v 1. v n R n we have for each 1 j n that v j v n max 1 j n v j. i=1


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1 1. Limits and Continuity It is often the case that a nonlinear function of nvariables x = (x 1,..., x n ) is not really defined on all of R n. For instance f(x 1, x 2 ) = x 1x 2 is not defined when x x 2 1 = ±x 2. However, 1 x2 2 I will adopt a convention from the vector calculus notes of Jones and write F : R n R m regardless, meaning only that the source of F is some subset of R n. While a bit imprecise, this will not cause any big problems and will simplify many statements. I will often distinguish between functions f : R n R that are scalarvalued and functions F : R n R m, m 2 that are vectorvalued, using lowercase letters to denote the former and upper case letters to denote the latter. Note that any vectorvalued function F : R n R m may be written F = (F 1,..., F m ) where F j : R n R are scalarvalued functions called the components of F. For example, F : R 2 R 2 given by F (x 1, x 2 ) = (x 1 x 2, x 1 + x 2 ) is a vector valued function with components F 1 (x 1, x 2 ) = x 1 x 2 and F 2 (x 1, x 2 ) = x 1 + x 2. Definition 1.1. Let a R n be a point and r > 0 be a positive real number. The open ball of radius r about a is the set B(a, r) := {x R n : x a < r}. I will also use B (a, r) to denote the set of all x B(a, r) except x = a. Proposition 1.2. Let a, b R n be points and r, s > 0 be real numbers. Then B(a, r) B(b, s) if and only if a b s r. B(a, r) B(b, s) = if and only if a b s + r. Proof. Exercise. Both parts depend on the triangle inequality. Extending my above convention, I will say that a function F : R n R m is defined near a point a R n if there exists r > 0 such that F (x) is defined for all points x B(a, r), except possibly the center x = a. Below I will need the following inequality relating the magnitude of a vector to the sizes of its coordinates. Proposition 1.3. For any vector v = v 1. v n R n we have for each 1 j n that v j v n max 1 j n v j. Proof. Suppose that v J is the coordinate of v with largest absolute value. Then for any index j, we have n vj 2 vi 2 nvj. 2 Taking square roots throughout gives the inequalities in the statement of the lemma. Now we come to the main point. The idea of a limit is one of the most important in all of mathematics. In differential calculus, it is the key to relating nonlinear (i.e. hard) functions to linear (i.e. easier) functions. 1
2 2 Definition 1.4. Suppose that F : R n R m is a function defined near a point a R n. We say that F (x) has limit b R m as x approaches a, i.e. lim F (x) = b x a Rm, if for each ɛ > 0 there exists δ > 0 such that 0 < x a < δ implies F (x) b < ɛ. Notice that the final phrase in this definition can be written in terms of balls instead of magnitudes: for any ɛ > 0 there exists δ > 0 such that x B (a, δ) implies F (x) B(b, ɛ). A function might or might not have a limit as x approaches some given point a, but it never has more than one. Proposition 1.5 (uniqueness of limits). If F : R n R m is defined near a R n, then there is at most one point b R m such that lim x a F (x) = b. Proof. Suppose, in order to reach a contradiction, that F (x) converges to two different points b, b R m as x approaches a. Then the quantity ɛ := 1 2 b b is positive. So by the definition of limit, there exists a number δ 1 > 0 such that 0 < x a < δ 1 implies F (x) b < ɛ. Likewise, there exists a number δ 2 > 0 such that 0 < x a < δ 2 implies F (x) b < ɛ. So if I let δ = min{δ 1, δ 2 } be the smaller bound, then 0 < x a < δ implies that b b = (F (x) b) (F (x) b) F (x) b + F (x) b < ɛ + ɛ b b. Note that the in this estimate follows from the triangle inequality, and the < follows from my choice of ɛ. At any rate, no real number is smaller than itself, so I have reached a contradiction and conclude that F cannot have two different limits at a. Definition 1.6. We say that a function F : R n R m is continuous at a R n if F is defined near and at a and lim F (x) = F (a). x a If F is continuous at all points in its domain, we say simply that F is continuous. Now let us verify that many familiar scalarvalued functions are continuous. Proposition 1.7. The following are continuous functions. (a) The constant function f : R n R, given by f(x) = c for some fixed c R and all x R n. (b) The magnitude function f : R n R given by f(x) = x. (c) The addition function f : R 2 R given by f(x 1, x 2 ) = x 1 + x 2. (d) The multiplication function f : R 2 R given by f(x 1, x 2 ) = x 1 x 2. (e) The reciprocal function f : R R given by f(x) = 1/x. Proof. (a) Fix a point a R n. Given ɛ > 0, let δ > 0 be any positive number, e.g. δ = 1 (it won t matter). Then if x a < δ it follows that f(x) f(a) = c c = 0 < ɛ. So the constant function f(x) = c is continuous at any point a R n.
3 (b) Fix a point a R n. Given ɛ > 0, let δ = ɛ. Then if x a < δ, it follows that f(x) f(a) = x a x a < δ = ɛ. The < here follows from Problem (on Homework 1) in Shifrin. Hence f(x) = x is continuous at any point a R n. (c) Fix a point a = (a 1, a 2 ) R 2. Given ɛ > 0, let δ = ɛ/2. Then if x a < δ, it follows that f(x) f(a) = x 1 + x 2 a 1 a 2 x 1 a 1 + x 2 a 2 < δ + δ = ɛ. Hence f(x 1, x 2 ) = x 1 + x 2 is continuous at any point (a 1, a 2 ) R 2. (d) Fix a point a = (a 1, a 2 ) R 2. Given ɛ > 0, let δ = min{1, ɛ(1 + a 1 + a 2 ) 1 }. Then if x a < δ, it follows that f(x) f(a) = x 1 x 2 a 1 a 2 = (x 1 x 2 x 1 a 2 ) + (x 1 a 2 a 1 a 2 ) x 1 x 2 x 1 a 2 + x 1 a 2 a 1 a 2 = x 1 x 2 a 2 + a 2 x 1 a 1 < δ( x 1 + a 2 ) < δ( a a 2 ) ɛ. Notice that the final < follows from the fact that x 1 a 1 < δ 1. Hence f(x 1, x 2 ) = x 1 x 2 is continuous at any point a = (a 1, a 2 ) R 2. 3 (e) Homework exercise. Theorem 1.8. Linear transformations T : R n R m are continuous. This one requires a little warmup. If A is an m n matrix, let us define the magnitude of A to be the quantity m n A := a 2 ij. That is, we are measuring the size of A as if it were a vector in R mn. Be aware that elsewhere in mathematics (including Shifrin), there are other notions of the magnitude of a matrix. Lemma 1.9. Given any matrix A M m n and any vector x R n, we have Ax A x. Proof. Let row i denote the ith row of A. Then on the one hand, we have m n m A 2 = a 2 ij = row i 2. j=1 But on the other hand, the ith entry of Ax is row i x. Hence from the CauchySchwartz inequality, we obtain m m m Ax 2 = (row i x) 2 row i 2 x 2 = x 2 row i 2 = A 2 x 2. j=1
4 4 Proof of Theorem 1.8. Let A M m n be the standard matrix of the linear transformation T : R n R m and a R n be any point. Given ɛ > 0, I choose δ = ɛ. Then if x a < δ, A it follows that T (x) T (a) = T (x a) = A(x a) A x a < A δ = ɛ. Hence T is continuous at any point a R n Questions about limits of vectorvalued functions can always be reduced to questions about scalarvalued functions. Proposition Suppose that F : R n R m is a vectorvalued function F = (F 1,..., F m ) defined near a R n. Then the following are equivalent. (a) lim x a F (x) = b R m. (b) lim x a F (x) b = 0. (c) lim x a F j (x) = b j for 1 j m. Proof of Proposition (a) = (b) Suppose that lim x a F (x) = b and set f(x) := F (x) b. Given ɛ > 0, the definition of limit gives me δ > 0 such that 0 < x a < δ implies that F (x) b < ɛ. But this last inequality can be rewritten f(x) 0 < ɛ. Thus lim x a f(x) = 0, i.e. (b) holds. (b) = (a) Similar. (a) = (c) Suppose again that lim x a F (x) = b. Fix an index j between 1 and m and let ɛ > 0 be given. Since lim x a F (x) = b, there exists δ > 0 such that 0 < x a < δ implies that F (x) b < ɛ. Then by Proposition 1.3, I also have that So lim x a F j (x) = b j. F j (x) b j F (x) b < ɛ. (c) = (a) Suppose that lim x a F j (x) = b j for each 1 j m. Then given any ɛ > 0, there exist real numbers δ j > 0 such that 0 < x a < δ j implies that F j (x) b j < ɛ m. Taking δ = min{δ 1,..., δ m }, I infer that 0 < x a < δ implies F j (x) b j < ɛ m for all indices j. Thus by the lemma, F (x) b < m ɛ = ɛ. m So lim x a F (x) = b. The following theorem is sometimes paraphrased by saying that limits commute with continuous functions. Theorem 1.11 (composite limits). Let F : R n R m and G : R m R p be functions and a R n, b R m be points such that lim x a F (x) = b and G is continuous at b. Then lim G F (x) = G(lim F (x)) = G(b). x a x a Proof. Let ɛ > 0 be given. By continuity of G at b, there exists a number ɛ > 0 such that y b < ɛ implies G(y) G(b) < ɛ. Likewise, since lim x a F (x) = b, there exists δ > 0 such that 0 < x a < δ implies that F (x) b < ɛ. Putting these two
5 things together, I see that 0 < x a < δ further implies that G(F (x)) G(b) < ɛ. So lim x a G(F (x)) = G(b). Corollary Let F : R n R m and G : R m R p be continuous functions. Then G F is continuous. Proof. Note that G F is defined at a R n precisely when F is defined at a and G is defined at F (a). Then Since both functions are continuous wherever they are defined, we have lim G(F (x)) = G(lim F (x)) = G(F (a)). x a x a Hence G F is continuous at any point a R n where it is defined. Corollary Let f, g : R n R be functions with limits lim x a f(x) = b and lim x a g(x) = c at some point a R n. Then (a) lim x a f(x) = b. (b) lim x a f(x) + g(x) = b + c; (c) lim x a f(x)g(x) = bc; (d) lim x a 1 f(x) = 1 b, provided b 0. Hence a sum or product of continuous functions is continuous, as is the reciprocal of a continuous function. Actually, the corollary extends to dot products, magnitudes and sums of vectorvalued functions F, G : R n R m, too. I ll let you write down the statements of these facts. Proof. I prove (c). The other parts are similar. Let F : R R 2 be given by F (x) := (f(x), g(x)) and m : R 2 R be the multiplication function m(y 1, y 2 ) := y 1 y 2. Recall that m is continuous (Proposition 1.7). Moreover, Proposition 1.10 and our hypotheses about f and g imply that lim x a F (x) = (b, c). Hence I infer from Theorem 1.11 that lim f(x)g(x) = lim m(f (x)) = m(lim F (x)) = m(b, c) = bc. x a x a x a When used with the fact that functions can t have more than one limit at a given point, Theorem 1.11 leads to a useful criterion for establishing that a limit doesn t exist. Definition A parametrized curve is a continuous function γ : R R n. Corollary Given a function F : R n R m defined near a point a R n, suppose that γ 1, γ 2 : R R n are parametrized curves such that γ 1 (t) = γ 2 (t) = a if and only if t = 0. If the limits lim t 0 F γ 1 (t) and lim t 0 F γ 2 (t) are not equal, then lim x a F (x) does not exist. Proof. I will prove the contrapositive statement: suppose that lim x a F (x) = b exists and γ 1, γ 2 : R R n are parametrized curves with initial points γ j (0) = a but γ j (t) a for t a. Then the limits lim x a F (x) and lim t 0 F γ j (t) do not concern the value of F at a. So I may assume with no loss of generality that F (a) = b, i.e. that F is actually continuous at a. Theorem 1.11 then tells me that lim t 0 F γ 1 (t) = F (lim t 0 γ 1 (t)) = F (γ 1 (0)) = F (a) = b. 5
6 6 The second equality holds because γ 1 is continuous. Likewise, lim t 0 F γ 2 (t) = b. In particular, the two limits are the same. I remark that if γ : R R n is a continuous curve and F : R n R m is a function, then the composite function F γ : R R m is sometimes called the restriction of F to γ. One last fact about limits that will prove useful for us is the following. Theorem 1.16 (The Squeeze Theorem). Suppose that F : R n R m and g : R n R are functions defined near a R n. Suppose there exists r > 0 such that F (x) g(x) for all x B (a, r); lim x a g(x) = 0. Then lim x a F (x) = 0. Proof. Exercise.
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