Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi

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1 Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational. Since the rational numbers form a field, axiom (A5) guarantees the existence of a rational number r so that, by axioms (A4) and (A3), we have x = 0 + x = ( r + r) + x = r + (r + x). Both r and r + x are rational by assumption, so x is rational by axiom (A), contradicting that x is irrational. Similarly, axiom (M5) guarantees the existence of a rational number /r so that, by axioms (M4) and (M3), we have ( ) x = x = r r x = r (rx). Both /r and rx are rational by assumption, so x is rational by axiom (M), contradicting that x is irrational..2. Prove that there is no rational number whose square is 2. Solution. Assume the contrary, that there is a rational number p such that p 2 = 2. Then there are integers m and n with p = m/n and for which 3 is not a common factor. By assumption, m 2 = 2n 2, so m 2 is divisible by 3, and therefore m is divisible by 3. But then m 2 is divisible by 9. This means that 2n 2 is divisible by 9, so 4n 2 is divisible by 3. Thus n 2 is divisible by 3, and therefore n is divisible by 3. We have shown that 3 is a common factor of m and n, a contradiction..3. Prove Proposition.5.

2 Solution. Suppose x, y, and z are elements of a field and x 0. The axioms (M) imply ( ) y = y = x x y = x (xy) = x ( ) x (xz) = x z = z = z, proving part (a). Part (b) follows from part (a) by setting z =, and part (c) follows from part (a) by setting z = /x. Since x x =, replacing x with /x in part (c) gives part (d)..4. Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and β is an upper bound of E. Prove that α β. Solution. Since E is nonempty, there exists a point x E. Note that α is a lower bound of E and β is an upper bound of E by the definition of supremum and infimum. Thus α x by the definition of lower bounds and x β by the definition of upper bounds, and therefore α β by Definition.5(ii)..5. Let A be a nonempty set of real numbers which is bounded below. Let A be the set of all numbers x, where x A. Prove that inf A = sup ( A). Solution. By the least-upper-bound property, α = inf A exists. By the definition of infimum, α x for all x A, and if β > α, then there exists a y A such that β > y. Therefore Proposition.8(a) implies α x for all x A, and if γ < α, then there exists a y A such that γ < y. Thus α is an upper bound of A, and if γ < α, then γ is not an upper bound of A. By the definition of supremum, it follows that α = sup ( A), and therefore inf A = sup ( A)..6. Fix b >. (a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that (b m ) /n = (b p ) /q. Hence it makes sense to define b r = (b m ) /n. (b) Prove that b r+s = b r b s if r and s are rational. 2

3 (c) If x is real, define B(x) to be the set of all numbers b t, where t is rational and t x. Prove that b r = sup B(r) when r is rational. Hence it makes sense to define for every real x. b x = sup B(x) (d) Prove that b x+y = b x b y for all real x and y. Solution. Fix b >. (a) If m/n = p/q, then mq/nq = np/nq, and therefore mq = np. For simplicity, let α = (b m ) /n and β = (b p ) /q. Then α = (α nq ) /nq = (b mq ) /nq = (b np ) /nq = (β nq ) /nq = β, so (b m ) /n = (b p ) /q. Thus b r = (b m ) /n is well-defined, since any two representations of r yield the same value. (b) Suppose r = m/n and s = p/q for some integers m, n, p, q with n > 0 and q > 0. Then r + s = (mq + np)/nq, and by part (a) and the corollary to Theorem.2, we have b r b s = ( b m/n) ( b p/q) = = ( b mq+np) /nq = b (mq+np)/nq = b r+s. ( b mq/nq) ( b np/nq) = (b mq ) /nq (b np ) /nq (c) Suppose r = p/q for some integers p and q > 0, and fix b s B(r). Then s is rational and s r, so b r s since b >. Multiplying through by b s yields b r b s, so b r is an upper bound of B(r). Since r r, we have b r B(r). Thus if γ < b r, then γ is not an upper bound of B(r). Hence b r = sup B(r). The definition b x = sup B(x) for every real x makes sense because the equation b r = sup B(r) holds for every rational r by the argument above. (d) Suppose x and y are real, r and s are rational, r x, and s y. Then r + s x + y, so b r b s = b r+s b x+y. Thus taking the supremum over all rationals r x and s y gives b x b y b x+y. Now suppose t is rational and t x+y. Then there exist rational numbers r and s such that t = r + s, r x, and s y. Thus b t = b r+s = b r b s b x b y, and taking the supremum over all rationals t x + y gives b x+y b x b y. Hence b x+y = b x b y. 3

4 .8. Prove that no order can be defined in the complex field that turns it into an ordered field. Solution. By Definition.27, we have i 0, and by Theorem.28 and Proposition.8(a,d), we have i 2 = < 0. This contradicts Proposition.8(d), and therefore no order can be defined in the complex field that turns it into an ordered field..9. Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = c but b < d. Prove that this turns the set of all complex numbers into an ordered set. Does this ordered set have the least-upper-bound property? Solution. The real numbers form an ordered set, so one and only one of the following holds: a < c, a = c, a > c. Similarly, either b < d, b = d, or b > d. There are 5 possible cases: () If a < c, then z < w. (2) If a > c, then z > w. (3) If a = c and b < d, then z < w. (4) If a = c and b > d, then z > w. (5) If a = c and b = d, then z = w. Thus one and only one of z < w, z = w, and z > w holds, so Definition.5(i) holds. Let z, z 2, z 3 be complex numbers satisfying z < z 2 and z 2 < z 3. Then one and only one of the following holds, () Re(z ) < Re(z 2 ), (2) Re(z ) = Re(z 2 ) and Im(z ) < Im(z 2 ), and similarly, one and only one of the following holds, () Re(z 2 ) < Re(z 3 ), (2) Re(z 2 ) = Re(z 3 ) and Im(z 2 ) < Im(z 3 ). This yields 2 possible cases: () If Re(z ) < Re(z 2 ) or if Re(z 2 ) < Re(z 3 ), then Re(z ) < Re(z 3 ). (2) If Re(z ) = Re(z 2 ) = Re(z 3 ), then Im(z ) < Im(z 2 ) < Im(z 3 ). 4

5 Thus z < z 3 in all cases, so Definition.5(ii) holds. Hence the set of all complex numbers is an ordered set. Now let E = {in : n Z}. Then E is a nonempty subset of C which is bounded above by. Suppose α + iβ = sup E. Then α 0 since α is an upper bound. But if α > 0, then α/2 is a smaller upper bound of E, contradicting that α + iβ = sup E. Thus α = 0, and so iβ = sup E. But for any β, there exists an integer n > β, contradicting that iβ is an upper bound of E. Hence sup E does not exist, so the least-upper-bound property does not hold. 5

6 Real Analysis Math 3AH Rudin, Chapter #2 Dominique Abdi 2.. Prove that the empty set is a subset of every set. Solution. Assume the contrary, that there is a set E such that the empty set is not a subset of E. Then there is an element x such that x / E, but this contradicts that the empty set is empty. Hence E A complex number z is said to be algebraic if there are integers a 0,..., a n, not all zero, such that a 0 z n + a z n + + a n z + a n = 0. Prove that the set of all algebraic numbers is countable. Solution. Let p be a polynomial of degree n with integer coefficients and let Z p := {z C : p(z) = 0}. By the fundamental theorem of algebra, Z p has at most n elements. Now define the set { n } P n (Z) := a k z k : a k Z, a n 0, k=0 consisting of all polynomials of degree n with integer coefficients, and let E n := {(a 0, a,..., a n ) : a k Z, a n 0}. Example 2.5 can be easily modified to show that the set of non-zero integers is countable. Thus by Example 2.5 and Theorem 2.3, E n is countable. Define n f : E n P n (Z), f(a 0, a,..., a n ) := a k z k. Every polynomial of degree n is uniquely determined by its n + coefficients, so f is one-to-one and onto. Hence P n (Z) is countable, and by Theorem 2.2, the set P (Z) := P n (Z), n=0 k=0

7 consisting of all polynomials with integer coefficients, is countable. Since the algebraic numbers are defined to be the roots of polynomials with integer coefficients, the set A of all algebraic numbers can be written as A = Z p, p P (Z) a union of a countable collection of finite (hence at most countable) sets. By the corollary to Theorem 2.2, A is at most countable. But for any n Z, p(z) = z n is a polynomial with integer coefficients for which z = n is a root, so Z A. We have shown that A is an at most countable set which contains an infinite subset. Hence A is infinite and at most countable, and thus A is countable Prove that there exist real numbers which are not algebraic. Solution. Assume that every real number is algebraic. Then R is countable by Exercise 2. This contradicts the corollary to Theorem 2.43, which states that R is uncountable Is the set of all irrational real numbers countable? Solution. Assume that the set Q c of all irrational real numbers is countable. Since R = Q Q c and Q is countable by the corollary to Theorem 2.3, it follows that R is countable by the corollary to Theorem 2.2. This contradicts the corollary to Theorem 2.43, which states that R is uncountable Construct a bounded set of real numbers with exactly three limit points. Solution. Define the sets { } { n E 0 := n : n N, E := n } { } 2n : n Z, E 2 := : n N, n and E := E 0 E E 2. Then E is bounded since E (0, 2). We now show that E = {0,, 2}. Fix ɛ > 0 and choose N such that N > /ɛ. Then 0 < /n < ɛ for all n N, and in particular, n ( ɛ, ɛ), n ( ɛ, + ɛ), n 2n n (2 ɛ, 2 + ɛ) for all n N. This shows that an open interval of arbitrarily small radius ɛ > 0 centered at 0,, or 2 intersects E at infinitely many points, so {0,, 2} E. 2

8 Now suppose x / {0,, 2}. If x < 0, then (2x, 0) E =, so x / E. If x > 0, then there are three possible cases: Case : If 0 < x <, then choose k N such that k + < x k, and define δ to be the smallest non-zero element of the set { x, x k +, x } k, x. Case 2 : If < x < 2, then choose k N such that k k < x k k +, and define δ to be the smallest non-zero element of the set { x, x k k, x k } k +, x 2. Case 3 : If x > 2, then define δ := x 2. In all three cases, we have (x δ, x+δ) E = by the choice of δ, so x / E. Hence E = {0,, 2} Let E be the set of all limit points of a set E. Prove that E is closed. Prove that E and E have the same limit points. (Recall that E = E E.) Do E and E always have the same limit points? Solution. By Definition 2.26 and Theorem 2.27(b), to prove that E is closed, it suffices to show that E contains its limit points. Let x be a limit point of E and let N r (x) be a neighborhood of x. Then there exists a point y N r (x) E with x y. Let r := r d(x, y) and note that N r (y) E contains a point z E since y is a limit point of E. Moreover, x z since d(x, z) > 0. The triangle inequality gives d(x, z) d(x, y) + d(y, z) < d(x, y) + r = r, so z N r (x). Thus x is a limit point of E, so therefore (E ) E and hence E is closed. We now show that E and E have the same limit points. If x E and r > 0, then there exists a point y N r (x) E with x y. Since E E, we have y N r (x) E, and therefore x (E). Conversely, suppose that x (E) and r > 0. Then there exists a point y N r (x) E with x y. If y / E, then y E since E = E E. Setting r := r d(x, y), then there exists a z E with z y and z x (since d(x, z) > 0). In either case, N r (x) contains a point of E distinct from x, so x E. Thus E = (E). 3

9 The sets E and E need not have the same limit points. Consider the set E defined in the solution to Exercise 2.5. We showed that E = {0,, 2}. If x E, then (x, x + ) E = {x}, and if x / E, then for δ := min { x y : y E }, we have (x δ, x + δ) E =. Thus (E ) =, and therefore E (E ) Let A, A 2, A 3,... be subsets of a metric space. (a) If B n = n i= A i, prove that B n = n i= A i, for n =, 2, 3,.... (b) If B = i= A i, prove that B i= A i. Show, by an example, that the inclusion can be proper. Solution. (a) Fix x n i= A i. Then x A i = A i A i for some i. If x A i, then x B n since A i B n B n. If x A i, then for every r > 0, there is a point y A i B n with x y such that d(x, y) < r. Thus x B n B n, so n i= A i B n. Fix x / n i= A i. Then ( n ) c n x A i = (A i ) c. i= Thus there exist r,..., r n > 0 such that N ri (x) (A i ) c for each i. Let r := min {r,..., r n }. Then N r (x) n i= N r i (x), so N r (x) n i= (A i) c. In particular, N r (x) contains no points of any A i, and therefore no points of B n. If N r (x) contains a point y B n, then letting r := r d(x, y), we see that N r (y) contains a point z B n by the definition of limit points, and z N r (y) N r (x) by the triangle inequality, contradicting that N r (x) B n =. Thus i= x B c n (B n) c = (B n B n) c = (B n ) c, so therefore B n n i= A i, and by the result proved above, B n = n i= A i. (b) Fix x i= A i. Then x A i = A i A i for some i. If x A i, then x B since A i B B. If x A i, then for every r > 0, there is a point y A i B with x y such that d(x, y) < r. Thus x B B, so B i= A i. For all i N, let A i := {/i}. Then A i = A i since finite sets are closed by the corollary to Theorem However, B = {/i : i N}, and therefore { } B = i : i N {0} {/i : i N} = A i, and the inclusion is proper since 0 / i= A i. 4 i=

10 2.8. Is every point of every open set E R 2 a limit point of E? Answer the same question for closed sets in R 2. Solution. Yes, every point of every open set E R 2 a limit point of E. Let E R 2 be open and fix x E. Then x is an interior point of E, so there exists an r > 0 such that N r (x) E. Let N s (x) be an arbitrary neighborhood of x. If s r, then N r (x) N s (x), so N r (x) N s (x) E. If s < r, then N s (x) N r (x) E, so N s (x) = N s (x) E. In either case, N s (x) E is itself a neighborhood of x in R 2 and therefore contains infinitely many points, and in particular, some point y x. Thus x is a limit point of E. The same result does not hold for closed sets in R 2. A set consisting of a single point x R 2 is closed since its complement is open, but finite sets have no limit points by the corollary to Theorem Let E denote the set of all interior points of a set E. (a) Prove that E is always open. (b) Prove that E is open if and only if E = E. (c) If G E and G is open, prove that G E. (d) Prove that the complement of E is the closure of the complement of E. (e) Do E and E always have the same interiors? (f) Do E and E always have the same closures? Solution. (a) Fix x E. Then there exists an r > 0 such that N r (x) E. If d(x, y) < r and r = r d(x, y), then d(y, z) < r implies d(x, z) d(x, y) + d(y, z) < r, so y E. This shows that N r (x) E, so x is an interior point of E. Thus E is open. (b) If E = E, then E is open by part (a). If E is open, then every point of E is an interior point, so E E. Since every interior point of a set must be contained in the set, the reverse inclusion holds as well, so E = E. (c) Suppose G E and G is open, and fix x G. Then x is an interior point of G, so there exists an r > 0 such that N r (x) G E. Thus x is an interior point of E, so G E. 5

11 (d) Fix x (E ) c. Then x is not an interior point of E, so every neighborhood N r (x) intersects E c. Thus x is in E c or x is a limit point of E c, so x E c (E ) c = E. Now suppose x / (E ) c. Then x E, so there exists an r > 0 such that N r (x) E. Thus x E and x is not a limit point of E c, so x E (E ) c = (E c ) c (E ) c = (E E ) c = (E) c. (e) No, E and E do not always have the same interiors. For example, note that Q = because every neighborhood of a rational number contains irrational numbers. However, Q = R, so (Q) = R since R is open. (f) No, E and E do not always have the same closures. For example, note that Q = because every neighborhood of a rational number contains irrational numbers. Therefore Q = = since the empty set is closed, but Q = R Let X be an infinite set. For p X and q X, define { (if p q) d(p, q) = 0 (if p = q). Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact? Solution. Parts (a) and (b) of Definition 2.5 follow immediately from the definition of d. If p = q, then part (c) is trivial. If p q, then for every r X, either p r or q r. Thus d(p, r) + d(q, r), and since d(p, q), part (c) holds in this case as well. Thus d is a metric. Fix E X and x E. Then N /2 (x) = {y X : d(x, y) < /2} = {x}, so every point is a neighborhood of itself. By Theorem 2.9, every point is open, and since E = {x}, x E every set is open by Theorem 2.24(a). If E X, then E c is open by what we just proved, so therefore every set is closed by the corollary to Theorem Suppose K X is compact. Then the open cover {{x} : x K} has a finite subcover, so K must be finite. Conversely, if K is finite, say K = {x,..., x n }, then given any open cover {G α } α A of K, there are indices α i such that x i G αi for all i =,..., n. Thus {G α,..., G αn } is a finite subcover of K, so K is compact. Hence the compact subsets of X are the finite subsets. 6

12 2.. For x R and y R, define d (x, y) = (x y) 2, d 2 (x, y) = x y, d 3 (x, y) = x 2 y 2, d 4 (x, y) = x 2y, d 5 (x, y) = x y + x y. Determine, for each of these, whether it is a metric or not. Solution. (a) Since d (2, 0) = (2 0) 2 = 4, d (2, ) = (2 ) 2 =, d (, 0) = ( 0) 2 =, it follows that d (2, 0) = 4 > 2 = d (2, ) + d (, 0), so d (x, y) is not a metric by Definition 2.5(c). (b) Since x y > 0 if x y, we have d 2 (x, y) = x y > 0 if x y. But if x = y, then d 2 (x, y) = 0. Thus Definition 2.5(a) holds. Since x y = y x, we have so Definition 2.5(b) holds. d 2 (x, y) = x y = y x = d 2 (y, x), By the triangle inequality, we have x z x y + y z x y + 2 x y y z + y z = ( x y + y z ) 2, and taking square roots gives d 2 (x, z) d 2 (x, y) + d 2 (y, z), so Definition 2.5(c) holds. Hence d 2 (x, y) is a metric. (c) Since d 3 (, ) = 2 ( ) 2 = = 0, Definition 2.5(a) fails, so d 3 (x, y) is not a metric. 7

13 (d) Since d 4 (2, ) = 2 2() = 2 2 = 0, Definition 2.5(a) fails, so d 4 (x, y) is not a metric. (e) Since x y defines a metric, we have d 5 (x, y) = x y + x y 0, with equality holding if and only if x = y. Thus Definition 2.5(a) holds. Since x y = y x, we have d 5 (x, y) = so Definition 2.5(b) holds. x y y x = + x y + y x = d 5(y, x), Definition 2.5(c) is satisfied if and only if d 5 (x, y) + d 5 (y, z) d 5 (x, z) = x y y z x z + + x y + y z + x z 0. Let a := x y, b := y z, and z := x z. Then taking ( + a)( + b)( + c) as the common denominator in the fraction above, the numerator is a + ab + ac + abc + b + ba + bc + abc c ac cb abc Hence d 5 (x, y) is a metric. = a + b c + 2ab + abc (2 + c) ab Let K R consist of 0 and the numbers /n, for n =, 2, 3,.... Prove that K is compact directly from the definition (without using the Heine-Borel theorem). Solution. Let {G α } α A be an open cover of K. Then there is an index α 0 A such that 0 G α0. Since G α0 is an open set containing 0, it contains an open interval ( ɛ, ɛ) for some ɛ > 0. Let N be the largest integer such that N /ɛ. Note that for all n > N, we have 0 < /n < ɛ, so /n G α0, and for all positive n N, there exist indices α n A such that /n G αn. Then {G α0, G α,..., G αn } is a finite subcover of K, so K is compact Construct a compact set of real numbers whose limit points form a countable set. 8

14 Solution. Define a set E := {/m + /n : m, n N} {/n : n N} {0}. Then E = {/n : n N} {0}, so E is countable and E E, so E is closed. Moreover, E [0, 2], so E is bounded. By the Heine-Borel theorem, it follows that E is compact Give an example of an open cover of the segment (0, ) which has no finite subcover. Solution. For every n N, define G n := (/n, ). Then each G n is an open interval in R, and n= G n = (0, ), so {G n } is an open cover of (0, ). Suppose {G n } has a finite subcover C of (0, ). Then there is a largest integer N such that G N C. But then G G n C n = (/N, ), so every ɛ satisfying 0 < ɛ < /N is an element of (0, ) but is not covered by C. This contradicts that C is a cover, so therefore {G n } has no finite subcover of (0, ) Show that Theorem 2.36 and its Corollary become false (in R, for example) if the word compact is replaced by closed or by bounded. Solution. For every n N, define E n := [n, ) R. Then E n is closed, and for any finite subcollection C, let N be the largest integer such that E N C. Then E E n C n = [N, ), but n= E n =. Thus Theorem 2.36 and its corollary are false if compact is replaced by closed. For every n N, define E n := (0, /n) R. Then E n is bounded, and for any finite subcollection C, let N be the largest integer such that E N C. Then E E n C n = (0, /N), but n= E n =. Thus Theorem 2.36 and its corollary are false if compact is replaced by bounded Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = p q. Let E be the set of all p Q such that 2 < p 2 < 3. Show that E is closed and bounded in Q, but that E is not compact. Is E open in Q? Solution. Fix x E c. Since 2 and 3 are irrational, either x 2 > 3 or x 2 < 2. Assume without loss of generality that x 2 > 3 (the proof of the other case is similar). Let r := x 3 and note that N r (x) = { p Q : 3 < p 2 < (x + r) 2} E c, proving that x is an interior point of E c. Thus E c is open, and by the corollary to Theorem 2.23, E is closed. 9

15 Since E [0, 2], it follows that E is bounded. For every integer n 0, define G n := ( 2 + /n, 3) Q. If p E, then 2 < p 2 < 3, so there is an integer N such that ( 2 + /N) 2 < p. Thus p G N, and therefore {G n } 0 is an open cover of E. If C is a finite subcollection of {G n } 0, then there is a largest integer k such that G k C. Choose p Q such that 2 < p < 2 + /k. Then p / G G n C n = ( 2 + /k, 3) Q, so C is not a cover of E. Thus {G n } 0 contains no finite subcover of E, and therefore E is not compact. Yes, E is open in Q. Since E = ( 2, 3) Q, E is open by Theorem (a) If A and B are disjoint closed sets in some metric space X, prove that they are separated. (b) Prove the same for disjoint open sets. (c) Fix p X, δ > 0, define A to be the set of all q X for which d(p, q) < δ, define B similarly, with > in place of <. Prove that A and B are separated. (d) Prove that every connected metric space with at least two points is uncountable. Hint: Use (c). Solution. (a) If A and B are closed, then A = A and B = B, so A B = A B = and A B = A B =. (b) If A and B are open, then A c and B c are closed. Since A B c, we have A B c by Theorem 2.27(c), and similarly, B A c. Thus A B = = A B. (c) The set A is open since A = N δ (p). Fix x B and let r := δ d(p, x). Then r > 0 and N r (x) B by the triangle inequality, so B is open. Moreover, A B = since d(p, q) cannot be both greater than δ and less than δ. By part (b), A and B are separated. (d) Suppose X is a metric space which is at most countable. Fix x X. Then there exists an r > 0 such that {y X : d(x, y) = r} =, for otherwise, we would have an injective function from (0, ) into X. The sets A := {y X : d(x, y) < r} and B := {y X : d(x, y) > r} separate X by part (c). 0

16 2.20. Are closures and interiors of connected sets always connected? (Look at subsets of R 2.) Solution. The closure of a connected set is always connected. Suppose E is connected and E A B, where A B = = A B. Then E B = without loss of generality, for otherwise, A and B separate E. Thus E A, and by Theorem 2.27(c), E A. Since A B =, it follows that E B =. Thus E is not a union of two nonempty separated sets. The interior of a connected set need not be connected. Define Then E is connected, but E := { x R 2 : d(x, ±) }. E = { x R 2 : d(x, ) < } { x R 2 : d(x, ) < }, which is a union of two nonempty separated sets by Exercise 2.9(b) Let A and B be separated subsets of some R k, suppose a A, b B, and define p(t) = ( t)a + tb for t R. Put A 0 = p (A), B 0 = p (B). [Thus t A 0 if and only if p(t) A.] (a) Prove that A 0 and B 0 are separated subsets of R. (b) Prove that there exists t 0 (0, ) such that p(t 0 ) / A B. (c) Prove that every convex subset of R k is connected. Solution. (a) Fix t A 0. Then p(t) A, and since A B =, p(t) / B. Thus there is an r > 0 such that N r (p(t)) B c. In particular, if s N r (t), then s / B 0. Assume that there exists an s N r (t) such that s B 0. If r := r s t, then there exists a point s N r (s) B 0. But t s t s + r < r by the triangle inequality, so therefore s N r (t), contradicting that N r (t) B0. c We have thus shown that N r (t) (B 0 ) c. The same argument applied to a point t B 0 shows that there is a neighborhood of t which is contained in (A 0 ) c. Thus A B = = A B.

17 (b) Assume the contrary, that p(t) A B for all t (0, ). Then we have [0, ] A 0 B 0, and by part (a), ( ) A 0 [0, ] ( ) A 0 [0, ] ( ) B 0 [0, ] = (A 0 B 0 ) [0, ] = [0, ] =, ( ) B 0 [0, ] = (A 0 B 0 ) [0, ] = [0, ] =. The sets A 0 [0, ] and B 0 [0, ] are nonempty since 0 A 0 [0, ] and B 0 [0, ], and they are disjoint since A 0 and B 0 are disjoint by part (a). Thus A 0 [0, ] and B 0 [0, ] separate [0, ], contradicting Theorem (c) Assume that there exists a convex set E R k such that E = A B for two nonempty separated sets A and B. Fix a A and b B. By the definition of convexity, p(t) := ( t)a + tb E = A B for every t (0, ). This contradicts part (b). Hence every convex set in R k is connected. 2

18 Real Analysis Math 3AH Rudin, Chapter 3 Dominique Abdi 3.. Prove that the convergence of {s n } implies convergence of { s n }. Is the converse true? Solution. We first prove the reverse triangle inequality from Exercise.3. If a, b C (or R k ), then a = a b + b a b + b, so a b a b. Interchanging a and b, we see that b a a b, and therefore a b a b. Now suppose s n s. Fix ɛ > 0 and N such that s n s < ɛ for all n N. The reverse triangle inequality then yields sn s sn s < ɛ for all n N, so s n s. The converse is not true. Consider the sequence s n := ( ) n. This sequence diverges, but since s n = ( ) n = for all n N, we have sn. ( 3.2. Calculate lim n2 + n n). n Solution. By elementary algebra, we have ( n n2 + n n) = n2 + n + n =. + /n + Since /n 0, Theorem 3.3 and the continuity of the square root function on the positive real numbers yields ( lim n2 + n n) = lim = n n + /n + 2.

19 3.3. If s = 2, and s n+ = 2 + s n (n =, 2, 3,...), prove that {s n } converges, and that s n < 2 for n =, 2, 3,.... Solution. First, note that s = 2 < = s 2 and s 2 = Suppose s < s 2 < < s n < 2. Then s n = 2 + s n < 2 + s n = s n < 2. Moreover, s n < 2 implies that s n < 2, and therefore 2 + s n < 4. Taking square roots, we have s n+ = 2 + s n < 2, and therefore s < < s n < s n+ < 2. By induction, {s n } is a monotonically increasing sequence which is bounded above by 2, so {s n } converges by Theorem Find the upper and lower limits of the sequence {s n } defined by s = 0; Solution. We can prove by induction that s 2m = s 2m ; s 2m+ = s 2m. s 2m = 2 2 m and s 2m+ = 2 m for all m N. Since the subsequence of even terms converges to /2 and the subsequence of odd terms converges to, both /2 and are subsequential limits of {s n }. If x / {/2, }, then let r := 2 min { x, x /2 } and note that s n N r (x) for at most finitely many n N. Thus /2 and are the only subsequential limits of {s n }, so therefore lim sup s n = and lim inf s n = n n For any two real sequences {a n }, {b n }, prove that lim sup n (a n + b n ) lim sup n a n + lim sup b n, n provided the sum on the right is not of the form. 2

20 Solution. Assume that the sum on the right is not of the form. Suppose one of the terms on the right is, say lim sup a n =. Then a n, and for any M > 0, there are only finitely many n N such that b n > M. Thus a n + b n, so lim sup(a n + b n ) =. Now suppose one of the terms on the right is, say lim sup a n =. Then for any M > 0, there are infinitely many n N such that a n > M, but only finitely many n N such that b n < M. Thus there are infinitely many n N such that a n + b n > M, so lim sup(a n + b n ) =. Finally, suppose α := lim sup a n and β := lim sup b n are both finite and fix ɛ > 0. By Theorem 3.7(b), there exist N and N 2 such that n N implies a n < α + ɛ/2, and n N 2 implies b n < β + ɛ/2. Let N := max {N, N 2 }. Then n N implies a n + b n < α + β + ɛ. But ɛ > 0 is arbitrary, so therefore a n + b n α + β = lim sup n a n + lim sup b n n for all but finitely many n. Thus every subsequential limit of {a n + b n } is bounded above by lim sup a n + lim sup b n, and in particular, by Theorem 3.7(a). lim sup n (a n + b n ) lim sup n a n + lim sup b n n 3.8. If a n converges, and if {b n } is monotonic and bounded, prove that an b n converges. Solution. By Theorem 3.4, {b n } converges to some number b. Define { b b n if {b n } is monotonically increasing, c n := b n b if {b n } is monotonically decreasing. Then {c n } is monotonically decreasing and lim c n = 0. Moreover, for all N N, N a n b n = n= N N a n c n + b n= if {b n } is decreasing, for example, and similarly if {b n } is increasing. By Theorem 3.42, a n c n converges, and therefore a n b n converges. n= a n Suppose {p n } is a Cauchy sequence in a metric space X, and some subsequence {p ni } converges to a point p X. Prove that the full sequence {p n } converges to p. 3

21 Solution. Fix ɛ > 0. Choose N such that n i N implies d(p ni, p) < ɛ/2, and choose N 2 such that m, n N 2 implies d(p m, p n ) < ɛ/2. Now define N := max {N, N 2 }. Then n N implies for some n i N. Thus lim p n = p. d(p n, p) d(p n, p ni ) + d(p ni, p) < ɛ 3.2. Prove the following analogue of Theorem 3.0(b): If {E n } is a sequence of closed nonempty and bounded sets in a complete metric space X, if E n E n+, and if lim n diam E n = 0, then E n consists of exactly one point. Solution. For every n N, choose a point x n E n. Fix ɛ > 0 and choose N such that diam E N < ɛ. Then m, n N implies x m, x n E N since E n E n+ for all n, so therefore d(x m, x n ) diam E N < ɛ. Thus {x n } is a Cauchy sequence, and since X is complete, {x n } converges to some point x X. For every N N, {x n } N E N and E N is closed, so therefore x E N. Hence x E n. If x, y E n and x y, then d(x, y) > 0. Then diam E n d(x, y) > 0 for all n, contradicting that lim diam E n = 0. 4

22 Real Analysis Math 3AH Rudin, Chapter 4 Dominique Abdi 4.. Suppose f is a real function defined on R which satisfies lim [f(x + h) f(x h)] = 0 h 0 for every x R. Does this imply that f is continuous? Solution. No. Define f : R R by Then f(x) := but f is not continuous at x = 0. { 0 if x 0, if x = 0.. lim [f(h) f( h)] = 0, h If f is a continuous mapping of a metric space X into a metric space Y, prove that f(e) f(e) for every set E X. Show, by an example, that f(e) can be a proper subset of f(e). Solution. Fix x E and a sequence {x n } E such that lim x n = x. Then {f(x n )} is a sequence in f(e), and since f is continuous, lim f(x n ) = f(x). If f(x) / f(e) f(e) = f(e), then there is a neighborhood N r (f(x)) which does not intersect f(e). By Theorem 4.8, f (N r (f(x))) is open, so there is a neighborhood N s (x) which does not intersect E. This contradicts that lim x n = x, so f(x) f(e) and therefore f(e) f(e). Define f : N R by f(n) = /n. Since N = N, we have { } { } f(n) = n : n N n : n N {0} = f(n).

23 4.3. Let f be a continuous real function on a metric space X. Let Z(f) (the zero set of f) be the set of all p X at which f(p) = 0. Prove that Z(f) is closed. Solution. Let x be a limit point of Z(f) and let {x n } such that lim x n = x. Then Z(f) be a sequence f(x) = lim n f(x n) = lim n 0 = 0, so x Z(f). Thus Z(f) contains its limit points, so Z(f) is closed Let f and g be continuous mappings of a metric space X into a metric space Y, and let E be a dense subset of X. Prove that f(e) is dense in f(x). If g(p) = f(p) for all p E, prove that g(p) = f(p) for all p X. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Solution. Fix y f(x) and choose x X such that f(x) = y. Since E is dense in X, there is a sequence {x n } X such that lim x n = x. Consider the sequence {f(x n )} f(e). Since f is continuous, lim f(x n ) = f(x) = y, so y f(e) f(e) = f(e). Thus f(x) f(e), so f(e) is dense in f(x). Fix p X and choose a sequence {p n } E such that lim p n = p. Since f and g are continuous and f(p n ) = g(p n ), we have proving that f(p) = g(p) for all p X. f(p) = lim n f(p n) = lim n g(p n) = g(p), 4.5. If f is a real continuous function defined on a closed set E R, prove that there exist continuous real functions g on R such that g(x) = f(x) for all x E. (Such functions g are called continuous extensions of f from E to R.) Show that the result becomes false if the word closed is omitted. Extend the result to vector-valued functions. Hint: Let the graph of g be a straight line on each of the segments which constitute the complement of E (compare Exercise 29, Chap. 2). The result remains true if R is replaced by any metric space, but the proof is not so simple. Solution. Since E is closed, E c is the union of an at most countable collection of disjoint open intervals C = {I n } by Exercise If (, c) is one of these intervals for some c R, then define g(x) := f(c) for all x (, c), and similarly if (c, ) is one of the intervals. On the remaining intervals I n = (a n, b n ), define ( ) f(bn ) f(a n ) g(x) := f(a n ) + (x a n ). b n a n 2

24 Finally, define g(x) := f(x) for all x E. Then clearly g extends f to R and g is continuous on E and on E c. It remains to show that g is continuous at the endpoints of the intervals I n C. Let x be the endpoint of some interval I n C. If {x j } is a sequence in E converging to x, then lim g(x j ) = g(x) = f(x). If {x j } is a sequence in E c converging to x = a n, then [ lim g(x j) = lim f(a n ) + j j ( f(bn ) f(a n ) b n a n ) ] (x j a n ) = f(a n ), and similarly if x = b n. Thus g is continuous on R. If the word closed is omitted, let E := (0, ), and define f(x) := /x for all x E. If f has a continuous extension g to R, then g is continuous on [, ], and therefore bounded by the intermediate value theorem. However, f is not bounded in any neighborhood of x = 0, so therefore g is not bounded either, a contradiction. If f : E R k is continuous and E R is closed, then f has a continuous extension g : R R k. Each component function f j : R R of f = (f,..., f k ) extends to a continuous function g j : R R by the proof above, and it follows that g = (g,..., g k ) is a continuous extension of f to R If f is defined on E, the graph of f is the set of points (x, f(x)), for x E. In particular, if E is a set of real numbers, and f is real-valued, the graph of f is a subset of the plane. Suppose E is compact, and prove that f is continuous on E if and only if its graph is compact. Solution. Suppose f is continuous and let {U α V β } be an open cover of the graph of f. Then {U α } is an open cover of E, so there exist α,..., α m such that {U αi } m is a finite subcover of E. Similarly, {V β } is an open cover of f(e), and since f(e) is compact by Theorem 4.4, there exist β,..., β n such that {V βj } n is a finite subcover of f(e). Thus { Uαi V βj : i m, j n } is a finite subcover of {U α V β }, proving that the graph of f is compact. Now suppose f is not continuous. Then there is a sequence {x n } E such that lim x n = x E, but f(x n ) does not converge to f(x). Thus there is an ɛ > 0 such that d(f(x n ), f(x)) ɛ for infinitely many n, and by restricting to a subsequence, we can assume without loss of generality that d(f(x n ), f(x)) ɛ for all n. If the graph of f is compact, then the sequence {(x n, f(x n ))} contains a convergent subsequence by Theorem 3.6(a). That is, (x nk, f(x nk )) (x, f(x)) for some x E, contradicting that d(f(x n ), f(x)) ɛ for all n. The graph of f is therefore not compact. 3

25 4.7. If E X and if f is a defined on X, the restriction of f to E is the function g whose domain of definition is E, such that g(p) = f(p) for p E. Define f and g on R 2 by: f(0, 0) = g(0, 0) = 0, f(x, y) = xy 2 /(x 2 + y 4 ), g(x, y) = xy 2 /(x 2 + y 6 ) if (x, y) (0, 0). Prove that f is bounded on R 2, that g is unbounded in every neighborhood of (0, 0), and that f is not continuous at (0, 0); nevertheless, the restrictions of both f and g to every straight line in R 2 are continuous! Solution. First, note that 0 (x y 2 ) 2 = x 2 2xy 2 + y 4 implies x 2 + y 4 2xy 2. Thus f(x, y) = xy 2 x 2 + y 4 xy 2 2xy 2 = 2, so f is bounded on R 2. Along the curve y = x /3, we have lim g(x, x /3 ) = lim x 5/3 x 0 x 0 2x 2 = lim x 0 2x /3 =, proving that g is unbounded in every neighborhood of (0, 0). Along the curve y = x /2, we have lim f(x, x 2 x 0 x/2 ) = lim x 0 2x 2 = 0 = f(0, 0), 2 so f is not continuous at (0, 0). It is clear that f and g are continuous away from (0, 0) and along the axes, so it suffices to consider lines of the form y = cx for some fixed constant c R. Since lim f(x, cx) = lim x 0 x 0 lim g(x, cx) = lim x 0 x 0 cx 3 x 2 + c 4 x 4 = lim x 0 cx 3 x 2 + c 6 x 6 = lim x 0 cx + c 4 x 2 = 0, cx + c 6 x 4 = 0, the restrictions of both f and g to every straight line in R 2 are continuous Let f be a real uniformly continuous function on the bounded set E in R. Prove that f is bounded on E. Show that the conclusion is false if boundedness of E is omitted from the hypothesis. 4

26 Solution. Fix ɛ > 0 and choose δ > 0 such that f(x), f(y) < ɛ whenever x y < δ. The collection {N δ (x) : x E} is an open cover of E. In fact, this is also an open cover of E, for otherwise, there is a limit point p of E with the property that p / N δ (x) for all x E, a contradiction. Since E is closed and bounded, it is compact by the Heine-Borel theorem, so there exist points x,..., x n E such that {N δ (x j )} n is a finite subcover of E. But then for every x E, there is a j such that x x j < δ, and therefore f(x) f(x j ) < ɛ. That is, {N ɛ (f(x j ))} n is a finite cover of f(e), so diam f(e) nɛ. Thus f is bounded on E. If E = R and f(x) := x on E, then f is uniformly continuous since for every ɛ > 0, x y < ɛ implies that f(x) f(y) < ɛ. However, f is not bounded Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every ɛ > 0 there exists a δ > 0 such that diam f(e) < ɛ for all E X with diam E < δ. Solution. Suppose f is uniformly continuous, fix ɛ > 0, and choose δ > 0 such that d(x, y) < δ implies d(f(x), f(y)) < ɛ. Given a set E with diam E < δ, we have d(x, y) < δ for all x, y E. Thus d(f(x), f(y)) < ɛ for all x, y E, so therefore diam f(e) < ɛ. Conversely, fix ɛ > 0 and choose δ > 0 such that diam f(e) < ɛ for all E X with diam E < δ. If d(x, y) < δ, let E := {x, y}. Then diam E < δ and f(e) = {f(x), f(y)}. By assumption, diam f(e) < ɛ, that is, d(f(x), f(y)) < ɛ, proving that f is uniformly continuous Complete the details of the following alternate proof of Theorem 4.9: If f is not uniformly continuous, then for some ɛ > 0 there are sequences {p n }, {q n } in X such that d X (p n, q n ) 0 but d Y (f(p n ), f(q n )) > ɛ. Use Theorem 2.37 to obtain a contradiction. Solution. Suppose f is not uniformly continuous. Then there exists an ɛ > 0 such that for any n N, there are points p n, q n X such that d X (p n, q n ) < /n, but d Y (f(p n ), f(q n )) > ɛ. Thus {p n } and {q n } are sequences in X such that d X (p n, q n ) 0, but d Y (f(p n ), f(q n )) > ɛ. By Theorem 4.4, f(x) is compact. If the sequences {p n } and {q n } are both finite, then p n = q n for all n sufficiently large, contradicting that d Y (f(p n ), f(q n )) > ɛ for all n. Therefore the set E consisting of the sequences {p n } and {q n } is infinite, so it has a limit point x by Theorem It follows that there are 5

27 increasing sequences of positive integers {n k } and {n k } such that lim p n k = x and lim q n k = x. Then lim f(p n k ) = f(x) = lim f(q n ) k k k since f is continuous, and d X (p nk, q n k ) 0. Choose k sufficiently large so that d Y (f(p nk ), f(x)) < ɛ 2 and d Y (f(q n k ), f(x)) < ɛ 2. Then d Y (f(p nk ), f(q n k )) d Y (f(p nk ), f(x)) + d Y (f(q n k ), f(x)) < ɛ, contradicting that for all n. d Y (f(p nk ), f(q n k )) > ɛ 4.. Suppose f is a uniformly continuous mapping of a metric space X into a metric space Y and prove that {f(x n )} is a Cauchy sequence in Y for every Cauchy sequence {x n } is in X. Use this result to give an alternative proof of the theorem stated in Exercise 3. Solution. Fix ɛ > 0 and choose δ > 0 such that d(f(x), f(y)) < ɛ whenever d(x, y) < δ. Now choose N such that d(x m, x n ) < δ for all m, n N. Then d(f(x m ), f(x n )) < ɛ for all m, n N, so {f(x n )} is a Cauchy sequence. Let E be a dense subset of a metric space X, and let f be a uniformly continuous real function defined on E. For every x X, there is a sequence {x n } E such that lim x n = x. Since every convergent sequence is Cauchy, the result above shows that {f(x n )} is a Cauchy sequence in R, and thus converges. If lim x n = x and lim y n = x, then lim f(x n) = f(x) = lim f(y n) n n by the continuity of f, so we can extend f to X by defining f(x) = lim f(x n ) for any sequence {x n } E converging to x. Moreover, by Theorem 4.2, the extension of f is continuous on X A uniformly continuous function of a uniformly continuous function is uniformly continuous. State this more precisely and prove it. 6

28 Solution. Let X, Y, Z be metric spaces and suppose f : X Y and g : Y Z are uniformly continuous. Then g f is uniformly continuous. Proof. Fix ɛ > 0 and choose η > 0 such that d Y (y, y 2 ) < η implies d Z (g(y ), g(y 2 )) < ɛ. Now choose δ > 0 such that d X (x, x 2 ) < δ implies d Y (f(x ), f(x 2 )) < η. Then d X (x, x 2 ) < δ implies d Y (f(x ), f(x 2 )) < η, which in turn implies Thus g f is uniformly continuous. d Z (g(f(x ), g(f(x 2 ))) < ɛ Let I = [0, ] be the closed unit interval. Suppose f is a continuous mapping of I into I. Prove that f(x) = x for at least one x I. Solution. The function g : I I defined by g(x) := f(x) x is continuous by Theorem 4.4. It suffices to show that g(x) = 0 for some x I. If g(0) = 0 or g() =, then there is nothing to prove. Otherwise, g(0) > 0 since 0 < f(0), and g() < 0 since 0 f() <. By the intermediate value theorem, there is a point x I such that g(x) = 0. 7

29 Real Analysis Math 3AH Rudin, Chapter 5 Dominique Abdi 5.. Let f be defined for all real x, and suppose that f(x) f(y) (x y) 2 for all real x and y. Prove that f is constant. Solution. If x y, dividing by x y and taking the limit as y x gives lim f(x) f(y) y x x y lim x y = 0. y x Thus f (x) exists and f (x) = 0 for every x R. By Theorem 5.(b), f is constant Suppose f (x) > 0 in (a, b). Prove that f is strictly increasing in (a, b), and let g be its inverse function. Prove that g is differentiable, and that g (f(x)) = f (x) (a < x < b). Solution. Fix x, y (a, b) with x < y. By the mean value theorem, there is an x 0 (a, b) with f(y) f(x) = f (x 0 )(y x) > 0. Thus f is strictly increasing. Fix p = f(x) in the range of f, let {p n } be a sequence in the range of f converging to p, and choose x n (a, b) such that f(x n ) = p n. Then g(p n ) g(p) g(f(x n )) g(f(x)) x n x lim = lim = lim n p n p n f(x n ) f(x) n f(x n ) f(x) = f (x) Suppose g is a real function on R, with bounded derivative (say g M). Fix ɛ > 0, and define f(x) = x + ɛg(x). Prove that f is one-to-one if ɛ is small enough. (A set of admissible values of ɛ can be determined which depends only on M.)

30 Solution. First, note that f is differentiable by Theorem 5.3 because g is differentiable. Moreover, f (x) = + ɛg (x). If ɛ < /M, then ɛg (x) < g (x) M, so f (x) > 0. By the mean value theorem, if x < y, then there is an x 0 (x, y) such that f(y) f(x) = f (x 0 )(y x) > 0, so f is strictly increasing, and therefore one-to-one If C 0 + C C n n + C n n + = 0, where C 0,..., C n are real constants, prove that the equation C 0 + C x + + C n x n + C n x n = 0 has at least one real root between 0 and. Solution. Define f(x) := C 0 x + C 2 x2 + + C n n + xn+. Then f is a polynomial, and is therefore differentiable, and Since f(0) = 0 and f (x) = C 0 + C x + + C n x n + C n x n. f() = C 0 + C C n n + C n n + = 0 by assumption, by the mean value theorem, there is an x (0, ) such that 0 = f() f(0) = f (x) Suppose f is defined and differentiable for every x > 0, and f (x) 0 as x +. Put g(x) = f(x + ) f(x). Prove that g(x) 0 as x +. Solution. Fix ɛ > 0 and choose N such that x > N implies f (x) < ɛ. Then for any x N, by the mean value theorem, there is an x 0 (x, x + ) such that g(x) = f(x + ) f(x) = f (x 0 )(x + x) = f (x 0 ) < ɛ. Thus g(x) 0 as x +. 2

31 5.6. Suppose (a) f is continuous for x 0, (b) f (x) exists for x > 0, (c) f(0) = 0, (d) f is monotonically increasing. Put g(x) = f(x) (x > 0) x and prove that g is monotonically increasing. Solution. Since f is differentiable and x > 0, g is differentiable by Theorem 5.3. It suffices to prove that g is non-negative by Theorem 5.(a). Moreover, since g (x) = xf (x) f(x) x 2, it suffices to prove that xf (x) f(x). Fix x > 0. By the mean value theorem, there is an x 0 (0, x) such that f(x) = f(x) f(0) = f (x 0 )(x 0) = xf (x 0 ) xf (x), where the first equality follows by (c), and the last inequality follows by (d) Suppose f (x), g (x) exist, g (x) 0, and f(x) = g(x) = 0. Prove that f(t) lim t x g(t) = f (x) g (x). (This also holds for complex functions.) Solution. f (x) g (x) = lim f(t) f(x) t x t x t x g(t) g(x) = lim f(t) f(x) t x g(t) g(x) = lim f(t) t x g(t) = f(x) g(x) Suppose f is continuous on [a, b] and ɛ > 0. Prove that there exists δ > 0 such that f(t) f(x) f (x) t x < ɛ whenever 0 < t x < δ, a x b, a t b. (This could be expressed by saying that f is uniformly differentiable on [a, b] if f is continuous on [a, b].) Does this hold for vector-valued functions too? 3

32 Solution. Fix ɛ > 0. By Theorem 4.9, f is uniformly continuous on [a, b], so there exists a δ > 0 such that x y < δ implies f (x) f (y) < ɛ. Fix x [a, b], and without loss of generality, fix t [a, b] with 0 < t x < δ. By the mean value theorem, there is an x 0 (x, t) such that f(t) f(x) t x = f (x 0 ). Then 0 < x x 0 < δ, so f(t) f(x) f (x) t x = f (x 0 ) f (x) < ɛ. Now suppose f is continuous on [a, b] for some function f: R n R, f = (f,..., f n ). Then each component f j is differentiable, f = (f,..., f n), and each f j is continuous on [a, b]. Fix ɛ > 0, and by the result above, choose δ j > 0 such that f j (t) f j (x) f t x j(x) < ɛ n whenever 0 < t x < δ j, a x b, a t b, for j =,..., n. Then ( f(t) f(x) f (x) t x = f (t) f (x) f t x (x),..., f n(t) f n (x) f t x n(x)) /2 n ( ) 2 = fj (t) f j (x) f t x j(x) j= /2 n ( ) 2 < ɛ = ɛ n j= whenever 0 < t x < δ j, a x b, a t b Let f be a continuous real function on R, of which it is known that f (x) exists for all x 0 and that f (x) 3 as x 0. Does it follow that f (0) exists? Solution. Fix ɛ > 0 and choose δ > 0 such that f (x) 3 < ɛ whenever 0 < x < δ. Now fix t R such that 0 < t < δ. By the mean value theorem, there is an x 0 (0, t) (or x 0 (t, 0) if t < 0) such that f(t) f(0) t = f (x 0 ). Then 0 < x 0 < δ, so therefore f(t) f(0) 3 t = f (x 0 ) 3 < ɛ. 4

33 This proves that so therefore f (0) exists. f(t) f(0) lim = 3, t 0 t 5.0. Suppose f and g are complex differentiable functions on (0, ), f(x) 0, g(x) 0, f (x) A, g (x) B as x 0, where A and B are complex numbers, B 0. Prove that f(x) lim x 0 g(x) = A B. Compare with Example 5.8. Hint: { } f(x) f(x) g(x) = x A x g(x) + A x g(x). Apply Theorem 5.3 to the real and imaginary parts of f(x)/x and g(x)/x. Solution. Define f := Re(f), f 2 := Im(f), g := Re(g), and g 2 := Im(g). Then so therefore f = Re(f ), f 2 = Im(f ), g = Re(g ), g 2 = Im(g ), f (x) Re(A), f 2 Im(A), g Re(B), g 2 Im(B) as x 0. Moreover, f i 0 and g i 0 for i =, 2 as x 0. By Theorem 5.3, ( f(x) lim x 0 x = lim f (x) + i f ) 2(x) f (x) f 2 (x) = lim + i lim x 0 x x x 0 x x 0 x = lim x 0 f (x) + i lim x 0 f 2(x) = Re(A) + i Im(A) = A, Since B 0, it follows from Theo- and similarly, g(x)/x B as x 0. rem 4.4(c) that f(x) lim x 0 g(x) = lim f(x) x 0 x x g(x) = A B. 5.. Suppose f is defined in a neighborhood of x, and suppose f (x) exists. Show that f(x + h) + f(x h) 2f(x) lim h 0 h 2 = f (x). Show by an example that the limit may exist even if f (x) does not. Hint: Use Theorem

34 Solution. Since f (x) exists, so does f (x), and by Theorem 5.2, f is continuous at x. Therefore f(x+h) f(x) and f(x h) f(x) as h 0, so the numerator and denominator of the fraction f(x + h) + f(x h) 2f(x) h 2 both converge to 0 as h 0. By Theorem 5.3, f(x + h) + f(x h) 2f(x) f (x + h) f (x h) lim h 0 h 2 = lim h 0 2h f (x + h) f (x) + f (x) f (x h) = lim h 0 = 2 lim h 0 = f (x). Define f : R R by 2h f (x + h) f (x) h f(x) := + 2 lim f (x) f (x h) h 0 h { /x if x 0, 0 if x = 0. Then f is not continuous at x = 0, so f (0) and f (0) do not exist by Theorem 5.2. However, for any h 0, we have f(h) + f( h) 2f(0) h 2 = 0, so the fraction above converges to 0 as h If f(x) = x 3, compute f (x), f (x) for all real x, and show that f (3) (0) does not exist. Solution. If x > 0, then f(x) = x 3, so f (x) = 3x 2 and f (x) = 6x. If x < 0, then f(x) = x 3, so f (x) = 3x 2 and f (x) = 6x. Since we have f (0) = 0, and since f(t) t 3 lim = lim t 0 t t 0 sgn(t) t = lim sgn(t) t 0 t 2 = 0, f (t) 3 t 2 lim = lim t 0 t t 0 sgn(t) t = lim sgn(t)3 t = 0, t 0 we have f (0) = 0. However, f (3) (0) does not exist because f (t) 6t lim = lim t 0+ t t 0 t = 6, lim f (t) 6t = lim = 6, t 0 t t 0 t shows that the limit of f (t)/t as t 0 does not exist. 6

35 Math 3C: Homework Solutions (From Rudin, Chapter 7) Problem 4: The continuity of Φ follows from the continuity of x and y, which follows from the Weierstrass M-test (Theorem 7.0, Rudin) and the continuity and boundedness of f. To show that Φ maps the Cantor set onto I 2, we follow the hint. Note that t o as given belongs to the Cantor set (its ternary expansion contains no s). We have so that 3 k t 0 = 3 k i (2a i ) i= k = 2 3 k i a i + i= 3 j (2a j+k ) j=0 3 j (2a j+k ) (mod 2), j=0 ( ) f(3 k t 0 ) = f 3 j (2a j+k ) j=0 by the 2-periodicity of f. Now suppose a k = 0; then 0 3 j (2a j+k ) ( 3 = f ) 3 j (2a j+k ) = f(3 k t 0 ) = 0 = a k. j=0 j=0 (The lower bound follows from setting a i = 0 for i > k, and the upper bound follows from setting a i = for i > k.) Similarly, if a k =, we have 2 3 j=0 ( 3 j (2a j+k ) = f The hint and surjectivity of Φ follow immediately. j=0 ) 3 j (2a j+k ) = f(3 k t 0 ) = = a k. Problem 5: Such an f must be constant on [0, ). Fix arbitrary nonnegative numbers x and y. Let ɛ > 0 be arbitrary, and choose δ > 0 s.t. x y < δ f n (x) f n (y) < ɛ x, y [0, ], n N. Now choose n N s.t. x y n < δ and x n, y n. Then x n y n < δ, so f(x) f(y) = f n ( x n ) f n ( y n) < ɛ.