MATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n

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1 MATH 43 FINAL EXAM Math 43 fial exam, 3 May 28. The exam starts at 9: am ad you have 5 miutes. No textbooks or calculators may be used durig the exam. This exam is prited o both sides of the paper. Good luck! () (2 marks) Let X = (, ] R. State whether each of the followig statemets about X is true or false, givig a brief reaso for each aswer. (a) X is bouded. True. For all x X, x. (b) X ca be writte as a coutable uio of ope sets. False. Ay uio of ope sets is ope, but X is ot ope. (c) X is compact. False. X is ot closed (it does ot cotai the cluster poit ), so by the Heie-Borel Theorem caot be compact. (d) There is a poit x X at which the fuctio f(x) = log(x) + x 5 8x 4 3 achieves its supremum o X (that is, f(x ) = sup{f(x) : x X}). True. Sice log(x) as x, we have also that f(x) as x. So there exists N such that f(x) < f() if x <. Therefore, sup x X f(x) = sup x [,] f(x) ad this is attaied by f sice [, ] is a compact set. (2) (2 marks) Let A R. Recall that a fuctio f : A R is said to satisfy a Lipschitz coditio o A if there is some M R such that f(x) f(y) M x y for all x, y A. (a) Let N. Show that the fuctio f : [, ] R defied by f (x) = satisfies a Lipschitz coditio o [, ]. (Hit: you may wish to use the fact that for all a, b >, ( a + b)( a b) = a b.) We eed to show that there exists M such that x + f (x) f (y) M x y

2 for all x, y [, ]. Usig the hit, if x y, we have f (x) f (y) = x + y + = (x+ ) (y+ ) x+ +. Sice x, y, we get f y+ (x) f (y) (x y)/(2/ ) = M x y where M = /2. (b) Show that the sequece of fuctios {f } coverges uiformly o [, ] to the fuctio f(x) = x. We eed to show that for all ε > there exists N N such that if > N the f (x) f(x) < ε for all x [, ]. We have f (x) f(x) = x + x = x+ x / x+ + x =. So give ε >, we choose N with / N < ε, ad the if > N ad x [, ], we have f (x) f(x) < N < ε as required. (c) Show that f(x) = x does ot satisfy a Lipschitz coditio o [, ]. There are several ways of doig this. Oe way is to observe that if x was Lipschitz, the x x = x would be bouded for x (, ), but this is false. (d) Now suppose A R ad f : A R are fuctios such that there exists M R such that f (x) f (y) M x y for all N ad all x, y A. Suppose the sequece of fuctios {f } coverges uiformly o A to a fuctio f : A R. Show that f satisfies a Lipschitz coditio o A. Why does this ot cotradict your aswer to part (c)? To prove the statemet, let x, y A. Let ε >. Choose N such that f (z) f(z) < ε for all z A. The f(x) f(y) f(x) f (x) + f (x) f (y) + f (y) f(y) by the triagle iequality. But by the choice of, the first ad third terms are < ε, while f (x) f (y) M x y by hypothesis. So f(x) f(y) 2ε+M x y. But sice this holds for all ε >, we must have f(x) f(y) M x y as required. This does ot cotradict part (c) because although the f were Lipschitz, they do ot have a commo boud M, so their uiform limit does ot ecessarily have to be Lipschitz. I fact, parts (a)-(c) show that a uiform limit of Lipschitz fuctios eed ot be Lipschitz. [TURN OVER] 2

3 (3) (2 marks) Let f : R R be a fuctio. (a) State what it meas for f to be uiformly cotiuous o R. f is uiformly cotiuous o R if ad oly if for all ε > there exists δ > such that for all x, y R, if x y < δ the f(x) f(y) < ε. (b) State the Mea Value Theorem. Let f : [a, b] R be cotiuous o [a, b] ad differetiable o (a, b). The there exists x (a, b) such that f(b) f(a) b a = f (x ). (c) Suppose that f : R R is a differetiable fuctio ad that the derivative f is bouded. Show that f is uiformly cotiuous o R. Let f : R R be differetiable ad suppose there exists M > with f (x) M for all x R. The if a < b, the f(b) f(a) b a = f (x ) M for some x (a, b). So f(b) f(a) M b a. Therefore, give ε >, if δ < ε/m the b a < δ implies f(b) f(a) < ε. So f is uiformly cotiuous. (d) Show that f(x) = e x2 is uiformly cotiuous o R. I view of the previous problem, it suffices to show that the derivative of f is bouded. We have f (x) = 2xe x2 for all x R. Therefore, f (x) = 2 x e x2. Now, there are may ways of showig this is bouded, for example e x2 + x 2 for all x, by the power series expasio of e x2. So it suffices to show that bouded. If x the 2 x +x 2 2 x (4) (2 marks) Let f : [a, b] R be a bouded fuctio. (a) State what it meas for f to be Riema itegrable. 2 x +x 2 is 2 while if x the also 2 x +x 2 2 x 2. Defie a partitio P of [a, b] to be a sequece of poits x < x <... < x with x = a ad x = b. For a partitio P, defie P = max(x i x i ). Defie a Cauchy sum of f for the partitio P to be a sum of the form i f(q i)(x i x i ) where q i [x i, x i ] for each i. The f : [a, b] R is Riema itegrable if ad oly if there exists a umber L such that for all ε > there exists δ >, such that wheever P is a partitio with P < δ, we have S(f, P ) L < ε, where S(f, P ) is ay Cauchy sum of f for the partitio P. (b) Show that if f, g : [a, b] R are Riema itegrable, the so is f + g. 3

4 This is difficult to do from the above defiitio, so istead we use a theorem from the textbook, which says that a fuctio f is Riema itegrable if ad oly if there exists a sequece of partitios P j such that Osc(f, P j ) as j, where Osc(f, P j ) = i (sup q [x i,x i ] f(q) if q [xi,x i ] f(q))(x i x i ). We have sup q [xi,x i ](f(q) + g(q)) sup q [xi,x i ] f(q) + sup q [xi,x i ] g(q) ad if q [xi,x i ](f(q) + g(q)) if q [xi,x i ] f(q) + if q [xi,x i ] g(q). From these two facts it follows that for ay partitio P, Osc(f + g, P ) Osc(f, P ) + Osc(g, P ) ad so if Osc(f, P j ) ad Osc(g, P j ) ted to as j, so does Osc(f + g, P j ) (we have left out some details). This theorem ca also be proved by various other similar methods. (c) Show that the fuctio f : [, ] R defied by, x Q f(x) =, x / Q is ot Riema itegrable. This time, the best characterizatio of Riema itegrability to use is the statemet that f is Riema itegrable if ad oly if if P S + (f, P ) = sup P S (f, P ) where S + (f, P ) ad S (f, P ) deote the upper ad lower Riema sums respectively. For every partitio P = {x < x <... < x } of [, ] ad each i, [x i, x i ] cotais a poit of Q ad a poit of R \ Q. So S + (f, P ) = while S (f, P ) =. Therefore, if P S + (f, P ) = ad sup P S (f, P ) =, so f is ot Riema itegrable. (d) Now let f : R R be ay cotiuous fuctio. Defie F (x) = f(x + t)dt. Show that F is cotiuous o R. We use the defiitio of cotiuity. Let x R. To show that F is cotiuous at x, let ε >. We eed to fid δ > such that if x x < δ the F (x) F (x ) < ε. We have F (x) F (x ) = (f(x + t) f(x + t))dt f(x + t) f(x + t) dt. Sice f is cotiuous o R, it is uiformly cotiuous o the compact set [x 2, x + 2]. Therefore, we may choose a δ < such that if a, b [x 2, x + 2] 4

5 with a b < δ the f(a) f(b) < ε. Now if x x < δ the for all t [, ], we have x + t, x + t [x 2, x + 2] ad so if x x < δ the f(x + t) f(x + t) < ε. So F (x) F (x ) < εdt = ε as required. (5) (2 marks) Cosider the power series f(x) = k= (4k + )! x4k+. (a) Prove that the series coverges absolutely ad uiformly o [ a, a] for all a >. Deduce that this power series defies a C fuctio f : R R. The easiest way to prove this is to observe that for all x [ a, a] ad all r < s, we have s k=r x4k+ (4k+)! = s k=r x 4k+ (4k+)! s a 4k+ k=r 4s+ a b (4k+)! b=4r+. The b! expressio 4s+ a b b=4r+ is the tail of a power series defiig the umber e a. So b! give ε >, there exists M N such that if u, v > M the v a b b=u < ε. b! Thus, s k=r x4k+ < ε if r, s > M. So the give series coverges absolutely (4k+)! ad uiformly o [ a, a] as claimed. The power series defies a C fuctio by Theorem of the textbook. (b) Prove that for all x R. f(x) + f (x) + f (x) + f (x) = e x We are allowed to differetiate the power series term-by-term at ay x for which it coverges. We have f(x) = x + 5! x5 + 9! x9 + f (x) = + 4! x4 + 8! x8 + f (x) = + 3! x3 + 7! x7 + f (x) = + 2! x2 + 6! x6 + Absolute covergece guaratees that we may rearrage these series as we wish, ad the sum is the series for e x, as required. (c) Show that lim x f(x) = ad lim x f(x) =. 5

6 If x > the f(x) x ad if x < the f(x) x sice the series for f cotais oly odd powers of x. This is eough to show the required properties. (d) Show that f : R R is a bijectio. To show that f is oto, if c R the there exists a < with f(a) < c ad there exists b > with f(b) > c, because we showed i the previous part that f i ubouded from above ad from below. Sice f is cotiuous, the itermediate value theorem ow implies that there exists z with a < z < b ad f(z) = c. To show that f is oe-to-oe, observe from above that f (x) > for all x, sice the expressio for f (x) cotais oly eve powers of x. So f is strictly icreasig. Therefore, if a < b the f(a) < f(b), which implies that f is oe-tooe. Thus, f is a bijectio. [END.] 6