DIFFERENTIATION TSOGTGEREL GANTUMUR n Abstract. We discuss differentiation in R and the inverse function theorem. Contents 1. Continuity of scalar functions1 2. Differentiability of scalar functions4 3. Continuity of vector functions8 4. Differentiability of vector functions 10 5. Functions of several variables: Continuity 11 6. Differentiability 15 7. Second order derivatives 19 8. Inverse functions: Single variable case 23 9. The inverse function theorem 27 1. Continuity of scalar functions Let us recall first a definition of continuous functions. Intuitively, a continuous function f sends nearby points to nearby points, i.e, if x is close to y then f(x) is close to f(y). Definition 1.1. Let K ⊂ R be a set. A function f : K ! R is called continuous at y 2 K if for any " > 0 there exists δ > 0 such that x 2 (y − δ; y + δ) \ K implies jf(x) − f(y)j < ". In order for f to be continuous at y, first, the value f(x) must be getting closer and closer to some number, say α 2 R, as x tends to y, and second, that number α must be equal to the value f(y). The first requirement alone leads to the notion of the limit of a function. Definition 1.2. Let K ⊂ R be a set, and let f : K ! R be a function. We say that f(x) converges to α 2 R as x ! y 2 R, and write f(x) ! α as x ! y; or lim f(x) = α; (1) x!y if for any " > 0 there exists δ > 0 such that jf(x) − αj < " whenever 0 < jx − yj < δ and x 2 K. One can write lim f(x), K 3 x ! y, etc., to explicitly indicate the domain K. x2K; x!y Remark 1.3. Note that the point y is not required to be in K, and even if y 2 K, the existence and the value of the limit lim f(x) does not depend on the value f(y), since we x!y never consider x = y due to the condition 0 < jx − yj. In other words, we can replace K by K n fyg with no effect on the existence and the value of the limit. Example 1.4. Let K = [0; 1), and let f : K ! R be a function. Take y = 2, and α 2 R. Then as long as δ ≤ 1, there is no x satisfying 0 < jx − yj < δ and x 2 K. Hence by convention, when δ ≤ 1, we must assume that the implication x 2 K and 0 < jx − yj < δ ) jf(x) − αj < " Date: 2016/10/29. 1 2 TSOGTGEREL GANTUMUR is true for any " > 0, because there is no x for which we need to check the condition. This means that by Definition 1.2, any possible function f : [0; 1) ! R would have a limit as x ! 2, and this limit can be an arbitrary number α 2 R. We could have removed this inconvenience by modifying our definition to require that the set fx 2 K : 0 < jx − yj < δg is nonempty for any δ > 0, but we have not done so because any consideration of situations where y is \far away" from K as in this example would only lead to useless and trivial statements. The following lemma expresses the limit of a function in terms of the limit of a sequence. Lemma 1.5. Let K ⊂ R be a set, and let f : K ! R be a function. Let y 2 R and α 2 R. Then f(x) ! α as x ! y if and only if f(xn) ! α as n ! 1 for every sequence fxng ⊂ K n fyg converging to y. Proof. Suppose that f(x) ! α as x ! y, and let fxng ⊂ K n fyg be a sequence converging to y. We want to show that f(xn) ! α as n ! 1. Let " > 0 be arbitrary. Then by definition, there exists δ > 0 such that 0 < jx − yj < δ and x 2 K imply jf(x) − αj < ". Since xn ! y as n ! 1, there is N such that jxn − yj < δ whenever n > N. Hence we have jf(xn) − αj < " whenever n > N. As " > 0 is arbitrary, we conclude that f(xn) ! α as n ! 1. To prove the other direction, assume that f(x) does not converge to α as x ! y, i.e., that there is some " > 0, such that for any δ > 0, there exists some x 2 K with 0 < jx−yj < δ and 1 jf(x) − αj ≥ ". In particular, taking δ = n , we infer the existence of a sequence fxng ⊂ K 1 satisfying 0 < jxn − yj < n , with jf(xn) − αj ≥ " for all n. Thus we have a sequence fxng ⊂ K n fyg converging to y, with f(xn) 6! α as n ! 1. An immediate corollary is that continuous functions are precisely the ones that send con- vergent sequences to convergent sequences. This is sometimes called the sequential criterion of continuity. Corollary 1.6. Let K ⊂ R be a set. Then f : K ! R is continuous at y 2 K if and only if f(xn) ! f(y) as n ! 1 for every sequence fxng ⊂ K n fyg converging to x. Proof. The second condition is equivalent to f(x) ! f(y) as x ! y by Lemma 1.5, and hence we only need to show that continuity at y is equivalent to f(x) ! f(y) as x ! y. Let us explicitly write the definitions of these two concepts side by side to compare. • Continuity of f at y: For any " > 0 there exists δ > 0 such that jf(x) − f(y)j < " whenever jx − yj < δ and x 2 K. • Convergence of f(x) to f(y) as x ! y: For any " > 0 there exists δ > 0 such that jf(x) − f(y)j < " whenever 0 < jx − yj < δ and x 2 K. We see that the only difference is in whether we allow x = y. Hence it is immediate that continuity implies the convergence property. Now if the convergence property is satisfied, then everything for continuity is there, except the condition jf(x) − f(y)j < " when x = y. But this is trivially true, because jf(y) − f(y)j = 0. Exercise 1.7. Let K ⊂ R be a set. Show that f : K ! R is continuous at y 2 K if and only if f(xn) ! f(y) as n ! 1 for every sequence fxng ⊂ K converging to y. 2 Example 1.8. (a) Let f : R ! R be the function given by f(x) = x for x 2 R. Then f is continuous at every point y 2 R, because given any sequence fxng ⊂ R converging to y, 2 2 we have f(xn) = xn ! y = f(y) as n ! 1. (b) Let g : R ! R be the function given by g(x) = jxj for x 2 R. Then f is continuous at every point y 2 R, because given any sequence fxng ⊂ R converging to y, we have f(xn) = jxnj ! jyj = f(y) as n ! 1. DIFFERENTIATION 3 (c) We define the Heaviside step function θ : R ! R by ( 1 for x > 0; θ(x) = (2) 0 for x ≤ 0: It is clear that θ is continuous at every x 2 R n f0g . Our intuition tells us that θ is 1 1 not continuous at x = 0. Indeed, let xn = n and yn = − n for n 2 N. Then we have xn ! 0 and yn ! 0, but θ(xn) ! 1 and θ(yn) ! 0 as n ! 1. Since 1 6= 0, the sequential criterion of continuity implies that θ is not continuous at x = 0. (d) The Dirichlet function h : R ! R is defined by ( 1 for x 2 ; h(x) = Q (3) 0 for x 2 R n Q: For any x 2 R, we can find two sequences fxng ⊂ Q and fyng ⊂ R n Q satisfying xn ! x and yn ! x as n ! 1. Since h(xn) = 1 and h(yn) = 0, we have h(xn) ! 1 and h(yn) ! 0, and hence we conclude that h is not continuous at any point x 2 R. Exercise 1.9. In each of the following cases, verify if the value f(0) can be defined so that the resulting function f is continuous in R. Choose from the following phrases to best describe the situation in each case: jump discontinuity, removable singularity, blow up or pole, essential/oscillatory singularity. 1 1 jxj sin x (a) f(x) = (b) f(x) = (c) f(x) = (d) f(x) = x jxj x x 1 1 1 1 (e) f(x) = sin x (f) f(x) = x sin x (g) f(x) = x sin x Now we want to introduce a way to compare asymptotic magnitudes of two functions. Definition 1.10 (Little `o' notation). Let K ⊂ R be a set, let y 2 R, and let f : K ! R and g : K ! R be functions. Then we write f(x) = o(g(x)) as x ! y; (4) to mean that f(x) ! 0 as x ! y: (5) g(x) Furthermore, for h : K ! R, the notation f(x) = h(x) + o(g(x)) as x ! y; (6) is understood to be f(x) − h(x) = o(g(x)) as x ! y: (7) 3 2 x3 Example 1.11. (a) We have x = o(x ) as x ! 0, because x2 ! 0 as x ! 0. (b) We have sin x 6= o(x) as x ! 0, because sin x 6! 0 as x ! 0.