2 Sequences, Continuity, and Limits

Transcription

1 2 Sequences, Continuity, and Limits In this chapter, we introduce the fundamental notions of continuity and limit of a real-valued function of two variables. As in ACICARA, the definitions as well as proofs of basic results will be given using sequences. There are, actually, two possible generalizations of real sequences that seem natural in the setting of two variables. First, functions defined on N with values in R 2, and second, functions defined on N 2 with values in R. As we shall see, for developing the notions of continuity and limit of a function of two variables, only the former is relevant, and it is studied in this chapter. The study of the latter will be taken up in Chapter 7. This chapter is organized as follows. Sequences in R 2 are introduced in Section 2.1 below and their fundamental properties, including the Bolzano Weierstrass Theorem and the Cauchy Criterion, are derived from the corresponding results for sequences in R. We also use the notion of sequence to introduce basic topological notions of closed and open sets, boundary points, and interior points, and also the closure and the interior of subsets of R 2. Section 2.2 deals with the notion of continuity, and it is shown here that continuous functions on path-connected subsets of R 2 or on closed and bounded subsets of R 2 possess several nice properties. An important result known as the Implicit Function Theorem is also proved in this section. Finally, in Section 2.3 we introduce limits of functions of two variables. The definition is given using sequences, while most of the basic properties are proved using a simple observation that the existence of limit of a function at a point is equivalent to the continuity of an associated function at that point. 2.1 Sequences in R 2 A sequence in R 2 is a function from N to R 2. Typically, a sequence in R 2 is denoted by ( (x n, y n ) ), ( (u n, v n ) ), etc. The value of a sequence ( (x n, y n ) ) at n N is given by the element (x n, y n ) of R 2, and this element is called the S.R. Ghorpade and B.V. Limaye, A Course in Multivariable Calculus and Analysis, Undergraduate Texts in Mathematics, DOI / _2, Springer Science + Business Media, LLC

2 44 2 Sequences, Continuity, and Limits nth term of that sequence. In case the terms of a sequence ( (x n, y n ) ) lie in a subset D of R 2, then we say that ( (x n, y n ) ) is a sequence in D. The notions of boundedness and convergence extend readily from the setting of sequences in R to sequences in R 2. Let ( (x n, y n ) ) be a sequence in R 2. We say that ( (x n, y n ) ) is bounded if there is α R such that (x n, y n ) α for all n N. The sequence ( (x n, y n ) ) is said to be convergent if there is (x 0, y 0 ) R 2 that satisfies the following condition: For every ǫ > 0 there is n 0 N such that (x n, y n ) S ǫ (x 0, y 0 ) for all n n 0, that is, x n x 0 < ǫ and y n y 0 < ǫ for all n n 0. In this case, we say that ( (x n, y n ) ) converges to (x 0, y 0 ) or that (x 0, y 0 ) is a limit of ( (x n, y n ) ), and write (x n, y n ) (x 0, y 0 ). If ( (x n, y n ) ) does not converge to (x 0, y 0 ), then we write (x n, y n ) (x 0, y 0 ); if ( (x n, y n ) ) is not convergent, then it is said to be divergent. A sequence ( (x n, y n ) ) in R 2 gives rise to two sequences (x n ) and (y n ) in R, and vice versa. It turns out that the properties of ( (x n, y n ) ) can be completely understood in terms of the properties of the sequences (x n ) and (y n ) in R. Proposition 2.1. Given a sequence ( (x n, y n ) ) in R 2, we have the following. (i) If ( (x n, y n ) ) is convergent, then it has a unique limit. (ii) ( (x n, y n ) ) is bounded both (x n ) and (y n ) are bounded. (iii) ( (x n, y n ) ) is convergent both (x n ) and (y n ) are convergent. In fact, for (x 0, y 0 ) R 2, we have (x n, y n ) (x 0, y 0 ) x n x 0 and y n y 0. Proof. Each of (i), (ii), and (iii) is immediate from the definitions. As noted in part (i) of Proposition 2.1, if ( (x n, y n ) ) is a convergent sequence in R 2, then it has a unique limit in R 2. The limit of ( (x n, y n ) ) is sometimes written as lim n (x n, y n ) or as lim n (x n, y n ). Examples 2.2. (i) If ( (x n, y n ) ) is a constant sequence in R 2, that is, if there is (x 0, y 0 ) R 2 such that (x n, y n ) = (x 0, y 0 ) for all n N, then clearly, ( (x n, y n ) ) is convergent and (x n, y n ) (x 0, y 0 ). (ii) If ( (x n, y n ) ) is the sequence in R 2 defined by (x n, y n ) := (1/n, 1/n) for all n N, then clearly, ( (x n, y n ) ) is convergent and (x n, y n ) (0, 0). (iii) The sequence ( (x n, y n ) ) in R 2 defined by (x n, y n ) := (1/n, ( 1) n ) for all n N is divergent, since the sequence (( 1) n ) in R is divergent. Basic properties of sequences in R 2 readily follow from the corresponding properties of sequences in R. For ease of reference, we recall the relevant results for sequences in R. For proofs, one may refer to pages of ACICARA. Fact 2.3. Let (a n ) and (b n ) be sequences in R, and let a, b, α, β R. (i) If a n a and b n b, then a n + b n a + b and a n b n ab.

3 2.1 Sequences in R 2 45 (ii) If a n a, then for any r R, we have ra n ra. (iii) If a 0 and a n 0 for all n N, then (1/a n ) (1/a). (iv) Let a n a. If there is l N such that a n α for all n l, then a α. Likewise, if there is m N such that a n β for all n m, then a β. (v) If a n a and a n 0 for all n N, then a 1/k n a 1/k for any k N. (vi) (Sandwich Theorem in R) If (b n ) and (c n ) are sequences such that b n a and c n a, and if there is m N such that b n a n c n for all n m, then a n a. A few of these facts yield the result that sums, dot products, and scalar multiples of sequences in R 2 converge, respectively, to the sums, dot products, and scalar multiples of the corresponding limits. Proposition 2.4. Let ( (x n, y n ) ) and ( (u n, v n ) ) be sequences in R 2, and let (x 0, y 0 ), (u 0, v 0 ) R 2. (i) If (x n, y n ) (x 0, y 0 ) and (u n, v n ) (u 0, v 0 ), then (x n, y n )+(u n, v n ) (x 0, y 0 ) + (u 0, v 0 ) and (x n, y n ) (u n, v n ) (x 0, y 0 ) (u 0, v 0 ). (ii) If (x n, y n ) (x 0, y 0 ), then for any r R, r(x n, y n ) r(x 0, y 0 ). Proof. Immediate consequence of part (iii) of Proposition 2.1 together with parts (i) and (ii) of Fact 2.3. Analogues of properties of sequences in R that depend on order relations, are considered in Exercise 2. Subsequences and Cauchy Sequences Let ( (x n, y n ) ) be a sequence in R 2. If n 1, n 2,... are positive integers such that n k < n k+1 for each k N, then the sequence ( (x nk, y nk ) ), whose terms are (x n1, y n1 ), (x n2, y n2 ),..., is called a subsequence of ( (x n, y n ) ). The sequence ( (xn, y n ) ) is said to be Cauchy if for every ǫ > 0 there is n 0 N such that x n x m < ǫ and y n y m < ǫ for all n, m n 0. It is clear that ( (x n, y n ) ) is Cauchy if and only if both (x n ) and (y n ) are Cauchy sequences in R. Let us recall the following basic facts about sequences in R. For proofs, one may refer to pages 45, 56, and 58 of ACICARA. Fact 2.5. Let (a n ) be a sequence in R. Then we have the following. (i) (a n ) is convergent = (a n ) is bounded. (ii) (Bolzano Weierstrass Theorem in R) If (a n ) is bounded, then (a n ) has a convergent subsequence. (iii) (a n ) is convergent (a n ) is bounded and every convergent subsequence of (a n ) has the same limit. (iv) (Cauchy Criterion in R) (a n ) is Cauchy (a n ) is convergent. These facts, in turn, lead to the following results.

4 46 2 Sequences, Continuity, and Limits Proposition 2.6. Given a sequence ( (x n, y n ) ) in R 2, we have the following. (i) ( (x n, y n ) ) is convergent = ( (x n, y n ) ) is bounded. (ii) (Bolzano Weierstrass Theorem) If ( (x n, y n ) ) is a bounded sequence, then ( (x n, y n ) ) has a convergent subsequence. (iii) ( (x n, y n ) ) is convergent ( (x n, y n ) ) is bounded and every convergent subsequence of ( (x n, y n ) ) has the same limit. (iv) (Cauchy Criterion) ( (x n, y n ) ) is Cauchy ( (x n, y n ) ) is convergent. Proof. Clearly, (i) is an immediate consequence of parts (ii) and (iii) of Proposition 2.1 and part (i) of Fact 2.5. To prove (ii), suppose ( (x n, y n ) ) is bounded. Then (x n ) is a bounded sequence in R and hence by part (ii) of Fact 2.5, (x n ) has a convergent subsequence, say (x nk ). Now, (y n ) is a bounded sequence in R and hence so is (y nk ). So, by part (ii) of Fact 2.5, (y nk ) has a convergent subsequence, say ( y nkj ). Clearly, ( (xnkj, y nkj ) ) is a convergent subsequence of ( (x n, y n ) ). This proves (ii). Next, if ( (x n, y n ) ) is convergent, then it is clear that it is bounded and every convergent subsequence of ( (x n, y n ) ) has the same limit. To prove the converse, suppose ( (xn, y n ) ) is bounded. By (ii), ( (x n, y n ) ) has a convergent subsequence. Suppose (x 0, y 0 ) is the (same) limit for every convergent subsequence of ( (x n, y n ) ). If (x n, y n ) (x 0, y 0 ), then there are ǫ > 0 and positive integers n 1 < n 2 < such that max{ x nk x 0, y nk y 0 } ǫ for all k N. Now, ( (x nk, y nk ) ) is bounded and hence by (ii), it has a convergent subsequence. Moreover, this subsequence must converge to (x 0, y 0 ). This is a contradiction. Thus (iii) is proved. Finally, (iv) follows from part (iii) of Proposition 2.1, part (iv) of Fact 2.5, and our earlier observation that ( (x n, y n ) ) is Cauchy if and only if both (x n ) and (y n ) are Cauchy sequences in R. The result in part (iv) of Proposition 2.6 is sometimes referred to as the Cauchy completeness of R 2. A similar result holds for R n. Closure, Boundary, and Interior Let D R 2. We say that D is closed if every convergent sequence in D converges to a point of D. The set of all points in R 2 that are limits of convergent sequences in D is called the closure of D and is denoted by D. It is clear that D is closed if and only if D = D. A point of R 2 is said to be a boundary point of D if there is a sequence in D that converges to it and also a sequence in R 2 \ D that converges to it. The set of all boundary points of D in R 2 is called the boundary of D (in R 2 ), and is denoted by D. It is easy to see that D = (R 2 \ D), that is, the boundary of a set coincides with the boundary of its complement. A relation between the closure and the boundary is described by the following. Proposition 2.7. Given any D R 2, we have D = D D.

5 2.1 Sequences in R 2 47 Proof. Let (x, y) D. Then there is a sequence in D converging to (x, y). Further, if (x, y) D, then the constant sequence ( (x n, y n ) ) defined by (x n, y n ) = (x, y) for all n N gives a sequence in R 2 \ D converging to (x, y), and so in this case, (x, y) D. It follows that D D D. On the other hand, if (x, y) D D, then it is clear, using either a constant sequence or the definition of D, that (x, y) D, and so D D D. Proposition 2.8. Let D be a nonempty subset of R 2 such that D R 2. Then D is nonempty. Proof. Since D is nonempty, there is some (x 0, y 0 ) D, and since D R 2, there is some (x 1, y 1 ) R 2 \ D. Consider the line segment joining these two points, that is, consider L := {t [0, 1] : (1 t)(x 0, y 0 ) + t(x 1, y 1 ) D}. Then L is a nonempty subset of R bounded above by 1. Let t := supl and (x, y ) := (1 t )(x 0, y 0 ) + t (x 1, y 1 ). We claim that (x, y ) is a boundary point of D. To see this, let (t n ) be a sequence in L such that t n t. Let (x n, y n ) := (1 t n )(x 0, y 0 ) + t n (x 1, y 1 ) for n N. Clearly ( (x n, y n ) ) is a sequence in D that converges to (x, y ). Further, if t < 1, then we can find s n R for n N such that s n t and t < s n 1, and we let (u n, v n ) := (1 s n )(x 0, y 0 ) + s n (x 1, y 1 ) for n N, whereas if t = 1, then we let (u n, v n ) := (x 1, y 1 ) for n N. In any case, we see that ( (u n, v n ) ) is a sequence in R 2 \ D that converges to (x, y ). This proves the claim. Let D be a subset of R 2 and let (x 0, y 0 ) be any point of R 2. We say that (x 0, y 0 ) is an interior point of D if (x 0, y 0 ) D and (x 0, y 0 ) is not a boundary point of D. It is easy to see that (x 0, y 0 ) is an interior point of D if and only if there is r > 0 such that S r (x 0, y 0 ) D. The interior of D is defined to be the set of all interior points of D. Clearly, the interior of D is a subset of D. We say that D is open if every point of D is an interior point of D. The following proposition shows the connection between the notions of an open set and a closed set. Proposition 2.9. Let D R 2. Then D is closed if and only if R 2 \D is open. Proof. First, suppose D is a closed set. Let (x 0, y 0 ) R 2 \ D. If (x 0, y 0 ) is not an interior point of R 2 \ D, then there is a sequence ( (x n, y n ) ) in the complement of R 2 \ D, that is, in D, such that (x n, y n ) (x 0, y 0 ), and so (x 0, y 0 ) D = D, which is a contradiction. This proves that R 2 \ D is an open set. Conversely, suppose R 2 \ D is open. Let ( (x n, y n ) ) be any sequence in D such that (x n, y n ) (x 0, y 0 ) for some (x 0, y 0 ) R 2. Then (x 0, y 0 ) cannot be an interior point of R 2 \D. But since R 2 \D is open, it follows that (x 0, y 0 ) R 2 \ D, that is, (x 0, y 0 ) D. This proves that D is closed. Example Let α, β R with α > 0 and β > 0. Consider the sets D 1 := {(x, y) R 2 : x α and y β}, D 2 := {(x, y) R 2 : x < α and y β}, D 3 := {(x, y) R 2 : x α and y < β}, and D 4 := {(x, y) R 2 : x < α and y < β}. In view of part (iv) of Fact 2.3, we readily see that D 1 is

6 48 2 Sequences, Continuity, and Limits closed, D 4 is open, whereas D 2 and D 3 are neither closed nor open. Further, for each i = 1, 2, 3, 4, the closure of D i is D 1, the interior of D i is D 4, and the boundary of D i is the set {(x, y) R 2 : x = α and y = β}. Remark The notions discussed in this section concerning sequences in R 2, closed sets, closure, boundary points, boundary, interior points, interior, and open sets admit a straightforward extension to R 3 and more generally, to R n for any n N. To avoid a notational conflict, one may denote a sequence in R n by (x k ), where the parameter k runs through N and x k R n for each k N. It may be instructive to formulate precise analogues of the notions and results in this section for R n and write down proofs of analogous results in the general case. This may also be a good opportunity to review the results in this section. 2.2 Continuity Let D be a subset of R 2 and let (x 0, y 0 ) be any point in D. A function f : D R is said to be continuous at (x 0, y 0 ) if for every sequence ( (x n, y n ) ) in D such that (x n, y n ) (x 0, y 0 ), we have f(x n, y n ) f(x 0, y 0 ). If f is not continuous at (x 0, y 0 ), then we say that f is discontinuous at (x 0, y 0 ). When f is continuous at every (x 0, y 0 ) D, we say that f is continuous on D. Examples (i) If D is any subset of R 2 and f : D R is a constant function on D, that is, if there is c R such that f(x, y) = c for all (x, y) D, then clearly, f is continuous on D. (ii) If f : R 2 R is the norm function given by f(x, y) := x 2 + y 2 for (x, y) R 2, then f is continuous on R 2. To see this, let (x 0, y 0 ) R be any point and let ( (x n, y n ) ) be a sequence in R 2 such that (x n, y n ) (x 0, y 0 ). Then by part (iii) of Proposition 2.1, the sequences (x n ) and (y n ) in R are such that x n x 0 and y n y 0. Hence, by parts (i) and (v) of Fact 2.3, we see that x 2 n + y2 n x y2 0. Thus f is continuous on R2. (iii) Consider the coordinate functions p 1, p 2 : R 2 R defined by p 1 (x, y) := x and p 2 (x, y) := y for (x, y) R 2. Then by part (iii) of Proposition 2.1, we immediately see that p 1 and p 2 are continuous on R 2. (iv) Let D R 2 and let us fix (x 0, y 0 ) D. Consider D 1 := {x R : (x, y 0 ) D} and D 2 := {y R : (x 0, y) D}. Notice that the set D 1 depends on y 0, whereas D 2 depends on x 0. Given any f : D R, let φ : D 1 R and ψ : D 2 R be functions of one variable defined by φ(x) := f(x, y 0 ) for x D 1 and ψ(y) := f(x 0, y) for y D 2. These functions will play a useful role in the study of the function f of two variables around the point (x 0, y 0 ). If f is continuous at (x 0, y 0 ), then

7 2.2 Continuity 49 φ is continuous at x 0 and ψ is continuous at y 0. To see this, let (x n ) be a sequence in D 1 such that x n x 0. Then (x n, y 0 ) (x 0, y 0 ) and hence f(x n, y 0 ) f(x 0, y 0 ), that is, φ(x n ) φ(x 0 ). Thus φ is continuous at x 0. Similarly, ψ is continuous at y 0. Let us recall that the sign of a continuous function of one variable is preserved in a neighborhood of that point. More precisely, we have the following. For a proof, one may refer to page 68 of ACICARA. Fact Let E R, c E, and let φ : E R be continuous at c. If φ(c) > 0, then there is δ > 0 such that φ(x) > 0 for all x E (c δ, c + δ). Likewise, if φ(c) < 0, then there is δ > 0 such that φ(x) < 0 for all x E (c δ, c + δ). A similar result holds for functions of two variables. Lemma Let D R 2, (x 0, y 0 ) D, and let f : D R be a function that is continuous at (x 0, y 0 ). If f(x 0, y 0 ) > 0, then there is δ > 0 such that f(x, y) > 0 for all (x, y) D S δ (x 0, y 0 ). Likewise, if f(x 0, y 0 ) < 0, then there is δ > 0 such that f(x, y) < 0 for all (x, y) D S δ (x 0, y 0 ). Proof. First, suppose f(x 0, y 0 ) > 0. If there is no δ > 0 with the desired property, then for each n N, we can find (x n, y n ) D S 1/n (x 0, y 0 ) such that f(x n, y n ) 0. Now (x n, y n ) (x 0, y 0 ), and since f is continuous at (x 0, y 0 ), we have f(x n, y n ) f(x 0, y 0 ). Hence, by part (iv) of Fact 2.3, f(x 0, y 0 ) 0, which is a contradiction. The proof when f(x 0, y 0 ) < 0 is similar. Proposition Let D R 2, (x 0, y 0 ) D, r R, and let f, g : D R be continuous at (x 0, y 0 ). Then f + g, rf, and fg are continuous at (x 0, y 0 ). In case f(x 0, y 0 ) 0, there is δ > 0 such that f(x, y) 0 for all (x, y) D S δ (x 0, y 0 ), and the function 1/f : D S δ (x 0, y 0 ) R is continuous at (x 0, y 0 ). In case there is δ > 0 such that f(x, y) 0 for all (x, y) D S δ (x 0, y 0 ), the function f 1/k : D S δ (x 0, y 0 ) R is continuous at (x 0, y 0 ) for every k N. Proof. The continuity of f+g, rf, and fg at (x 0, y 0 ) follows readily from parts (i) and (ii) of Fact 2.3. In case f(x 0, y 0 ) 0, we have either f(x 0, y 0 ) > 0 or f(x 0, y 0 ) < 0. Thus, by Lemma 2.14, there is δ > 0 such that f(x, y) 0 for all (x, y) D S δ (x 0, y 0 ). Now, by part (iii) of Fact 2.3, we see that the function 1/f : D S δ (x 0, y 0 ) R is continuous at (x 0, y 0 ). Finally, the assertion about the continuity of f 1/k at (x 0, y 0 ) is a direct consequence of part (v) of Fact 2.3. As in the case of functions of one variable, we can easily deduce from Proposition 2.15 the following. Suppose D R 2 and f, g : D R are continuous at (x 0, y 0 ) D. Then the difference f g is continuous at (x 0, y 0 ). Also, if g(x 0, y 0 ) 0, then the quotient f/g is continuous at (x 0, y 0 ). Further, if there is δ > 0 such that f(x) 0 for all x D S δ (x 0, y 0 ), then for every

8 50 2 Sequences, Continuity, and Limits positive rational number r, the function f r is continuous at (x 0, y 0 ). Similarly, if f(x 0, y 0 ) > 0, then for every negative rational number r the function f r is continuous at (x 0, y 0 ). Examples (i) Using Proposition 2.15 and the above remarks, we see that every polynomial function on R 2 is continuous and every rational function is continuous wherever it is defined, that is, if p(x, y) and q(x, y) are polynomials in two variables and if D := {(x, y) R 2 : q(x, y) 0}, then the rational function f : D R defined by f(x, y) := p(x, y)/q(x, y) for (x, y) D is continuous on D. Moreover, if E = {(x, y) R 2 : p(x, y) 0 and q(x, y) > 0}, then for any m, n N, the algebraic function g : E R defined by g(x, y) := p(x, y) 1/m /q(x, y) 1/n for (x, y) E, is continuous on E. (ii) Consider f : R 2 R defined as follows. xy f(x, y) := x 2 + y 2 if (x, y) (0, 0), 0 if (x, y) = (0, 0). Then f is not continuous at (0, 0). To see this, consider a sequence in R 2 approaching (0, 0) along the line y = x; for example, the sequence ( (1/n, 1/n) ). Then (1/n, 1/n) (0, 0), but f(1/n, 1/n) 1/2 f(0, 0). (iii) Consider a variant of the function in (ii), namely, f : R 2 R given by x 2 y f(x, y) := x 2 + y 2 if (x, y) (0, 0), 0 if (x, y) = (0, 0). Then f is continuous at (0, 0). To see this, note that for any (x, y) R 2, we have x 2 x 2 +y 2 and consequently, f(x, y) y. Hence if ( (x n, y n ) ) is any sequence in R 2 with (x n, y n ) (0, 0), then y n 0, and as a result, f(x n, y n ) 0 = f(0, 0). (iv) Consider a variant of the function in (iii), namely, f : R 2 R given by x 2 y f(x, y) := x 4 + y 2 if (x, y) (0, 0), 0 if (x, y) = (0, 0). Then f(x, y) approaches 0 along every line passing through the origin [indeed, f(0, y) = 0 and f(x, mx) = mx/(x 2 + m 2 ) 0 as x 0]. However, f is not continuous at (0, 0). To see this, consider a sequence in R 2 approaching (0, 0) along the parabola y = x 2 ; for example, the sequence ( (1/n, 1/n 2 ) ). Then (1/n, 1/n 2 ) (0, 0), but f(1/n, 1/n 2 ) 1/2 f(0, 0). (v) Consider a variant of the function in (iv), namely, f : R 2 R given by

9 2.2 Continuity 51 x 3 y f(x, y) := x 4 + y 2 if (x, y) (0, 0), 0 if (x, y) = (0, 0). Then f is continuous at (0, 0). To see this, use the A.M.-G.M. Inequality (given, for example, on page 12 of ACICARA) to obtain 2 x 2 y x 4 + y 2, and hence f(x, y) x /2 for all (x, y) R 2. Thus, if ( (x n, y n ) ) is any sequence in R 2 with (x n, y n ) (0, 0), then we see that x n 0 and as a result, f(x n, y n ) 0 = f(0, 0). Composition of Continuous Functions We now show that the composition of continuous functions is continuous. It may be noted that for functions of two variables, three types of composites are possible. Thus, the following result is stated in three parts. Proposition Let D R 2, (x 0, y 0 ) D, and let f : D R be continuous at (x 0, y 0 ). (i) Suppose E R is such that f(d) E. If g : E R is continuous at f(x 0, y 0 ), then g f : D R is continuous at (x 0, y 0 ). (ii) Suppose E R, t 0 E, and x, y : E R are such that (x(t), y(t)) D for all t E and (x(t 0 ), y(t 0 )) = (x 0, y 0 ). If x, y are continuous at t 0, then F : E R defined by F(t) := f(x(t), y(t)) is continuous at t 0. (iii) Suppose E R 2, (u 0, v 0 ) E, and x, y : E R are such that (x(u, v), y(u, v)) D for all (u, v) E and (x(u 0, v 0 ), y(u 0, v 0 )) = (x 0, y 0 ). If x, y are continuous at (u 0, v 0 ), then F : E R defined by F(u, v) := f(x(u, v), y(u, v)) is continuous at (u 0, v 0 ). Proof. (i) Suppose E and g satisfy the hypotheses in (i). Let ( (x n, y n ) ) be a sequence in D such that (x n, y n ) (x 0, y 0 ). By the continuity of f at (x 0, y 0 ), we obtain f(x n, y n ) f(x 0, y 0 ). Now (f(x n, y n )) is a sequence in f(d), and hence by the continuity of g at f(x 0, y 0 ), we obtain g (f(x n, y n )) g (f(x 0, y 0 )). So g f : D R is continuous at (x 0, y 0 ). (ii) Suppose E, t 0, and the functions x, y satisfy the hypotheses in (ii), and F is as defined in (ii). Let (t n ) be a sequence in E such that t n t 0. By the continuity of x and y at t 0, we obtain x(t n ) x(t 0 ) and y(t n ) y(t 0 ). Thus, by part (iii) of Proposition 2.1, ( x(t n ), y(t n ) ) is a sequence in D that converges to (x 0, y 0 ). Hence by the continuity of f at (x 0, y 0 ), we obtain f(x(t n ), y(t n )) f(x 0, y 0 ), that is, F(t n ) F(t 0 ). So F is continuous at t 0. (iii) Suppose E, (u 0, v 0 ), and the functions x, y satisfy the hypotheses in (iii), and F is as defined in (iii). Let (u n, v n ) be a sequence in E such that (u n, v n ) (u 0, v 0 ). By the continuity of x and y at (u 0, v 0 ), we obtain x(u n, v n ) x(u 0, v 0 ) and y(u n, v n ) y(u 0, v 0 ). Thus, by part (iii) of Proposition 2.1, ( x(u n, v n ), y(u n, v n ) ) is a sequence in D that converges to (x 0, y 0 ). Hence by the continuity of f at (x 0, y 0 ), we obtain f(x(u n, v n ), y(u n, v n )) f(x 0, y 0 ), that is, F(u n, v n ) F(u 0, v 0 ). So F is continuous at (u 0, v 0 ).

10 52 2 Sequences, Continuity, and Limits Examples (i) By part (i) of Proposition 2.17, f : R 2 R defined by f(x, y) := sin(xy) is continuous at each (x 0, y 0 ) R 2, and g : R 2 R defined by g(x, y) := cos(x + y) is continuous at each (x 0, y 0 ) R 2. (ii) By part (ii) of Proposition 2.17, if f(x, y) is any polynomial in two variables, then F : R R defined by F(t) := f(e t, sin t) for t R is continuous at every t 0 R. (iii) By part (iii) of Proposition 2.17, if f(x, y) is any polynomial in two variables, then F : R 2 R defined by F(u, v) := f(sin(uv), cos(u + v)) for (u, v) R 2 is continuous at every (u 0, v 0 ) R 2. (iv) Consider the functions that give the polar coordinates of a point in R 2 other than the origin. (See Section 1.3 and, in particular, Fact 1.26.) More precisely, consider r : R 2 R and θ : R 2 \ {(0, 0)} R defined by ( ) x cos r(x, y) := 1 if y 0, r(x, y) x 2 + y 2 and θ(x, y) := ( ) x cos 1 if y < 0. r(x, y) Then, as seen already in Example 2.12 (ii), the function r is continuous on R 2. Also, we know that cos 1 : [ 1, 1] R is a continuous function of one variable. (See, for example, page 252 of ACICARA.) Consequently, by Proposition 2.15 and part (i) of Proposition 2.17, we see that the function θ is continuous at every (x 0, y 0 ) R 2 for which y 0 0. Also, θ is continuous on the positive x-axis. To see this, note that if (x 0, 0) R 2 with x 0 > 0 and if ( (x n, y n ) ) is any sequence in R 2 \ {(0, 0)} converging to (x 0, 0), then ( θ(x n, y n ) = cos 1 x n x 2 n + yn) 2 ( ) x0 cos 1 = cos 1 (1) = 0, x 0 and hence θ(x n, y n ) 0. However, at points on the negative x-axis, the function θ is discontinuous. To see this, fix (x 0, 0) R 2 with x 0 < 0. Clearly, we can find sequences ( (x n, y n ) ) and ( (u n, v n ) ) in R 2 \ {(0, 0)} converging to (x 0, 0) such that y n 0 and v n < 0 for all n N. Now, whereas θ(x n, y n ) = cos 1 ( θ(u n, v n ) = cos 1 ( x n x 2 n + y 2 n u n u 2 n + v 2 n ) ) ( ) cos 1 x0 = cos 1 ( 1) = π, x 0 ( ) cos 1 x0 = cos 1 ( 1) = π. x 0 Thus, θ is discontinuous at every point of {(x, y) R 2 : x < 0 and y = 0}. In fact, given any x 0 < 0, we can take x n = u n = x 0 for all n N in

11 2.2 Continuity 53 the above argument, and this shows that the function from (, 0] to R given by y θ(x 0, y) is discontinuous at 0. On the other hand, the functions that give the rectangular coordinates of a point in the (polar) plane are continuous. More precisely, the functions x, y : R 2 R defined by x(r, θ) := r cosθ and y(r, θ) := r sinθ are continuous on R 2. Piecing Continuous Functions on Overlapping Subsets An effective way to construct continuous functions of one variable is to piece together two continuous function defined on overlapping subsets that intersect at a single point, provided their values agree at the common point of intersection. (See, for example, Proposition 3.5 of ACICARA.) We now obtain a similar result for functions of two variables. A precise statement is given below, and the key hypothesis in this result is illustrated in Figure 2.1. Proposition Let D 1 and D 2 be subsets of R 2 and let f 1 : D 1 R and f 2 : D 2 R be continuous functions such that f 1 (x, y) = f 2 (x, y) for all (x, y) D 1 D 2. Let D := D 1 D 2 and let f : D R be defined by { f 1 (x, y) if (x, y) D 1, f(x, y) := f 2 (x, y) if (x, y) D 2. If D i is closed in D, that is, D i D = D i for i = 1, 2, then f is continuous. Proof. Since f 1 and f 2 agree on D 1 D 2, it is clear that f is well defined. Assume now that each D i is closed in D for i = 1, 2. Fix (x 0, y 0 ) D. Let ( (xn, y n ) ) be a sequence in D such that (x n, y n ) (x 0, y 0 ). In case there is n 1 N such that (x n, y n ) D 1 for all n n 1, then (x 0, y 0 ) D 1 since D 1 is closed in D; further, by the continuity of f 1 on D 1, we obtain f(x n, y n ) = f 1 (x n, y n ) f 1 (x 0, y 0 ) = f(x 0, y 0 ). Similarly, in case there is n 2 N such that (x n, y n ) D 2 for all n n 2, then (x 0, y 0 ) D 2 and f(x n, y n ) f(x 0, y 0 ). In the remaining case, there are two subsequences ( (x lk, y lk ) ) and ( (x mk, y mk ) ) of ( (x n, y n ) ) such that (x lk, y lk ) D 1 and (x mk, y mk ) D 2 for all k N, and moreover, N = {l 1, l 2,... } {m 1, m 2,... }. Clearly, (x lk, y lk ) (x 0, y 0 ) and (x mk, y mk ) (x 0, y 0 ). Now, since each D i is closed in D, we have (x 0, y 0 ) D 1 D 2 ; further, since each f i is continuous at (x 0, y 0 ), we have f (x lk, y lk ) = f 1 (x lk, y lk ) f 1 (x 0, y 0 ) = f(x 0, y 0 ) and f (x mk, y mk ) = f 2 (x mk, y mk ) f 2 (x 0, y 0 ) = f(x 0, y 0 ). Since N = {l 1, l 2,... } {m 1, m 2,... }, it follows that f(x n, y n ) f(x 0, y 0 ). This proves that f is continuous at (x 0, y 0 ). Examples (i) Consider the semiopen rectangles D 1 := (0, 1] [ 1, 1] and D 2 := [1, 2) [ 1, 1]. (See Figure 2.1.) Note that neither D 1 nor D 2 is closed in R 2, but each D i is closed in D := D 1 D 2 for i = 1, 2. Thus the hypothesis of Proposition 2.19 is satisfied, and continuous functions on D 1 and D 2 that agree on D 1 D 2 = {1} [ 1, 1] extend to a continuous function on D.

12 54 2 Sequences, Continuity, and Limits y 1 D 1 D 2 y x D 1 D 2 0 x 1 2 Fig Illustration of the conditions D 1 D = D 1 and D 2 D = D 2 in Proposition 2.19 that are satisfied in Example 2.20(i) and violated in Example 2.20(ii). (ii) Let D 1 be the open disk B 1 (0, 0) and let D 2 the closure of the disk B 1 (2, 0), that is, D 1 := {(x, y) R 2 : x 2 + y 2 < 1} and D 2 := {(x, y) R 2 : (x 2) 2 + y 2 1}. (See Figure 2.1 (ii).) Consider f 1 : D 1 R and f 2 : D 2 R defined by f 1 (x, y) := 0 for all (x, y) D 1 and f 2 (x, y) := 1 for all (x, y) D 2. Clearly, f 1 and f 2 are continuous. Moreover, D 1 D 2 = and hence f : D 1 D 2 R as given in Proposition 2.19 is well defined. But f is not continuous at (1, 0), since (x n, y n ) := ( 1 1 n, 0) (1, 0), whereas f(x n, y n ) = f 1 (x n, y n ) = 0 for all n N, and thus f(x n, y n ) 1 = f(1, 0). This shows that the hypothesis D i D = D i for i = 1, 2 in Proposition 2.19 cannot be dropped. An easy inductive argument shows that the result in Proposition 2.19 can be extended to piece together continuous functions not just on two overlapping sets, but on any finite number of sets, provided they agree on all pairwise intersections and each of the sets is closed in the union of all the sets. For our purpose, it will suffice to record the following special case of partitioning a set into four quadrants at a given point. Corollary Let D R and let f : D R be a function. Given any (x 0, y 0 ) D, let D 1 := {(x, y) D : x x 0 and y y 0 }, D 2 := {(x, y) D : x x 0 and y y 0 }, D 3 := {(x, y) D : x x 0 and y y 0 }, D 4 := {(x, y) D : x x 0 and y y 0 }, and f i = f Di for i = 1,...,4. Then f is continuous if and only if f i is continuous for each i = 1,...,4. Proof. If f is continuous, then clearly f i is continuous for each i = 1,...,4. To prove the converse, consider E 1 := D 1 D 2 and E 2 := D 3 D 4, and also g i := f Ei for i = 1, 2. Using Proposition 2.19, we see that the continuity of f 1 and f 2 implies the continuity of g 1, while the continuity of f 3 and f 4 implies the continuity of g 2. Further, the continuity of f follows from the continuity of g 1 and g 2 using Proposition 2.19 again.

13 2.2 Continuity 55 Characterizations of Continuity We have chosen to define continuity of a function at a point using sequences. Alternative definitions are possible, as is shown by the result below. Proposition Let D R 2, (x 0, y 0 ) D, and let f : D R be any function. Then the following are equivalent. (i) f is continuous at (x 0, y 0 ), that is, for every sequence ( (x n, y n ) ) in D such that (x n, y n ) (x 0, y 0 ), we have f(x n, y n ) f(x 0, y 0 ). (ii) For every ǫ > 0, there is δ > 0 such that f(x, y) f(x 0, y 0 ) < ǫ for all (x, y) D S δ (x 0, y 0 ). (iii) For every open subset V of R containing f(x 0, y 0 ), there is an open subset U of R 2 containing (x 0, y 0 ) such that f(u D) V, that is, f(x, y) V for all (x, y) U D. Proof. Assume that (i) holds. If (ii) does not hold, then there is ǫ > 0 such that for every δ > 0, there is (x, y) in D S δ (x 0, y 0 ) with the property that f(x, y) f(x 0, y 0 ) ǫ. Consequently, for each n N, there is (x n, y n ) in D S 1/n (x 0, y 0 ) such that f(x n, y n ) f(x 0, y 0 ) ǫ. But then (x n, y n ) (x 0, y 0 ) and f(x n, y n ) / f(x 0, y 0 ). This contradicts (i). Thus, (i) (ii). Next, assume that (ii) holds. Let V be an open subset of R containing f(x 0, y 0 ). Then there is ǫ > 0 such that (f(x 0, y 0 ) ǫ, f(x 0, y 0 ) + ǫ) V. By (ii), we can find δ > 0 such that f(x, y) f(x 0, y 0 ) < ǫ for all (x, y) D S δ (x 0, y 0 ). Thus, if we let U = S δ (x 0, y 0 ), then U is an open subset of R 2 containing (x 0, y 0 ) such that f(u D) V. Thus, (ii) (iii). Finally, assume that (iii) holds. Let ( (x n, y n ) ) be any sequence in D such that (x n, y n ) (x 0, y 0 ). Given any ǫ > 0, take V to be the open interval (f(x 0, y 0 ) ǫ, f(x 0, y 0 ) + ǫ) in R. By (iii), there is an open subset U of R 2 containing (x 0, y 0 ) such that f(u D) V. Since U is open, there is δ > 0 such that S δ (x 0, y 0 ) U. Further, since (x n, y n ) (x 0, y 0 ), there is n 0 N such that (x n, y n ) S δ (x 0, y 0 ) for all n n 0. Consequently, f(x n, y n ) is in (f(x 0, y 0 ) ǫ, f(x 0, y 0 ) + ǫ), that is, f(x n, y n ) f(x 0, y 0 ) < ǫ for all n n 0. Thus, f(x n, y n ) f(x 0, y 0 ), and so (iii) (i). This proves the equivalence of (i), (ii), and (iii). Corollary Let D R 2 be open in R 2 and let f : D R be any function. Then f is continuous on D if and only if for every open subset V of R, the set f 1 (V ) := {(x, y) D : f(x, y) V } is open in R 2. Proof. Follows easily from Proposition Example Clearly, the intervals (0, ), (, 0) and the set R \ {0} are open subsets of R. Thus, as a consequence of Corollary 2.23, we see that if f : R 2 R is continuous, then each of the sets {(x, y) R 2 : f(x, y) > 0}, {(x, y) R 2 : f(x, y) < 0}, and {(x, y) R 2 : f(x, y) 0} is open in R 2.

14 56 2 Sequences, Continuity, and Limits Continuity and Boundedness A bounded function need not be continuous. Consider, for example, the function f : R 2 R defined by { 1 if both x and y are rational, f(x, y) := 0 otherwise. Clearly, f is bounded but f is not continuous at any point of R 2. Also, a continuous function need not be bounded. For example, g : R 2 R and h : (0, 1) (0, 1) R defined by g(x, y) := x + y and h(x, y) := 1/(x + y) are both continuous, but neither g nor h is a bounded function. It may be noted that the domain of g is closed, but not bounded, whereas the domain of h is bounded, but not closed. The following result shows that the situation is nicer if the domain is closed as well as bounded. Proposition Let D R 2 be closed and bounded, and let f : D R be continuous. Then f is bounded, that is, f(d) := {f(x, y) : (x, y) D} is a bounded subset of R. Also, f(d) is a closed subset of R. As a consequence, if D is nonempty, then f attains its bounds, that is, there are (a, b), (c, d) D such that f(a, b) = sup f(d) and f(c, d) = inf f(d). Proof. Suppose f is not bounded above. Then for each n N, there is (x n, y n ) D such that f(x n, y n ) > n. Since D is bounded, by the Bolzano Weierstrass Theorem (part (ii) of Proposition 2.6), the sequence ( (x n, y n ) ) has a convergent subsequence, say ( (x nk, y nk ) ). Suppose (x nk, y nk ) (x 0, y 0 ). Then (x 0, y 0 ) D, since D is closed, and f(x nk, y nk ) f(x 0, y 0 ), since f is continuous. On the other hand, f(x nk, y nk ) > n k for each k N, and n k as k, which leads to a contradiction. Hence f must be bounded above. Similarly, it can be seen that f is bounded below. Thus f(d) is bounded. Next, suppose (z n ) is a sequence in f(d) such that z n r for some r R. Write z n = f(x n, y n ), where (x n, y n ) D for n N. As before, ( (x n, y n ) ) has a convergent subsequence, say ( (x nk, y nk ) ), which must converge to a point (x 0, y 0 ) of D. Since f is continuous at (x 0, y 0 ), z nk = f(x nk, y nk ) f(x 0, y 0 ), and hence r = f(x 0, y 0 ), which shows that r f(d). Thus f(d) is closed. Finally, if D is nonempty, then f(d) is a nonempty bounded subset of R and thus M := sup f(d) and m := inf f(d) are well defined. By the definition of supremum and infimum, for each n N, we can find (a n, b n ), (c n, d n ) D such that M 1 n < f(a n, b n ) M and m f(c n, d n ) < m+ 1 n. Consequently, f(a n, b n ) M and f(c n, d n ) m. Since f(d) is closed, M, m f(d), that is, f(a, b) = sup f(d) and f(c, d) = inf f(d) for some (a, b), (c, d) D. Remark Subsets of R 2 (and more generally, of R n ) that are both closed and bounded are often referred to as compact sets. Thus, the above proposition says that the continuous image of a compact set is compact. For more on compactness, see Exercise 17.

15 2.2 Continuity 57 Continuity and Monotonicity For functions of one variable, there is no direct relationship between continuity and monotonicity. Indeed, it suffices to consider the integer part function x [x] and the absolute value function x x to conclude that a monotonic function need not be continuous and a continuous function need not be monotonic. For functions of two variables, a similar situation prevails. In fact, using the product order on R 2, we have introduced in Chapter 1 two distinct notions: monotonicity and bimonotonicity. We will show below that neither of these implies or is implied by continuity. Examples (i) Consider f : [ 1, 1] [ 1, 1] R defined by f(x, y) := xy. Clearly, f is continuous but not monotonic on [ 1, 1] [ 1, 1]. Note, however, that f is bimonotonically increasing on [ 1, 1] [ 1, 1], since we have x 2 y 2 + x 1 y 1 x 2 y 1 x 1 y 2 = (x 2 x 1 )(y 2 y 1 ) for all (x 1, y 1 ), (x 2, y 2 ) R 2. (ii) Consider f : [ 1, 1] [ 1, 1] R defined by f(x, y) := (x + y) 3. Clearly, f is continuous. However, f is not bimonotonic on [ 1, 1] [ 1, 1]. To see this, observe that (x 1, y 1 ) := (0, 0) and (x 2, y 2 ) := (1, 1) are points of [ 1, 1] [ 1, 1] satisfying (x 1, y 1 ) (x 2, y 2 ) and f(x 1, y 1 ) + f(x 2, y 2 ) f(x 1, y 2 ) f(x 2, y 1 ) = = 6 > 0, whereas (u 1, v 1 ) := ( 1, 1) and (u 2, v 2 ) := (0, 0) are points of [ 1, 1] [ 1, 1] satisfying (u 1, v 1 ) (u 2, v 2 ) and f(u 1, v 1 ) + f(u 2, v 2 ) f(u 1, v 2 ) f(u 2, v 1 ) = = 6 < 0. (iii) Consider f : [ 1, 1] [ 1, 1] R defined by { 1 if x > 0 and y > 0, f(x, y) := 0 otherwise. It is easy to see that f is monotonically as well as bimonotonically increasing, but not continuous on [ 1, 1] [ 1, 1]. Continuity, Bounded Variation, and Bounded Bivariation In general, a function of bounded variation need not be continuous. Likewise for a function of bounded bivariation. In fact, Example 2.27 (iii) provides a common counterexample. We have seen earlier that a continuous function need not be monotonic or bimonotonic. The following example shows that it need not even be of bounded variation or of bounded bivariation. Example Consider f : [0, 1] [0, 1] R defined by { xy cos(π/2x) if x 0, f(x, y) := 0 if x = 0.

16 58 2 Sequences, Continuity, and Limits Clearly, f is continuous on (0, 1] [0, 1]. Moreover, since f(x, y) xy for all (x, y) [0, 1] [0, 1], it is readily seen that f is continuous at (0, y) for each y [0, 1]. Thus, f is continuous on [0, 1] [0, 1]. Next, given any even positive integer n, say n = 2k for some k N, if we consider the points x 0 = 0 = y 0 and x i := 1 n + 1 i and y i = 1 for i = 1,...,n, then we have (0, 0) = (x 0, y 0 ) (x 1, y 1 ) (x n, y n ) = (1, 1) and moreover, f(x i, y i ) = 0 if i is even and f(x i, y i ) = ±x i if i is odd. Thus n i=1 f(x i, y i ) f(x i 1, y i 1 ) = 1 n + 1 n + 1 n n k 2 = 1 i. Since the set { k i=1 (1/i) : k N} is not bounded above (as is shown, for example, on page 51 of ACICARA), it follows that f is not of bounded variation on [0, 1] [0, 1]. Furthermore, if we let n = 2k and x 0, x 1,...,x n be as above, but take m = 1, y 0 = 0, and y 1 = 1, then 0 = x 0 x 1 x n = 1 and 0 = y 0 y 1 = 1, and moreover, for any i 0, we have f(x i, 0) = 0, whereas f(x i, 1) = 0 if i is even and f(x i, 1) = ±x i if i is odd, and thus n i=1 j=1 m f(x i, y j ) + f(x i 1, y j 1 ) f(x i, y j 1 ) f(x i 1, y j ) = It follows, therefore, that f is not of bounded bivariation on [0, 1] [0, 1]. Remark Using Exercise 38, a refined version of the Jordan decomposition (Propositions 1.12 and 1.17) can be obtained for continuous functions. Namely, a continuous function of bounded variation is a difference of continuous monotonic functions, whereas a continuous function of bounded bivariation is a difference of continuous bimonotonic functions. Continuity and Convexity In general, a continuous function is neither convex nor concave. For example, consider D := [ 1, 1] [ 1, 1] and f : D R defined by f(x, y) := x 3 + y 3. Clearly, f is continuous. But f is neither convex nor concave. To see this, observe that ( 1 2, 2) 1 = 1 2 ( 1, 1) (0, 0) and ( 1 2, ) 1 2 = 1 2 (1, 1) (0, 0), but f ( 1 2, 2) 1 = 1 4 > 1 = 1 2 f( 1, 1) f(0, 0), and f ( 1 2, ) 1 2 = 1 4 < 1 = 1 2 f(1, 1) + 1 2f(0, 0). Moreover, a convex function need not be continuous. For example, if D := {(x, y) R 2 : x 2 + y 2 1} is the closed unit disk and f : D R is a variant of the norm function defined by { x2 + y f(x, y) := 2 if x 2 + y 2 < 1, 2 if x 2 + y 2 = 1, k i=1 i=1 1 i.

17 2.2 Continuity 59 then f is convex on D, but not continuous on D. Here, the continuity of f fails precisely at the boundary points of D. In fact, we will show that a convex function is always continuous at the interior points of its domain. First, we prove a couple of auxiliary results, which may also be of independent interest. Lemma Let a, b, c, d R with a < b and c < d. Then every real-valued convex function on the closed rectangle [a, b] [c, d] in R 2 is bounded. Proof. Let D := [a, b] [c, d] and let f : D R be any convex function. Define M := max{f(a, c), f(a, d), f(b, c), f(b, d)}. Let (x, y) D. Then there is s in [0, 1] such that x = (1 s)a+sb. Using the convexity of f on D, we see that f(x, y) (1 s)f(a, y) + sf(b, y). Further, there is t [0, 1] such that y = (1 t)c + td. Again, using the convexity of f on D, we obtain f(x, y) (1 s)[(1 t)f(a, c) + tf(a, d)] + s [(1 t)f(b, c) + tf(b, d)] (1 s)[(1 t)m + tm] + s [(1 t)m + tm] = M. It follows that M is an upper bound for f. Next, consider the center point (p, q) := ( a+b 2, ) c+d 2 of D and let (u, v) := (a + b x, c + d y). Clearly, (u, v) D and (p, q) = 1 2 (x, y) (u, v). Hence using the convexity of f, we obtain f(p, q) 1 2 f(x, y) f(u, v) 1 2f(x, y) + M, that is, f(x, y) m, where m := 2 (f(p, q) M). It follows that m is a lower bound for f. Lemma Let D be convex and open in R 2, and let f : D R be convex. Also, let [a, b] [c, d] be a closed rectangle contained in D, where a, b, c, d R with a < b and c < d. Then there is K R such that f(x, y) f(u, v) K ( x u + y v ) for all (x, y), (u, v) [a, b] [c, d]. Proof. Since D is open, there is δ > 0 such that [a δ, b+δ] [c δ, d+δ] D. By Lemma 2.30, there are m, M R such that m f(z, w) M for all (z, w) [a δ, b + δ] [c δ, d + δ]. Now, fix any (x, y), (u, v) [a, b] [c, d]. The case (x, y) = (u, v) is trivial, and so we will assume that (x, y) (u, v). Then l := x u + y v > 0, and we can consider z := u + δ l (u x) and w := v + δ l (v y). Since u x l, that is, l u x l, we have u δ z u + δ, and hence z [a δ, b + δ]. Similarly, w [c δ, d + δ]. In particular, (z, w) D. Moreover, it can be easily verified that u = δ l + δ x + l l + δ z and v = δ l + δ y + l l + δ w. Thus (u, v) = (1 t)(x, y) + t(z, w), where t := l/(l + δ). Since 0 < t < 1, using the convexity of f on D, we obtain f(u, v) (1 t)f(x, y) + tf(z, w). Further, since 0 < t < l/δ, we see that f(u, v) f(x, y) t [f(z, w) f(x, y)] l [M m] = K ( x u + y v ), δ where K := (M m)/δ. Similarly, f(x, y) f(u, v) K ( x u + y v ). This proves the desired inequality for f(u, v) f(x, y).

18 60 2 Sequences, Continuity, and Limits We are now ready to show that a convex function is continuous at all the interior points of its domain. This is an immediate consequence of the above lemma. (See also Exercise 10.) Proposition Let D be a convex subset of R 2 and let f : D R be a convex function. Then f is continuous at every interior point of D. In particular, if D is also open in R 2, then f is continuous on D. Proof. Let (x 0, y 0 ) be an interior point of D. Then there is r > 0 such that R := [x 0 r, x 0 +r] [y 0 r, y 0 +r] is contained in D. By Lemma 2.31, there is K R such that f(x, y) f(x 0, y 0 ) K ( x x 0 + y y 0 ) for (x, y) R. This implies that if ( (x n, y n ) ) is a sequence in D such that (x n, y n ) (x 0, y 0 ), then f(x n, y n ) f(x 0, y 0 ). Thus, f is continuous at (x 0, y 0 ). Continuity and Intermediate Value Property A result of fundamental importance in one-variable calculus is that continuous functions possess the intermediate value property (IVP). For ease of reference, we state this result below; see, for example, Proposition 3.13 of ACICARA. Fact (Intermediate Value Theorem) Let D be a subset of R and let φ : D R be a continuous function. Then φ has the IVP on every interval I D, that is, if a, b I with a < b and r R is between φ(a) and φ(b), then there is c [a, b] such that φ(c) = r; in particular, φ(i) is an interval in R. The following result may be viewed as an analogue of Fact 2.33 for realvalued continuous functions of two variables. Proposition 2.34 (Bivariate Intermediate Value Theorem). Let D be a subset of R 2 and let f : D R be a continuous function. Then f(e) is an interval in R for every path-connected subset E of D. In particular, f has the IVP on every 2-interval in D. Proof. Suppose E D is path-connected. Let z 1, z 2 f(e) and let r be any real number between z 1 and z 2. Then z 1 = f(x 1, y 1 ) and z 2 = f(x 2, y 2 ) for some (x 1, y 1 ), (x 2, y 2 ) E. Since E is path-connected, there is a path Γ joining (x 1, y 1 ) to (x 2, y 2 ) that lies in E. Let x, y : [α, β] R be continuous functions such that Γ is given by (x(t), y(t)), t [α, β]. Consider F : [α, β] R defined by F(t) := f(x(t), y(t)). By part (ii) of Proposition 2.17, F is continuous, and by Fact 2.33, F has the IVP on [α, β]. It follows that r = F(t 0 ) for some t 0 [α, β], and hence r f(e). This proves that f(e) is an interval in R. Finally, every 2-interval is path-connected (Example 1.5 (iv)), and so in view of Proposition 1.25, we see that f has the IVP on every 2-interval in D. The following example shows that the converse of the above result is not true, that is, the IVP does not imply continuity.

19 2.2 Continuity 61 Example Consider f : [0, 1] [0, 1] R defined by { cos(1/y) if y 0, f(x, y) := 0 if y = 0. Then f is not continuous on [0, 1] [0, 1], because, for example, (0, 1/nπ) (0, 0), but f(0, 1/nπ) = ( 1) n f(0, 0) = 0. Note, however, that f is continuous on [0, 1] (0, 1]. We show that f has the IVP on [0, 1] [0, 1]. Let r R be an intermediate value of f, that is, r is between f(x 1, y 1 ) and f(x 2, y 2 ) for some (x 1, y 1 ), (x 2, y 2 ) [0, 1] [0, 1]. If y 1 > 0 and y 2 > 0, then by the continuity of f on [0, 1] (0, 1] and Proposition 2.34, we see that r = f(x, y) for some (x, y) I (x1,y 1),(x 2,y 2). If y 1 = y 2 = 0, then f(x 1, y 1 ) = f(x 2, y 2 ) = 0 and there is nothing to prove. Thus we may assume, without loss of generality, that y 1 = 0 and y 2 > 0. Choose k N such that (1/kπ) < y 2. Now y 1 < (1/(k + 2)π) < (1/kπ) < y 2, and therefore cos(1/y) assumes every value between 1 and 1 as y varies from y 1 to y 2. It follows that r = f(x 1, y) for some y [y 1, y 2 ]. Thus f has the IVP on [0, 1] [0, 1]. Corollary Let D be a nonempty, path-connected, closed, and bounded subset of R 2 and let f : D R be a continuous function. Then the range f(d) of f is a closed and bounded interval in R. Proof. First, note that since D is nonempty, so is f(d). By Proposition 2.25, f(d) is bounded, and moreover, if m := inf f(d) and M := sup f(d), then f(d) [m, M] and m, M f(d). Further, by Proposition 2.34, f(d) is an interval in R. It follows that f(d) = [m, M]. Uniform Continuity The notion of uniform continuity for functions of one variable can be easily extended to functions of two variables. Let D be a subset of R 2. A function f ( : D R is said to be uniformly continuous on D if for any sequences (xn, y n ) ) and ( (u n, v n ) ) in D such that (x n, y n ) (u n, v n ) 0, we have f(x n, y n ) f(u n, v n ) 0. Specializing one of the two sequences to a constant sequence, we readily see that a uniformly continuous function is continuous. As in the case of functions of one variable, the converse is true if the domain is closed and bounded. Proposition Let D R 2 be a closed and bounded set. Then every continuous function on D is uniformly continuous on D. Proof. Suppose f : D R is continuous but not uniformly continuous on D. Then there are sequences ( (x n, y n ) ) and ( (u n, v n ) ) in D such that (x n, y n ) (u n, v n ) 0, but f(x n, y n ) f(u n, v n ) 0. The latter implies that there are ǫ > 0 and positive integers n 1 < n 2 < such that f(x nk, y nk ) f(u nk, v nk ) ǫ for all k N. Now, by the Bolzano Weierstrass

20 62 2 Sequences, Continuity, and Limits Theorem (part (ii) of Proposition 2.6), ( (x nk, y nk ) ) has a convergent subsequence, say ( (x nkj, y nkj ) ). If (x nkj, y nkj ) (x 0, y 0 ), then (u nkj, v nkj ) (x 0, y 0 ), because (x n, y n ) (u n, v n ) 0. Since f is continuous on D, we see that f(x nkj, y nkj ) f(u nkj, v nkj ) f(x 0, y 0 ) f(x 0, y 0 ) = 0. But this is a contradiction, since f(xnkj, y nkj ) f(u nkj, v nkj ) ǫ for all j N. Examples (i) Consider f : R 2 R defined by f(x, y) := x + y. Then it is clear that f is uniformly continuous on R 2. (ii) If D R 2 and f : D R is uniformly continuous, then for every fixed (x 0, y 0 ) D, the functions φ : D 1 R and ψ : D 2 R, defined as in Example 2.12(iv), are uniformly continuous. This follows from the definition of uniform continuity by specializing one of the coordinates in the two sequences to a constant sequence. (iii) Consider D R 2 and f : D R given by D := {(x, y) R 2 : x, y [0, 1] and (x, y) (0, 0)} and f(x, y) := 1 x + y. Then f is continuous on D but not uniformly continuous on D. To see the latter, consider the sequences ( (x n, y n ) ) and ( (u n, v n ) ) in D given by (x n, y n ) := (1/n, 0) and (u n, v n ) := (1/(n + 1), 0) for n N. We have (x n, y n ) (u n, v n ) = 1/n(n+1) 0, but f(x n, y n ) f(u n, v n ) = n (n + 1) = 1 0. Alternatively, we could use (ii) above and the fact that φ : (0, 1] R defined by φ(x) = f(x, 0) = 1/x is not uniformly continuous on (0, 1]. (See Example 3.18 (ii) on page 80 of ACICARA.) It may be noted here that the domain of f is bounded but not closed. (iv) Consider f : R 2 R defined by f(x, y) := x 2 + y 2. Then f is continuous on R 2, but not uniformly continuous on R 2. To see the latter, consider the sequences ( (x n, y n ) ) and ( (u n, v n ) ) in D given by (x n, y n ) := (n, 0) and (u n, v n ) := (n (1/n), 0) for n N. We have (x n, y n ) (u n, v n ) = 1/n 0, but f(x n, y n ) f(u n, v n ) = n 2 [n (1/n 2 )] = 2 (1/n 2 ) 0. Alternatively, we could use (ii) above and the fact that φ : R R defined by φ(x) = f(x, 0) = x 2 is not uniformly continuous on R. (See Example 3.18 (iii) on page 80 of ACICARA.) It may be noted here that the domain of f is closed but not bounded. On the other hand, the restriction of f to any bounded subset of R 2 is uniformly continuous. A criterion for the uniform continuity of a function of two variables that does not involve convergence of sequences can be given as follows. The result below may be compared with Proposition Proposition Let D R 2. Consider a function f : D R. Then f is uniformly continuous on D if and only if it satisfies the following ǫ-δ condition: For every ǫ > 0, there is δ > 0 such that (x, y), (u, v) D and (x, y) (u, v) < δ = f(x, y) f(u, v) < ǫ.