HW #1 SOLUTIONS. g(x) sin 1 x

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1 HW #1 SOLUTIONS 1. Let f : [a, b] R and let v : I R, where I is an interval containing f([a, b]). Suppose that f is continuous at [a, b]. Suppose that lim s f() v(s) = v(f()). Using the ɛ δ definition of limit, prove that lim t v(f(t)) = v(f()). Take an ɛ > 0. Since lim s f() v(s) = v(f()), there eists ɛ 1 > 0 such that if s I and s f() < ɛ 1 then v(s) v(f()) < ɛ. Now, f is continuous at, so there eists δ > 0 such that if t [a, b] is such that t < δ then f(t) f() < ɛ 1. Since I contains f([a, b]), we automatically have f(t) I, and also f(t) f() < ɛ 1 from above, so v(f(t)) v(f()) < ɛ as desired. This shows that lim t v(f(t)) = v(f()), i.e. that the composition v f is continuous at. g() 2. Let g : I R be a function, where I is an interval containing 0 in its interior. Suppose lim 0 = 0. Define { g() sin 1 f() = if 0, 0 if = 0 Prove that f (0) = 0. By definition, f f() f(0) g() sin (0) = lim 0 0 = lim 1 0. We have the inequalities 0 g() sin 1 for 0, since sin y 1 for all y R, in particular for y = 1. Since we are given that lim 0 g() also have lim 0 g() lim 0 g() sin 1 = 0. g() = 0, we g() sin = 0. Of course, lim 0 0 = 0, so by the Squeeze Theorem, lim 1 0 = 0, or Remark. We re using the easy-to-check fact that for a function h : (a, b) R and t (a, b), lim t h() = 0 if and only if lim t h() = 0. To see this, we need only unravel the definition of limit in this contet: lim t h() = 0 says that for all ɛ > 0, there eists δ > 0 such that if t < δ then h() 0 < ɛ, whereas lim t h() = 0 says that for all ɛ > 0, there eists δ > 0 such that if t < δ then h() 0 < ɛ. Note however that h() 0 = h() = h() 0 so the two statements are saying the same thing. 3. Let f, g : R R be differentiable functions satisfying f () > g () for all R and f(0) = g(0). Prove that f() > g() for > 0 and f() < g() for < 0. Let h : R R be defined by h() = f() g() for R. Then h(0) = f(0) g(0) = 0 and h () = 1

2 2 HW #1 SOLUTIONS f () g () > 0 if R. If > 0, then by the Mean Value Theorem, h() = h() h(0) = h (c)( 0) for some c between 0 and. But then h (c) = f (c) g (c) > 0, so h() = h (c) > 0. If < 0, then the same computation h() = h (c) for some c between 0 and shows that h() < 0 because h (c) > 0 and < 0.. Rudin Ch 5. Eercise #7. Suppose f (), g () eist, g () 0, and f() = g() = 0. Prove that This holds also for comple functions. f(t) lim t g(t) = lim t f(t) lim t g(t) = f () g (). f(t) f() t f(t) f(), and we are given that lim g(t) g() t t t both eist, hence the ratio lim t f(t) g(t) = f () g () (note that we need g () = 0). = f () and lim t g(t) g() t = g () 0 5. Rudin Ch 5. Eercise #13. Suppose a and c are real numbers, c > 0, and f is defined on [ 1, 1] by { f() = a sin( c ) (if 0), 0 (if = 0). Prove the following statements: (a) f is continuous if and only if a > 0. (b) f (0) eists if and only if a > 1. (c) f is bounded if and only if a 1 + c. (d) f is continuous if and only if a > 1 + c. (e) f (0) eists if and only if a > 2 + c. (f) f is bounded if and only if a 2 + 2c. (g) f is continuous if and only if a > 2 + 2c. ( ) 1/c 1 (a) Suppose a 0. Then for each positive integer k, consider a real number k = π (0, 1] [ 1, 1]. 2 ( ) +2πk a/c Then, f( k ) = k a sin( k c 1 ) = π sin( π 2 +2πk 2 + 2πk) = ( π 2 + 2πk) a/c. If a = 0, then lim k f( k ) = 1, whereas if a < 0 then lim k f( k ) =. In either case, we have lim k f( k ) f(0), whereas lim k k = 0 (c > 0). Therefore f is not continuous at 0 (recall that a function f defined on a neighborhood of 0 is continuous at 0 if and only if lim k f( k ) = f(0) for all sequences { k } k such that lim k k = 0). In fact, we have shown that f has neither the left nor the right limit at 0. Suppose now that a > 0. Then if 0, we have f() = a sin( c ) a, and for = 0 of course we have the same inequality (since a > 0, 0 a = 0 is defined also of course). Since lim 0 a = 0, we have lim 0 f() = 0 by the Squeeze Theorem.

3 HW #1 SOLUTIONS 3 (b) f f() f(0) (0) = lim 0 0 = a sin( c ) if the limit eists. ( ) 1/c If a 1, then lim 0 + a 1 sin( c 1 ) does not eist, as in part (a) by considering the sequence k = π. 2 +2πk Therefore we need a > 1. In this case, = a 1 sin( c ) a 1 and the latter tends to 0 a sin( c ) as 0, so f (0) eists by the Squeeze Theorem and f (0) = 0 in fact in this case. (c) Note first that f() = f( ) if [ 1, 1], i.e. f is an even function. For 0, f is certainly differentiable at since f is given by the epression a sin( c ) if > 0 on a neighborhood of not containing 0 and by ( ) a sin(( ) c ) on a neighborhood of not containing 0 and both epressions are products of compositions of differentiable functions, hence differentiable. We assume that a > 1 from part (b) to discuss the eistence of f on [ 1, 1] in the first place. For > 0, f () = a a 1 sin( c ) + a cos( c )( c c 1 ) = a a 1 sin( c ) c a c 1 cos( c ) (recall the remark above that f is given by a sin( c ) in a neighborhood of in [ 1, 1] not containing 0). Also for a > 1, f (0) eists and f (0) = 0. For < 0, f is again differentiable at because f() = f( ) epresses f as a composition of the differentiable function f (0,1] (this is the restriction of f on (0, 1]) together with the differentiable function on a neighborhood of not containing 0. Using the chain rule now, we have f () = f ( ), so that a a 1 sin( c ) c a c 1 cos( c ) if > 0 f () = 0 if = 0 c a c 1 cos( c ) a a 1 sin( c ) if < 0 At any rate, we then need only analyze the boundedness of the epression a a 1 sin( c ) c a c 1 cos( c ) on (0, 1] (for 0 < 1). Since a > 1, the first term has absolute value a a 1 sin( c ) a a 1, which tends to 0 as 0 + because a > 1. It is thus bounded. For the latter term, c a c 1 cos( c ) c a c 1, and this tends to 0 as 0 + if a 1 + c. In the case a < 1 + c, we take the sequence { k } k 1 defined by k = ( ) 1 1/c 2πk (0, 1], which tends to 0 as k, and yet f( k ) = c ( ) 1 (a c 1)/c 2πk as k, showing that f is not bounded on (0, 1] if a < 1 + c. (d) We assume f eists throughout [ 1, 1], i.e. a > 1. Assume first that a > 1 + c, then from f () = f ( ) again, we need only show that lim 0 + f () = f (0) = 0 to check contiuity of f (it is continuous on (0, 1]) because it is epressed on a neighborhood of a point in (0, 1] by a combination of products, addition, and composition of differentiable functions). lim 0 + f () = lim 0 +(a a 1 sin( c ) c a c 1 cos( c )), and the first term has limit lim 0 + a a 1 sin( c ) = 0 since a > 1 (Squeeze Theorem yet again here). For the latter term, note that lim 0 + c a c 1 cos( c ) = 0 again by the Squeeze Theorem because a > 1 + c. Assume now that f is continuous on [ 1, 1]. Then f ([ 1, 1]) is compact hence bounded, so part (c) implies that a 1 + c. We need only rule out the case a = 1 + c then to show that a > 1 + c. So suppose a = 1 + c. Then for > 0, f () = a c sin( c ) c cos( c ). Now take the sequence k = (2πk) 1/c, which has the properly that k 0 + as k, and f ( k ) = c for all k. That is, lim k f ( k ) = c 0 = f (0), so f is not continuous at 0. (e) We assume here that f is continuous throughout [ 1, 1], i.e. a > 1 + c by part (d). f (0) = lim 0 f () f (0) 0. For < 0, f () f (0) 0 = f ( ) f (0) 0 = f ( ), and for > 0, f () f (0) 0 =

4 HW #1 SOLUTIONS f (). Therefore f (0) eists if and only if the right hand limit lim f () 0 + = lim 0 +(a a 2 sin( c ) c a c 2 cos( c )) eists and if this right hand limit eists then its limit is f (0). Write g() = a a 2 sin( c ) c a c 2 cos( c ) then. Suppose a < 2 + c. Then take the sequence k = (2πk) 1/c. Then k 0 as k whereas g( k ) = c(2πk) (c+2 a)/c, so that lim k g( k ) =, so f (0) does not eist. Suppose a = 2 + c. Then take k = (πk) 1/c, so g( k ) = c( 1) k, and lim k g( k ) does not eist. Suppose a > 2 + c. Then the first term a a 2 sin( c ) a a 2 and this tends to 0 as 0. For the second term, c a c 2 cos( c ) c a c 2 and again this tends to 0 as 0. So by the Squeeze Theorem, f (0) eists. (f) Assume now that a > 2 + c. We know from the above computation in part (e) that f (0) = 0. f is epressed by differentiable function on [ 1, 0) (0, 1], so f is differentiable on it. Therefore with the assumption thta a > 2+c, f is differentiable on [ 1, 1]. We have f () = f ( ), so by the Chain Rule, f () = f ( ), i.e. f is an even function and it suffices to compute f d () for > 0, i.e. d (aa 1 sin( c ) c a c 1 cos( c )) = (a(a 1) a 2 c 2 a 2c 2 ) sin( c ) c( 2a + c + 1) a c 2 cos( c ). Note that both a(a 1) > 0 and ( 2a + c + 1) > 0 from a > 2 + c and c > 0. All the terms appearing here ecept for a 2c 2 sin( c ) tends to 0 as 0, so f is bounded on [ 1, 1] if and only if a 2c 2 sin( c ) is bounded on (0, 1]. By the same argument as in the previous parts, this happens eactly when a 2c 2 0, or when a 2 + 2c. (g) Suppose f is continuous on [ 1, 1]. Then f ([ 1, 1]) is compact since [ 1, 1] is compact (closed and bounded, Heine-Borel). So by part (f), f is bounded on [ 1, 1] and thus a 2 + 2c. We rule out the case a = 2 + 2c. If a = 2 + 2c, we have for > 0, f () = (a(a 1) a 2 c 2 a 2c 2 ) sin( c ) c( 2a + c + 1) a c 2 cos( c ). Set k = ( π 2 + 2πk) 1/c as usual, then f ( k ) = a(a 1)( π 2 + 2πk) (a 2)/c c 2, which does not tend to 0 as k. Since f (0) = 0 and lim k k = 0, f is not continuous at 0. Suppose a > 2 + 2c. Then all the terms in the epression above for f () for > 0 tends to 0 as 0 by the Squeeze Theorem, so f is continuous at 0. It s continuous everywhere else so f is continuous on [ 1, 1]. 6. Rudin Ch 5. Eercise #1. Let f be a differentiable real function defined in (a, b). Prove that f is conve if and only if f is monotonically increasing. Assume net that f () eists for every (a, b), and prove that f is conve if and only if f () 0 for all (a, b). Recall that a real function f defined on (a, b) is conve if and only if for all < y < z in (a, b), f(y) f() y We will prove this equivalence at the end of this solution. f(z) f(y) z y. Suppose that f is conve. Take < y in (a, b). Take sequences { n } n and {y n } n such that < n < y n < y for all n, lim n n = and lim n y n = y. Since f is assumed to be differentiable, we then have f () = lim n a n and f (y) = lim n b n if a n = f(n) f() n and b n = f(yn) f(y) y n y. By the conveity, a n = f(n) f() n f(yn) f(n) y n n f(y) f(y n) y y n = b n for all n 1. Therefore f () = lim n a n lim n b n = f (y). That is, f is monotonically increasing. Suppose now that f is monotonically increasing. Take < y < z in (a, b). By the Mean Value Theorem,

5 HW #1 SOLUTIONS 5 f(y) f() y = f (c) for some c (, y) and f(z) f(y) z y = f (d) for some d (y, z). Then < c < y < d < z, and in particular c < d. Therefore f (c) f (d), and so f is conve. Assuming that f () eists for all (a, b), note only that f is monotonically increasing if and only if f () 0 for all (a, b) by the Mean Value Theorem. One direction follows from Theorem Suppose that f is monotonically increasing. Then fi y (a, b) and let h : [y, b) R be defined by h() = f () f (y). Then h(y) = 0, h is differentiable since f is differentiable, and h() = h() 0 = h() h(y) = h (c)( y) for some c (y, ). But h (c) = f (c) 0 (we re fiing y), so that h() 0. That is, f () f (y) 0, or f () f (y). So f is monotonically increasing. Proof of equivalent definitions of conveity. We compare the two conditions on a function f : (a, b) R: i) For all t (0, 1) and, y (a, b), f(t + (1 t)y) tf() + (1 t)f(y) and ii) For all, y, z (a, b) such that < y < z, f(y) f() y f(z) f(y) z y. i) implies ii). Take < y < z in (a, b). We take here t = z y z, so that t (0, 1). Then y = t + (1 t)z, and so f(y) tf()+(1 t)f(z) = z y z f()+ y z f(z). Rearranging this inequality gives precisely f(y) f() y f(z) f(y) z y. ii) implies i). If t (0, 1) and < y are in (a, b), then < t + (1 t)y < y. So using the condition ii), we have f(t+(1 t)y) f() t+(1 t)y f(y) f(t+(1 t)y) y (t+(1 t)y). Rearranging this inequality gives precisely that f(t + (1 t)y) tf() + (1 t)f(y). 7. Rudin Ch 5, Eercise #15. Suppose a R 1, f is twice-differentiable real function on (a, ), and M 0, M 1, M 2 are the least upper bounds of f(), f (), f (), respectively, on (a, ). Prove that M 2 1 M 0 M 2. Does this hold for vector valued functions too? If (a, ) and h > 0, then by the theorem of Taylor, there is ζ (, + h) such that f( + h) = f() + f ()h + f (ζ) 2 h 2, or f ()h = f( + h) f() f (ζ) 2 h 2. Therefore f () h M 0 + M 0 + M2 2 h2. Or f () 2M0 h + M2 2 h. Fiing h > 0, we see here then that the set { f () (a, )} has 2M0 h + M2 2 h as an upper bound, or M 1 2M0 h + M2 2 h, or M 1h 2M0 M h2. This inequality certainly holds for h 0 as well since the RHS is nonnegative for h R. We have so far that M2 2 h2 M 1 h + 2M 0 0 for all h R. Considering the real vlaued function g(h) = M 2 2 h2 M 1 h + 2M 0 defined on R, a degree 2 polynomial, the fact that it is always nonnegative implies that it has at most 1 root, i.e. that its discriminant M1 2 ( M2 2 )(2M 0) 0, or M1 2 M 0 M 2. Let s consider now a vector valued, twice differentiable function f : (a, ) R n. We reduce the problem to the one-variable case by considering various projections of the curve f in R n onto the 1-dimensional subspaces of R n. To that end, take a direction u S n 1 = {(a 1, a 2,..., a n ) n i=1 a2 i = 1}. We then consider the projection of the curve f onto the line spanned by u, i.e. consider the function g : (a, ) R by g(t) = u f(t) = u f(t) cos θ where θ is the angle between u and f(t). Using the one-variable case, we have (sup t (a, ) g (t) ) 2 sup t (a, ) g(t) sup t (a, ) g (t). Note that g (t) = u f (t) and g (t) = u f (t), so by the Cauchy-Schwartz inequality we have g (t) u f (t) = f (t) and likewise g (t) f (t).

6 6 HW #1 SOLUTIONS We have thus shown that for all u S n 1, (sup t (a, ) u f (t) ) 2 M 0 M 2. Note that (sup t (a, ) f (t) ) 2 = sup t (a, ) f (t) 2. Now take any ɛ > 0. Then there eists ζ (a, ) such that sup t (a, ) f (t) ɛ < f (ζ). We now then choose u S n 1 to maimize the magnitude of the projection f (ζ) onto u, i.e. take u = f (ζ) f (ζ) in the direction parallel to f (ζ). Then we have the inequalities. ( sup f (t) ɛ) 2 f (ζ) 2 = u f (ζ) 2 ( sup u f (t) ) 2 M 0 M 2 t (a, ) t (a, ) Since ɛ > 0 was arbitrary, we have M 2 1 M 0 M 2. Now take { if 1 < < 0 f() = if 0 We show that M 0 = 1, M 1 =, and M 2 = for the above f defined on ( 1, ). It s easy to verify the following: { if 1 < < 0 f () = ( 2 +1) if 0 { 2 if 1 < < 0 f () = ( 2 +1) 3 if 0 The only care one needs to take in differentiating the above f is at 0, but this is handled in a straightforward manner by comparing the right-hand and the left-hand limits at 0. Writing = , it s clear that M 0 = 1. Likewise for M 1 = and M 2 =. 8. Rudin Ch 5, Eercise #18. Suppose f is a real function on [a, b], n is a positive integer, and f (n 1) eists for every t [a, b]. Let α, β, P be as in Taylor s theorem (5.15). Define for t [a, b], t β, differentiate Q(t) = f(t) f(β) t β f(t) f(β) = (t β)q(t) n-1 times at t = α, and derive the following version of Taylor s theorem: f(β) = P (β) + Q(n 1) (α) (β α) n. (n 1)! We ll verify the conclusion by induction on n. α) + f (α) 2! (β α) f (n 1) (α) (n 1)! (β α) n 1. Note that P (β) depends on n, as P (β) = f(α) + f (α)(β If n = 1, then the conclusion reads f(β) = P (β) + Q(α)(β α) = f(α) + Q(α)(β α), which just comes from the definition of Q(t) (and evaluating this at t = α).

7 HW #1 SOLUTIONS 7 Before we proceed to the inductive step, we develop a formula for derivatives of f(t) f(β) = (t β)q(t). Differentiating this once, we have f (t) = Q(t)+(t β)q (t), differentiating yet again, we have f (t) = 2Q (t)+(t β)q (t). It s easy to see by (yet another) induction that f (n) (t) = nq (n 1) (t) + (t β)q (n) (t) for n 1. Plugging in t = α, we have f (n) (α) = nq (n 1) (α) + (α β)q (n) (α). Now we resume the induction. Assuming that the conclusion holds for n, we have f(β) = P (β)+ Q(n 1) (α) (n 1)! (β α) n. From here, one simply substitutes Q (n 1) (α) = f (n) (α)+(β α)q (n) (α) n from the equation we derived in the previous paragraph. 9. Rudin Ch 5, Eercise #26. Suppose f is differentiable on [a, b], f(a) = 0, and there is a real number A such that f () A f() on [a, b]. Prove that f() = 0 for all [a, b]. Note that we may as well assume that A > 0, for if A 0 then 0 f () 0, or f () = 0 for all [a, b], in which case we have that for t [a, b], f(t) = f(t) 0 = f(t) f(a) = f (c)(t a) for some c [a, t], and since f (c) = 0 we have f(t) = 0, i.e. f() = 0 for all [a, b]. Assuming now that A > 0, set = b a n, where n Z+ is chosen such that b a n < 1 A (note that we can arrange for this by the Archimedean property). For i {0, 1, 2,..., n}, set t i = a+i, so that [a, b] = n i=1 [t i 1, t i ]. We show by induction on i that f is 0 on [t i 1, t i ]. Base case is when i = 1, or showing that f is 0 on [a, a + ]. Set M be the sup of the set { f() [a, a + ]}. Taking [a, a+ ], we have by the Mean Value Theorem there eists c [a, ] such that f() = f() f(a) = f (c) a A f(c) a A M. This being true for all [a, a + ], we have M A M. If M > 0, then dividing the inequality by M, we have 1 A, contradicting < 1 A. Therefore M = 0 and f() = 0 for all [a, a + ]. Assume now that f is 0 on [t 0, t 1 ] [t 1, t 2 ] [t i 1, t i ]. We now show that f is 0 on [t i, t i+1 ]. We know that f(t i ) = 0. The same argument applies: Let M be the sup of the set { f() [t i, t i+1 ]}. For any [t i, t i+1 ], we have f() = f() f(t i ) = f (c) t i A f(c) A M, for some c [t i, ]. This being true for all [t i, t i+1 ], we have M A M. Same argument as before, we must then have M = 0. Therefore f = 0 on [t i, t i+1 ]. 10. Rudin Ch 5, Eercise #27. Let φ be a real function defined on a rectangle R in the plane, given by a b, α y β. A solution of the initial-value problem y = φ(, y), y(a) = c (α c β) is, by definition, a differentiable function f on [a, b] such that f(a) = c, α f() β, and f () = φ(, f()) (a b).

8 8 HW #1 SOLUTIONS Prove that such a problem has at most one solution if there is a constant A such that whenever (, y 1 ) R and (, y 2 ) R. φ(, y 2 ) φ(, y 1 ) A y 2 y 1 Suppose g : [a, b] R is yet another solution of the above initial-value problem. Consider a function h : [a, b] R defined by h() = f() g() for [a, b]. Then for [a, b], h () = f () g () = φ(, f()) φ(, g()) A f() g() = A h(). Also, h(a) = f(a) g(a) = c c = 0. By problem #26 then we have h = 0 on [a, b]. We now classify all solutions to y = y 1/2, y(0) = 0. Certainly zero function is a solution to this differential equation. Suppose now that f is a solution to the differential equation on an interval I containing 0. Certainly f() 0 for all I, so that f () = f() implies that f is nondecreasing on I. For I and 0, we must thus have 0 f() f(0) = 0, or f() = 0. We can then just (uniquely by monotonicity of f) etend f to all of (, 0] I by defining it to be 0 on (, 0], therefore we assume that our function is defined f : (, a] R, where a 0. We can assume that a > 0 (if a = 0 the monotonicity argument above already implies that zero function is the only possible solution on (, 0]). Zero function defined on all of (, ) is of course a solution to the differential equation. Since f is nondecreasing, if f is not the zero function then f(a) = b > 0. Now let c = sup{ f() = 0} = S. Note that c eists because S has a as an upper bound and is nonempty because it contains 0. Then there is a sequence { n } n 1 in S such that lim n n = c, and by continuity of f we must have f(c) = 0. We have c < a. Take a point l such that c < l < a, and note that f(l) = s > 0. Now let K n = [c + l c n, l + a l n ] for n 2. Then K n is compact, f is defined on K n and f is never zero on K n. Therefore f(k n ) is compact and does not contain 0. Considering the function φ(y) = y 1/2 defined on K n, we see then that φ (y) = 1 2 y 1/2 is defined and continuous on f(k n ), so φ (f(k n )) is compact, i.e. there is a real number A (may depend on n) such that for all f(k n ), φ () A. Therefore if q, w f(k n ) then by the Mean Value Theorem, φ(q) φ(w) = φ (c) q w A q w, where c is between q and w. Also f(k n ) is compact and connected in R 1, so it is a closed interval (Theorem 2.7, say). If we use the rectangle R as in problem #27 as K n f(k n ), then f () = φ(f()) and φ(y 1 ) φ(y 2 ) A y 1 y 2 for all y 1, y 2 f(k n ), so the conditions of problem #27 are satisfied (note that φ(y) is independent of here, an autonomous case). Therefore the solution to this differential equation with the initial condition f(l) = s is unique, and it s readily checked that the mapping (+2 s l) 2 is a solution to this initial differential equation, so we must have f() = (+2 s l) 2 on K n. Now we take n = c + l c n for n 2, so that n K n and f( n ) = (n+2 s l) 2. Note that lim n n = c, and so by continuity of f we must then have 0 = f(c) = (c+2 s l) 2. s = f(l), so we must have c + 2 f(l) l = 0, or f(l) = (l c)2. Therefore { 0 if c f() = (l c) 2 if c