Solution of the 8 th Homework

Solution of the 8 th Homework Sangchul Lee December 8, 2014 1 Preinary 1.1 A simple remark on continuity The following is a very simple and trivial observation. But still this saves a lot of words in actual proofs. Lemma 1.1. Let f : X R be a function, and let x 0 X. Then the followings are equivalent: (i) f is continuous at x 0. I.e, x x0 ;x X f (x) = f (x 0 ). (ii) x x0 ; f (x) = f (x 0 ). Proof. This follows by just a few seconds of meditation. To be slightly precise, this follows because f (x 0 ) f (x 0 ) < ε for any ε > 0. 1.2 (optional) Weierstrass-Carathéodory formulation of differentiability Proposition 1.2. Let X R, let x 0 X be a it point of X, and let f : X R be a function. Then the followings are equivalent: (i) f is differentiable at x 0. (ii) There exists a function φ f : X R which is continuous at x 0 and satisfies Moreover, if these hold, then φ f (x 0 ) = f (x 0 ). f (x) = f (x 0 ) + φ f (x)( ) (1.1) It is loosely read as differentiability at x 0 holds if and only if the graph of a function is well approximated by a line (with finite slope) near x 0. What is good for this formulation is that various differentiation rules follows almost for free. Proof. (i) = (ii) : Define φ f : X R by Now, by the differentiability of f at x 0, we have φ f (x) = f (x) f (x 0 ) x x 0, x x 0 f (x 0 ), x = x 0. φ f (x) = = f (x 0 ) = φ f (x 0 ). This proves that φ f is continuous at x 0. Moreover, (1.1) holds for all x X by definition of φ f. ((1.1) is clearly true when x x 0. When x = x 0, both sides of (1.1) are zero.) Therefore (ii) follows. 1

(ii) = (i) : For x X \ {x 0 }, solving (1.1) in terms of φ f (x) gives φ f (x) = f (x) f (x 0). Taking it as x x 0, by the continuity of φ f at x 0, Therefore f is differentiable at x 0 and f (x 0 ) = φ f (x 0 ). = φ f (x) = φ f (x 0 ). 2 Solutions Exercise 1. Let X be a subset of R and let f : X R be a function. Then the following two statements are equivalent. (i) f is uniformly continuous on X. (ii) For any two equivalent sequences (a n ) n=m, (b n ) n=m, the sequences ( f (a n )) n=m, ( f (b n )) n=m are also equivalent sequences. Solution. (i) = (ii) : Let (a n ) n=m, (b n ) n=m be sequences in X which are equivalent. Then for any ε > 0, By the uniform convergence of f, there exists δ > 0 such that whenever x, y X and x y < δ we have f (x) f (y) < ε. By the equivalence, there exists N N such that whenever n N we have a n b n < δ. Combining two statements, it follows that whenever n N we have f (a n ) f (b n ) < ε. This proves that ( f (a n )) n=m, ( f (b n )) n=m are also equivalent sequences. (ii) = (i) : We prove contrapositive. Let us assume that f is not uniformly continuous. By negating Definition 3.31, There exists ε > 0 such that for any δ > 0 there exists x, y X such that x y < δ but f (x) f (y) ε. Now for each choice δ = n 1 (with n {1,2, }) pick such two elements a n, b n X (that is, a n b n < δ = n 1 but f (a n ) f (b n ) ε). Then (a n ) n=1, (b n) n=1 are equivalent but ( f (a n)) n=m, ( f (b n )) n=m cannot be equivalent. This proves the contrapositive as desired. Exercise 2. Give an example of a continuous function f : R (0, ) such that, for any real number 0 < ε < 1, there exists x R such that f (x) = ε. Remark. In view of the Maximum Principle, a continuous function f : R (0, ) must attain positive minimum on any finite closed interval [a, b] R. Consequently, if ε > 0 is small, any solution of f (x) = ε have large size. This in particular suggests that any continuous function f : R (0, ) satisfying x f (x) = 0 serves an example. 1 st Solution. Let f (x) = 2 x. Then clearly f is a continuous function with range (0, ). Moreover, for any 0 < ε < 1 we have f (log 2 ε) = ε. Therefore f satisfies all the desired properties. If you raise an objection by claiming that we have never learned both exponential function and logarithm in this course, here is another solution: 2 nd Solution. Let f : R (0, ) by f (x) = 1 1 + x. 2

Since f is defined by applying arithmetic operations on continuous functions such that denominator never vanishes, f defines a continuous function on R. Also, whenever 0 < ε < 1 we have ε 1 1 > 0 and hence f (ε 1 1 1) = 1 + (ε 1 1) = ε. Therefore f satisfies all the desired property. Exercise 3. Let X be a subset of R and let f : X R be a uniformly continuous function. Let x 0 be an adherent point of X. Then x x0 f (x) exists (and so it is a real number.) Solution. We divide the proof into two steps. Each step contains some important ideas which deserve their own right. Step 1. We prove the following lemma: Lemma. Let X be a subset of R and let f : X R be a uniformly continuous function. Then for every Cauchy sequence (a n ) n=m in X, the sequence ( f (a n )) n=m is also a Cauchy sequence. Proof of Lemma. Let ε > 0. Then There exists δ > 0 such that whenever x, y X and x y < δ we have f (x) f (y) < ε. There exists N m such that whenever j, k N we have a j a k < δ. Combining two statements, we have f (a j ) f (a k ) < ε whenever j, k N. This proves that ( f (a n )) n=m as desired. Remark. Before moving to the next step, we remark that this Lemma does not characterize uniform continuity. Indeed, f (x) = x 2 is not uniformly continuous, yet still mapping Cauchy sequences to Cauchy sequences. In fact, it is equivalent to the condition that f : X R is uniformly continuous on any bounded subsets of X. Step 2. To prove the original problem, we utilize Proposition 2.16. Indeed, let (a n ) n=0 and (b n) n=0 be any sequences in X that converge to x 0. Since ( f (a n )) n=0 and ( f (b n)) n=0 are Cauchy sequences by the Lemma above, they are also convergent. Let L = f (b n ). n Note that (a n ) n=0 and (b n) n=0 are equivalent, since we have n (a n b n ) = x 0 x 0 = 0. Then by Proposition 3.37 (Exercise 8.1), ( f (a n )) n=0 and ( f (b n)) n=0 are also equivalent. Thus f (a n) = ( f (a n ) f (b n )) + f (b n ) = 0 + L = L. n n n Therefore by Proposition 2.16, f converges to L at x 0. Exercise 4. Let X be a subset of R, and let f : X R be a uniformly continuous function. Assume that E is a bounded subset of X. Then f (E) is also bounded. 1 st Solution. We first present a proof by contradiction. Assume that f (E) is unbounded. Then for each n N, there exists a n E such that f (a n ) n. But since E is bounded, (a n ) n=0 is also bounded. Thus by the Bolzano-Weierstrass Theorem, there exists a subsequence (a j(n)) n=0 of (a n ) n=0 which is convergent. Let x 0 = n a j(n) denotes the it. Then by Lemma 2.13, x 0 is an adherent point of X. Thus by Corollary 3.40 (Exercise 8.3), we know that n f (a j(n) ) exists. But since f (a j(n) ) j(n) n, by choosing a sufficiently large n we get a contradiction. (Just pick n to be larger than a bound of ( f (a j(n) )) n=0.) 3

2 nd Solution. We present a direct proof. To make proof simple, we introduce some notations: Pick an arbitrary ε > 0. Then by the uniform continuity of f, we can choose δ > 0 such that whenever x, y X and x y < δ we have f (x) f (y) < ε. Let K Z be defined by K = {k Z : E [kδ, (k + 1)δ) }. Since E is bounded, K is a finite set. (Indeed, choose a bound M > 0 of E, and choose N N such that N δ > M. Then K { N,, N}.) Now we are ready. For each k K, pick an element x k E [kδ, (k + 1)δ). Since K is finite, L := ε + max{ f (x k ) : k K} is a well-defined positive real number. Moreover, for any x E, there exists k K such that x E [kδ, (k + 1)δ). Then x x k < δ and hence f (x) f (x k ) < ε. This shows that This proves that f (E) is also bounded. f (x) f (x) f (x k ) + f (x k ) ε + max{ f (x k ) : k K} = L. Exercise 5. Let X be a subset of R, let x 0 be a it point of X, and let f : X R be a function. If f is differentiable at x 0, then f is also continuous at x 0. Solution. We have ( ) = ( = ( ) f (x) f (x0 ) ( ) ) ( Therefore, by adding f (x 0 ) to both sides, it follows that f is continuous at x 0. ) ( ) = f (x 0 ) 0 = 0. Exercise 6. Let X be a subset of R, let x 0 be a it point of X, let f : X R be a function, and let L be a real number. Then the following two statements are equivalent. (i) f is differentiable at x 0 on X with derivative L. (ii) For every ε > 0, there exists a δ = δ(ε) > 0 such that, if x X satisfies < δ, then f (x) [ f (x 0 ) + L( )] ε. Solution. The proof is almost a tautology. Nevertheless we spell out every detail. (i) = (ii) : By definition, for every ε > 0, there exists δ = δ(ε) > 0 such that, if x X \ {x 0 } satisfies < δ, then L < ε. Now multiply both sides by. Then we have Since this continues to hold when x = x 0, we get (ii). f (x) [ f (x 0 ) + L( )] ε. (ii) = (i) : For every ε > 0, pick δ = δ(ε/2) > 0 as in (ii). Then whenever x X \ {x 0 } and < δ, we have f (x) [ f (x 0 ) + L( )] 1 2 ε. 4

Now divide both sides by. (This is possible because is positive.) Then we get L 1 x x 2 ε < ε. 0 This yields x x0 ;( )/( ) = L, which is equivalent to (i). Exercise 7. Let X be a subset of R, let x 0 be a it point of X, and let f : X R and g : X R be functions. (i) If f is constant, so that there exists c R such that f (x) = c, then f is differentiable at x 0 and f (x 0 ) = 0. (ii) If f is the identity function, so that f (x) = x, then f is differentiable at x 0 and f (x 0 ) = 1. (iii) If f, g are differentiable at x 0, then f + g is differentiable at x 0, and ( f + g) (x 0 ) = f (x 0 ) + g (x 0 ). (Sum Rule) (iv) If f, g are differentiable at x 0, then f g is differentiable at x 0, and ( f g) (x 0 ) = f (x 0 )g(x 0 ) + g (x 0 ) f (x 0 ). (Product Rule) (v) If f is differentiable at x 0, and if c R, then c f is differentiable at x 0, and (c f ) (x 0 ) = c f (x 0 ). (vi) If f, g are differentiable at x 0, then f g is differentiable at x 0, and ( f g) (x 0 ) = f (x 0 ) g (x 0 ). (vii) If g is differentiable at x 0, and if g(x) 0 for all x X, then 1/g is differentiable at x 0, and (1/g) (x 0 ) = g (x 0 )/(g(x 0 )) 2. (viii) If f, g are differentiable at x 0, and if g(x) 0 for all x X, then f /g is differentiable at x 0, and ( f /g) (x 0 ) = g(x 0) f (x 0 ) f (x 0 )g (x 0 ) (g(x 0 )) 2. (Quotient Rule) 1 st Solution. (i) Notice that = 0, x X \ {x 0 }. Taking it as x x 0 for x X \ {x 0 }, it follows that f is differentiable at x 0 and f (x 0 ) = 0. (ii) Notice that = 1, x X \ {x 0 }. Taking it as x x 0 for x X \ {x 0 }, it follows that f is differentiable at x 0 and f (x 0 ) = 1. (iii) Notice that [ f (x) + g(x)] [ f (x 0 ) + g(x 0 )] = f (x) f (x 0) + g(x) g(x 0), x X \ {x 0 }. Taking it as x x 0 for x X \ {x 0 }, we know from both the assumption and the Limit Laws that the right-hand side converges to f (x 0 ) + g (x 0 ). This proves the Sum Rule. (iv) Notice that f (x)g(x) f (x 0 )g(x 0 ) = f (x) f (x 0) g(x 0 ) + g(x) g(x 0) f (x), x X \ {x 0 }. Taking it as x x 0 for x X \ {x 0 }, we know from both the assumption and the Limit Laws that the right-hand side converges to f (x 0 )g(x 0 ) + g (x 0 ) f (x 0 ). This proves the Product Rule. (v) This follows by (i) and (iv). (vi) This follows by (iii) and (v), with the choice c = 1. (vii) Notice that [1/g(x)] [1/g(x 0 )] = g(x) g(x 0) 1 g(x)g(x 0 ), x X \ {x 0}. 5

Taking it as x x 0 for x X \ {x 0 }, we know from both the assumption and the Limit Laws that the right-hand side converges to g (x 0 )/g(x 0 ) 2. This proves (vii). (viii) This follows from (iv) and (vii). 2 nd Solution. We utilize the Weierstrass-Carathéodory formulation, Proposition 1.2. (i) Notice that (1.1) is satisfied with φ f (x) = 0. Therefore f is differentiable at x 0 with f (x 0 ) = φ f (x 0 ) = 0. (ii) Notice that (1.1) is satisfied with φ f (x) = 1. Therefore f is differentiable at x 0 with f (x 0 ) = φ f (x 0 ) = 1. (iii) Let φ f and φ g be as in (1.1) for f and g, respectively. Then f (x) + g(x) = [ f (x 0 ) + φ f (x)( )] + [g(x 0 ) + φ g (x)( )] = [ f (x 0 ) + g(x 0 )] + [φ f (x) + φ g (x)]( ), and hence φ(x) = φ f (x) + φ g (x) satisfies Proposition 1.2. This shows that f + g is differentiable at x 0 and ( f + g) (x 0 ) = φ(x 0 ) = φ f (x 0 ) + φ g (x 0 ) = f (x 0 ) + g (x 0 ). (iv) Let φ f and φ g be as in (1.1) for f and g, respectively. Then and hence f (x)g(x) = [ f (x 0 ) + φ f (x)( )][g(x 0 ) + φ g (x)( )] = f (x 0 )g(x 0 ) + [φ f (x)g(x 0 ) + φ g (x) f (x 0 ) + φ f (x)φ g (x)( )]( ), φ(x) = φ f (x)g(x 0 ) + φ g (x) f (x 0 ) + φ f (x)φ g (x)( ) satisfies Proposition 1.2. This shows that f g is differentiable at x 0 and (v) This follows by (i) and (iv). ( f g) (x 0 ) = φ(x 0 ) = φ f (x 0 )g(x 0 ) + φ g (x 0 ) f (x 0 ) + 0 = f (x 0 )g(x 0 ) + g (x 0 ) f (x 0 ). (vi) This follows by (iii) and (v), with the choice c = 1. (vii) Let φ g be as in (1.1) for g. Then ( 1 g(x) = 1 g(x 0 ) 1 g(x 0 ) 1 ) 1 = g(x) g(x 0 ) φ g (x) g(x)g(x 0 ) (), and hence φ(x) = φ g (x)/g(x)g(x 0 ) satisfies Proposition 1.2. This shows that 1/g is differentiable at x 0 and (1/g) (x 0 ) = φ(x 0 ) = φ g (x 0 ) g(x 0 ) 2 = g (x 0 ) g(x 0 ) 2. (viii) This follows from (iv) and (vii). Exercise 8. Let X, Y be subsets of R, let x 0 X be a it point of X, and let y 0 Y be a it point of Y. Let f : X Y be a function such that f (x 0 ) = y 0 and such that f is differentiable at x 0. Let g : Y R be a function that is differentiable at y 0. Then the function g f : X R is differentiable at x 0, and (g f ) (x 0 ) = g (y 0 ) f (x 0 ). Remark. This would have followed easily it it were true that g( f (x)) g( f (x 0 )) = g( f (x)) g( f (x 0)). Unfortunately, this is not always true as may vanish infinitely many times near x 0. We need to circumvent this technical issue. 6

1 st Solution. Let (a n ) n=0 be any sequence in X \ {x 0} which converges to x 0. Then we deduce from Exercise 8.5 that ( f (a n )) n=0 converges to f (x 0). Now we divide into two cases: Case 1) Assume that f (x 0 ) 0. Then for the choice ε = 1 2 f (x 0 ) > 0, there exists δ > 0 such that, whenever x X and < δ we have Then it follows from the reverse triangle inequality that f (x 0 )( ) ε. f (x 0 )( ) f (x 0 )( ) f (x 0 ) ε = ε since f (x 0 ) = 2ε. In particular, since 0, this implies that > 0 whenever < δ. Next, we pick N sufficiently large so that a n x 0 < δ whenever n N. Since a n X \ {x 0 }, it follows that f (a n ) f (x 0 ) > 0. Then our intuitive idea works and we have ( ) g( f (a n )) g( f (x 0 )) g( f (an )) g( f (x 0 )) f (a n ) f (x 0 ) = n a n x 0 n f (a n ) f (x 0 ) a n x 0 ( = n = g (y 0 ) f (x 0 ). g( f (a n )) g( f (x 0 )) f (a n ) f (x 0 ) ) ( n f (a n ) f (x 0 ) a n x 0 Case 2) Now assume that f (x 0 ) = 0. For ε = 42, pick δ > 0 such that, if y Y and y y 0 < δ then Then by the triangle inequality, we have g(y) g(y 0 ) g (y 0 )(y y 0 ) 42 y y 0. g(y) g(y 0 ) g (y 0 ) y y 0 + g(y) g(y 0 ) g (y 0 )(y y 0 ) ( g (y 0 ) + 42) y y 0. Also, choose N sufficiently large so that whenever n N, we have f (a n ) f (x 0 ) < δ. Then by noting that g( f (a n )) g( f (x 0 )) ( g (y 0 ) + 42) f (a n ) f (x 0 ), a n x 0 a n x 0 it follows from the squeezing theorem, together with n [ f (a n ) f (x 0 )]/(a n x 0 ) = f (x 0 ) = 0, we have g( f (a n )) g( f (x 0 )) = 0 = g (y 0 ) f (x 0 ). n a n x 0 Therefore, in any cases the Chain Rule follows. 2 nd Solution. Let φ f : X R and φ g : Y R, respectively, be as in the Weierstrass-Carathéodory formulation (Proposition 1.2) for f and g, respectively. Then g( f (x)) = g( f (x 0 )) + φ g ( f (x))[ ] = g( f (x 0 )) + φ g ( f (x))φ f (x)( ). Thus the function φ(x) = φ g ( f (x))φ f (x) is continuous at x 0 and satisfies (1.1). Thus it satisfies Proposition 1.2. This shows that g f is differentiable at x 0 and (g f ) (x 0 ) = φ(x 0 ) = φ g ( f (x 0 ))φ f (x 0 ) = g (y 0 ) f (x 0 ) ) Exercise 9. Let a < b be real numbers, and let f : (a, b) R be a function. If x 0 (a, b), if f is differentiable at x 0, and if f attains a local maximum or minimum at x 0, then f (x 0 ) = 0. Solution. We first consider local maximum case. Since x 0 is a local maximum point of f, there exists a sufficiently small 7

δ > 0 such that f attains a minimum on (x 0 δ, x 0 + δ) (a, b). Thenwe have 0, x 0 δ < x < x 0 0, x 0 < x < x 0 + δ. Thus taking right-it and left-it, we find that f (x 0 ) = 0 and f (x 0 ) = 0. + x x0 This proves that f (x 0 ) = 0. Local minimum case can be tackled in the same way. Exercise 10. Let a < b be real numbers, and let f : [a, b] R be a continuous function which is differentiable on (a, b). Assume that f (a) = f (b). Then there exists x (a, b) such that f (x) = 0. Solution. Utilizing the Maximum Principle, pick two points x 0, x 1 [a, b] such that f attains a global maximum at x 0 and a global minimum at x 1. If f (x 0 ) = f (x 1 ), then f must reduce to a constant function and hence the claim follows from Exercise 8.7.(i). (In this case, you can pick any point x (a, b).) In view of the previous observation, we may ssume f (x 0 ) f (x 1 ). This implies that either f (x 0 ) f (a) or f (x 1 ) f (a). In either cases, there exists a global extremum x of f, which is neither a nor b. Thus x is also a global extremum of f (a,b). Then x is a local extremum of f (a,b) and by the previous exercise, we have f (x) = 0. Exercise 11. Let X be a subset of R, let x 0 be a it point of X, and let f : X R be a function. If f is monotone increasing and if f is differentiable at x 0, then f (x 0 ) 0. If f is monotone decreasing and if f is differentiable at x 0, then f (x 0 ) 0. Solution. We first assume that f is monotone. If x X \ {x 0 }, then by dividing the cases based on whether x > x 0 or x < x 0, we find that 0 always holds. Now assume further that f is differentiable at x 0. Taking x x 0 in X \ {x 0 }, the inequality is preserved and hence we have f (x 0 ) = 0. For the monotone-decreasing case, this argument works exactly in the same way (but with inequalities reversed) to yield the corresponding proof. Exercise 12. Let a < b be real numbers, and let f : [a, b] R be a differentiable function. If f (x) > 0 for all x [a, b], then f is strictly monotone increasing. If f (x) < 0 for all x [a, b], then f is strictly monotone decreasing. If f (x) = 0 for all x [a, b], then f is a constant function. Solution. All these statements follows from the Mean Value Theorem in a direct manner. For example, assume first that f (x) > 0 for all x [a, b]. Then for any x < y in [a, b], the Mean Value Theorem says that there exists c [x, y] such that f (y) f (x) = f (c) > 0, y x 8

where the second inequality follows form the assumption. Multiplying y x to both sides, we get f (y) f (x) > 0, which implies that f is strictly increasing. The other two assertions are also proved in a similar manner. 9