Math 320-1: Midterm 2 Practice Solutions Northwestern University, Fall 2014

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1 Math 30-: Midterm Practice Solutions Northwestern University, Fall 04. Give an eample of each of the following. Justify your answer. (a) A function on (, ) which is continuous but not uniformly continuous. (b) A function which sends Cauchy sequences to Cauchy sequences but is not uniformly continuous. (c) A function f on R for which lim π f() eists but where f is not continuous at π. f(+h) f( h) (d) A function f on R for which lim h 0 h eists but f () does not. (The point here is that if f () eists, one can show that this limit does as well and equals f (), but the eistence of this limit is not enough to guarantee the eistence of f ().) (e) A function which is twice differentiable on R but not three-times differentiable. Solution. (a) The function f() = is continuous on (, ) since it is a quotient of never zero continuous functions, but is not uniformly on (, ) since it is unbounded. (Recall that a uniformly continuous function on a bounded interval must be bounded.) (b) The function f() = on R is not uniformly continuous but sends Cauchy sequences to Cauchy sequences since Cauchy means the same thing as convergent on R. (c) Define f() = for π and f(π) = 0. Then lim π f() =, which is not f(π) f is not continuous at π. (d) Define f() = for and f() = 00. Then for any h 0, f(+h) f( h) = = 0 f(+h) f( h) lim h 0 h = 0, but f is not differentiable at since it is not even continuous at. (e) The function defined by f() = 4 sin for 0 and f(0) = 0 works, as seen on Homework 5. Check the lutions to Homework 5 for the justification.. Show that the function f defined by f() = is continuous on (, ) by directly verifying 3 the ɛ-δ definition of continuity. Is this function uniformly continuous on (, )? The lution below is wrong since I computed f() f(a) incorrectly; it should be f() f(a) = 3 a a 3 = (a3 ) a( 3 ) ( 3 )(a 3 = a3 a 3 + a ) ( 3 )(a 3 ). But now the issue is that this correct epression becomes tough to bound correctly in fact, I would not epect that you be able to do this easily, certainly not in the time required on an eam. So, for the sake of getting across the idea I meant the problem to emphasize, go ahead and ignore my error and assume that f() f(a) = 3 a a 3 was in fact correct. Start with this and figure out how to come up with the appropriate bounds. So in actuality, we re not showing that f() = is continuous, but rather that whatever 3 function satisfies f() f(a) = 3 a a 3 is continuous. (I should have double-checked my work!) Thoughts. Before giving the proof, let s work out me details to see where the subtlety is. We want to make f() f(a) = 3 a a 3 = a 3 a 3

2 small. Here a is the point we are checking continuity, but varies we need a bound on this epression which only depends on a (which will be smaller than δ) and a. This means we need to make the denominator smaller by making the 3 term smaller. We have 3 = + + > since + + > given that (, ). Hence the problem boils down to finding a positive lower bound on. The point is that since the values of we will consider should be close to a and < a, being close enough to a will guarantee that > a. Indeed, we just need to be within (a ) away from a. (Draw a picture!) Proof. Let a (, ) and let ɛ > 0. a < δ. Then a < a Hence Thus we get a < a, and thus a Set δ = min { a, a a3 } > 0 and suppose that = a a <. 3 = + + > > a. f() f(a) = 3 a a 3 = a 3 a 3 < δ a a3 ɛ. We conclude that f is continuous at a, f is continuous on (, ) since a (, ) was arbitrary. We claim that f is not uniformly continuous on (, ). Indeed, if f was uniformly continuous on (, ) it would be uniformly continuous on any smaller interval as well, and on (, ). But f is unbounded on this interval since it gets arbitrarily large as approaches, is not uniformly continuous on (, ) since uniformly continuous functions on bounded intervals are bounded. Thus f is al not uniformly continuous on (, ). 3. Determine, with justification, the points at which the function f : R R defined by { ( ) Q f() = ( ) sin / Q is continuous. Yet another mistake on my part. The issue in the presented lution is when I claim that (a ) sin for a. In fact, if you graph the function g() = ( ) sin (a ) a ( ) you ll see that there are plenty of points where this is zero, giving various values of a where (a ) sin = (a ) a

3 is true. This function is continuous at all such points, but the problem is that such points cannot be determined precisely. So, the answer should be that f is continuous at any a satisfying (a ) sin a = (a ). I ll leave the incorrect lution as is since the basic idea is right, but be sure to note where my mistake is. Solution. First we claim that f is continuous at =. Indeed, let ɛ > 0 and set δ = min{ɛ, ɛ}. Suppose that < δ. Then if Q we have f() f() = < δ ɛ, while if / Q we have f() f() = sin < δ ɛ. Thus < δ implies f() f() < ɛ, f is continuous at as claimed. Second we claim that f is continuous nowhere else. Indeed, if a and a Q, taking a sequence y n of irrationals converging to a gives f(y n ) = (y n ) sin y n (a ) sin a (a ), while if a and a / Q, taking a sequence r n of rationals converging to a gives f(r n ) = (r n ) (a ) (a ) sin Thus either way, f is not continuous at a. a. 4. Show that a function f : R R is continuous if and only if the preimage of any open subset of R is open. Recall that the preimage f (U) of a subset U R is the set of all R which are sent to mething inside U: f (U) := { R f() U}. (This gives a characterization of continuity lely in terms of open sets. Technically this problem is not a good problem for the purposes of reflecting eam material since I ve said additional topics we spoke about, such as open sets, won t show up on eams; however, thinking about this problem might help to better understand the true meaning behind the ɛ-δ definition of continuity.) Proof. Suppose that f is continuous and let U R be an open set. We want to show that f (U) is open. To this end, let a f (U), that f(a) U. Since U is open, there eists ɛ > 0 such that (f(a) ɛ, f(a) + ɛ) U. Since f is continuous at a, there then eists δ > 0 such that But this can be reinterpreted as a < δ implies f() f(a) < ɛ. (a δ, a + δ) implies f() (f(a) ɛ, f(a) + ɛ,

4 and since (f(a) ɛ, f(a) + ɛ) U, this says that anything in (a δ, a + δ) is send to mething in U. Thus (a δ, a + δ) f (U), showing that f (U) is open. Conversely, suppose that the preimage of any open set under f is open. We want to show that f is continuous at any a R. Let a R and let ɛ > 0. Since U = (f(a) ɛ, f(a) + ɛ) is open, f (U) is open as well. Thus since a f (U), there eists δ > 0 such that (a δ, a + δ) f (U). This means that anything in (a δ, a + δ) is sent to mething in U = (f(a) ɛ, f(a) + ɛ), or in other words: a < δ implies f() f(a) < ɛ. Hence f is continuous at a as claimed. 5. Show that a function f is uniformly continuous on R if and only if for any sequences ( n ) and (y n ) such that ( n y n ) 0, we have (f( n ) f(y n )) 0 as well. (This gives a characterization of uniform continuity in terms of sequences.) Proof. Suppose that f is uniformly continuous on R and let ( n ) and (y n ) be sequences such that ( n y n ) 0. We must show that (f( n ) f(y n )) 0. Let ɛ > 0. Since f is uniformly continuous there eists δ > 0 such that y < δ implies f() f(y) < ɛ. Since ( n y n ) converges to 0, there eists N such that But then for n N we al have n y n < δ for n N. f( n ) f(y n ) < ɛ by the chose of δ. Thus (f( n ) f(y n )) converges to 0 as claimed. To prove the converse we instead prove its contrapositive: if f is not uniformly continuous, there eist sequences ( n ) and (y n ) such that ( n y n ) 0 but (f( n ) f(y n )) 0. If f is not uniformly continuous then there eists ɛ > 0 such that for any δ > 0 we can find, y R such that y < δ but f() f(y) ɛ. In particular, for each n applying this condition to δ = n gives n, y n satisfying n y n < n and f( n) f(y n ) ɛ. These number thus define sequences ( n ) and (y n ) such that ( n y n ) converges to 0, which we see by applying the squeeze theorem to 0 n y n < n, but (f( n ) f(y n )) does not converge to 0 since it is does not converge at all given that all terms are at least ɛ away from 0.

5 6. Suppose that a continuous function f : R R is uniformly continuous on (, 0] and uniformly continuous on [0, ). Show that f is uniformly continuous on R. (Here s the subtlety. For any ɛ > 0 the fact that f is uniformly continuous on the first interval gives an appropriate δ and the fact that it is uniformly continuous on the second gives an appropriate δ, at first glance it might seem that to satisfy the requirement of uniform continuity on all of R taking δ = min{δ, δ } would work. Certainly, if y < δ and both, y are nonpositive, then we can apply what we know about δ and if y < δ and both, y are nonnegative, we can apply what we know about δ ; but what happens if y < δ and one of, y is negative and the other positive? Do you see why this matters? How might the fact that f is continuous at 0 help to overcome this?) Proof. Let ɛ > 0. Since f is uniformly continuous on (, 0] there eits δ > 0 such that for, y (, 0] such that y < δ, f() f(y) < ɛ. Since f is uniformly continuous on [0, ) there eits δ > 0 such that for, y [0, ) such that y < δ, f() f(y) < ɛ. Finally, since f is continuous at 0 (since it is continuous on all of R) there eists δ 3 > 0 such that if < δ 3, then f() f(0) < ɛ. Set δ = min{δ, δ, δ 3 }, which is positive, and suppose that, y R satisfy y < δ. If and y are both nonpositive, then y < δ implies that f() f(y) < ɛ. If and y are both nonnegative, then y < δ implies that f() f(y) < ɛ. Now, consider the case where one of and y is negative and the other positive say that < 0 and y > 0. Since y < δ 3, we must have both < δ 3 and y < δ 3 because otherwise the distance between and y would be Thus the choice of δ 3 implies that + y δ 3. f() f(0) < ɛ and f(y) f(0) < ɛ, f() f(y) f() f(0) + f(0) f(y) < ɛ + ɛ = ɛ. Hence no matter where, y R are, we have f is uniformly continuous on R as claimed. y < δ implies f() f(y) < ɛ, 7. Determine where the function from Quiz 5 is differentiable, which was what that quiz was actually supposed to ask. (The TA misread the problem asked to determine where the function was continuous instead.) Proof. Included on the lutions to Quiz 5.

6 8. Suppose that f : R R is differentiable at a R and f (a) > 0. Show that there eists an interval around a such that f() f(a) f(y) for any to the left of a and y to the right of a in that interval. (Careful: don t assume that f is differentiable anywhere apart from a.) Proof. Since there eists δ > 0 such that f f() f(a) (a) = lim > 0, a a for (a δ, a + δ), f() f(a) a > 0. Thus for (a δ, a + δ), f() f(a) and a have the same sign; hence if < a < y, a < 0 and y a > 0 as desired. f() f(a) < 0 and f(y) f(a) > 0, giving f() < f(a) < f(y), 9. Wade, Prove that each of the following inequalities holds. (a) < e for all (b) log < 0.6 for all 4. (c) sin for all R. (d) sin e for all 0. Solution. (a) Let f() = e, which is differentiable everywhere. For any, by the Mean Value Theorem there eists c between and such that since and e c > for c. Thus f() f() = f (c)( ), (e ) (e ) = (e c )( ) 0 e e > 0.7, which gives the desired inequality. (b) (Note that here log means ln ; % of the time when a mathematician says log they mean natural log.) Let f() = log, which is differentiable at all > 0. For 4 by the Mean Value Theorem there eists c between 4 and such that f() f(4) = f (c)( 4), log (log 4 ) = ( c ) ( 4). c Since c > 4, c > c which implies that c < 0. Since > 4, the right side is negative, c log < log 4 < 0.6,

7 from which the desired inequality follows. (c) This is certainly true at = 0 since both sides are then 0. Assume that > 0. Applying the Mean Value Theorem to f() = sin gives that there eist c such that (sin ) (0 0) = ( sin c cos c )( 0). Since > 0 and sin c cos c, the right side is 0 sin = since > 0. If < 0, we instead apply the Mean Value Theorem to g() = sin + to get (sin + ) (0 0) = ( sin c cos c + )( 0) for me c. Then 0 < 0 and sin c cos c + 0, the right side is 0 and hence sin + 0, giving sin = since < 0. (d) Applying the Mean Value Theorem to f() = e + sin gives that for any 0 there eists c > 0 such that e + sin ( 0) = (e c + cos c). Since c > 0, e c > > cos c, the right side is positive and hence e + sin > as required. 0. Suppose that f, g : R R are functions such that f() f(y) g() g(y) y for any, y R. If g is differentiable with bounded derivative on all of R, show that f is constant. (Note: I really like this problem, since it brings together much of what we covered when speaking about differentiability.) Proof. Say that M > 0 is a bound on g, meaning that g () M for all R. By the Mean Value Theorem, for any y R there eists c between and y such that Thus for any, y R we have Hence g() g(y) = g (c) y, g() g(y) M y. f() f(y) g() g(y) y M y y. f() f(y) y M y for any, y R. Fiing y and taking the limit as y thus gives lim f() f(y) y y = 0, since y 0 as y. This implies that f() f(y) lim = 0, y y f (y) eists and equals 0. Since y R was arbitrary, f (y) = 0 for all y R, which implies that f is constant as claimed.