F (y)(x y) + R 2 (x, y), with (1.62) R 2 (x, y) C n x y 2 sup 0 s 1, α =2 F (α) (y + s(x y)) F (α) (y). Example. If we take F (x) as in (1.8), so DF (x) is as in (1.9), then ( ) D 2 sin x1 sin x F (x) = 2 cos x 1 cos x 2. cos x 1 cos x 2 sin x 1 sin x 2 The results (1.61) (1.62) are useful for extremal problems, i.e., determining where a sufficiently smooth function F : O R has local maxima and local minima. Clearly if F C 1 (O) and F has a local maximum or minimum at x 0 O, then DF (x 0 ) = 0. In such a case, we say x 0 is a critical point of F. To check what kind of critical point x 0 is, we look at the n n matrix A = D 2 F (x 0 ), assuming F C 2 (O). By Proposition 1.2, A is a symmetric matrix. A basic result in linear algebra is that if A is a real, symmetric n n matrix, then R n has an orthonormal basis of eigenvectors, {v 1,..., v n }, satisfying Av j = λ j v j ; the real numbers λ j are the eigenvalues of A. We say A is positive definite if all λ j > 0, and we say A is negative definite if all λ j < 0. We say A is strongly indefinite if some λ j > 0 and another λ k < 0. Equivalently, given a real, symmetric matrix A, (1.63) for some C > 0, all v R n, and (1.64) A positive definite v Av C v 2, A negative definite v Av C v 2, A strongly indefinite v, w R n, nonzero, such that v Av C v 2, w Aw C w 2, for some C > 0. In light of (1.44) (1.45), we have the following result. Proposition 1.6. Assume F C 2 (O) is real valued, O open in R n. Let x 0 O be a critical point for F. Then (i) D 2 F (x 0 ) positive definite F has a local minimum at x 0, (ii) D 2 F (x 0 ) negative definite F has a local maximum at x 0, (iii) D 2 F (x 0 ) strongly indefinite F has neither a local maximum nor a local minimum at x 0. In case (iii), we say x 0 is a saddle point for F. The following is a test for positive definiteness.

32 32 Proposition 1.7. Let A = (a ij ) be a real, symmetric, n n matrix. For 1 l n, form the l l matrix A l = (a ij ) 1 i,j l. Then (1.65) A positive definite det A l > 0, l {1,..., n}. Regarding the implication, note that if A is positive definite, then det A = det A n is the product of its eigenvalues, all > 0, hence is > 0. Also in this case, it follows from the hypothesis on the left side of (1.65) that each A l must be positive definite, hence have positive determinant, so we have. The implication is easy enough for 2 2 matrices. If A is symmetric and det A > 0, then either both its eigenvalues are positive (so A is positive definite) or both are negative (so A is negative definite). In the latter case, A 1 = (a 11 ) must be negative, so we have in this case. We prove for n 3, using induction. The inductive hypothesis implies that if det A l > 0 for each l n, then A n 1 is positive definite. The next lemma then guarantees that A = A n has at least n 1 positive eigenvalues. The hypothesis that det A > 0 does not allow that the remaining eigenvalue be 0, so all the eigenvalues of A must be positive. Thus Proposition 1.7 is proven, once we have the following. Lemma 1.8. In the setting of Proposition 1.7, if A n 1 is positive definite, then A = A n has at least n 1 positive eigenvalues. Proof. Since A is symmetric, R n has an orthonormal basis v 1,..., v n of eigenvectors of A; Av j = λ j v j. If the conclusion of the lemma is false, at least two of the eigenvalues, say λ 1, λ 2, are 0. Let W = Span(v 1, v 2 ), so w W = w Aw 0. Since W has dimension 2, R n 1 R n satisfies R n 1 W 0, so there exists a nonzero w R n 1 W, and then w A n 1 w = w Aw 0, contradicting the hypothesis that A n 1 is positive definite. Remark. Given (1.65), we see by taking A A that if A is a real, symmetric n n matrix, (1.66) A negative definite ( 1) l det A l > 0, l {1,..., n}. We return to higher order power series formulas with remainder and complement Proposition 1.5. Let us go back to (1.39) (1.40) and note that the integral in (1.40) is 1/(k + 1) times a weighted average of ϕ (k+1) (s) over s [0, t]. Hence we can write r k (t) = 1 (k + 1)! ϕ(k+1) (θt), for some θ [0, 1],

33 if ϕ is of class C k+1. This is the Lagrange form of the remainder; see Appendix D for more on this, and for a comparison with the Cauchy form of the remainder. If ϕ is of class C k, we can replace k + 1 by k in (1.39) and write (1.67) ϕ(t) = ϕ(0) + ϕ (0)t (k 1)! ϕ(k 1) (0)t k k! ϕ(k) (θt)t k, for some θ [0, 1]. Pluging (1.67) into (1.43) for ϕ(t) = F (tx) gives (1.68) F (x) = J k 1 1 J! F (J) (0)x J + 1 k! J =k F (J) (θx)x J, for some θ [0, 1] (depending on x and on k, but not on J), when F is of class C k on a neighborhood B r (0) of 0 R n. Similarly, using the α-multi-index notation, we have, as an alternative to (1.53) (1.54), (1.69) F (x) = α k 1 1 α! F (α) (0)x α + α =k 1 α! F (α) (θx)x α, for some θ [0, 1] (depending on x and on α, but not on α), if F C k (B r (0)). Note also that 33 (1.70) 1 2 J =2 F (J) (θx)x J = 1 2 n j,k=1 2 F x k x j (θx)x j x k = 1 2 x D2 F (θx)x, with D 2 F (y) as in (1.58), so if F C 2 (B r (0)), we have, as an alternative to (1.59), (1.71) F (x) = F (0) + DF (0)x x D2 F (θx)x, for some θ [0, 1]. We next complemant the multi-index notations for higher derivatives of a function F by a multi-linear notation, defined as follows. If k N, F C k (U), and y U R n, set (1.72) D k F (y)(u 1,..., u k ) = t1 tk F (y + t 1 u t k u k ), t1 = =t k =0 for u 1,..., u k R n. For k = 1, this formula is equivalent to the definition of DF given at the beginning of this section. For k = 2, we have (1.73) D 2 F (y)(u, v) = u D 2 F (y)v,

34 34 with D 2 F (y) on the right as in (1.58). Generally, (1.72) defines D k F (y) as a symmetric, k-linear form in u 1,..., u k R n. We can relate (1.72) to the J-multi-index notation as follows. We start with (1.74) t1 F (y + t 1 u t k u k ) = F (J) (y + Σt j u j )u J 1, and inductively obtain (1.75) t1 tk F (y + Σt j u j ) = hence (1.76) D k F (y)(u 1,..., u k ) = J 1 = = J k =1 J 1 = = J k =1 J =1 In particular, if u 1 = = u k = u, (1.77) D k F (y)(u,..., u) = Hence (1.68) yields J =k F (J 1+ +J k ) (y + Σt j u j )u J 1 1 uj k k, F (J 1+ +J k ) (y)u J 1 1 uj k k. F (J) (y)u J. 1 (1.78) F (x) = F (0)+DF (0)x+ + (k 1)! Dk 1 F (0)(x,..., x)+ 1 k! Dk F (θx)(x,..., x), for some θ [0, 1], if F C k (B r (0)). In fact, rather than appealing to (1.68), we can note that ϕ(t) = F (tx) = ϕ (k) (t) = t1 tk ϕ(t + t t k ) = D k F (tx)(x,..., x), and obtain (1.78) directly from (1.67). We can also use the notation D j F (y)x j = D j F (y)(x,..., x), t1 = =t k =0 with j copies of x within the last set of parentheses, and rewrite (1.78) as (1.79) F (x) = F (0) + DF (0)x + + Note how (1.78) and (1.79) generalize (1.71). Convergent power series and their derivatives 1 (k 1)! Dk 1 F (0)x (k 1) + 1 k! Dk F (θx)x k. Here we consider functions given by convergent power series, of the form (1.80) F (x) = α 0 b α x α. We allow b α C, and take x = (x 1,..., x n ) R n, with x α given by (1.38). Here is our first result.