MATH 351 Solutions: TEST 3-B 23 April 2018 (revised)

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1 MATH Soluions: TEST -B April 8 (revised) Par I [ ps each] Each of he following asserions is false. Give an eplici couner-eample o illusrae his.. If H: (, ) R is coninuous, hen H is unbounded. Le H() = (, ).. If f: R R and g: R R are uniformly coninuous funcions, hen fg is uniformly coninuous on R. Le f() = g() = for all real. Then f and g are uniformly coninuous. However, (fg)() = which is no uniformly coninuous on R.. Given wo funcions f: [, ] R and g: [, ] R such ha each is nowhere coninuous on [, ], hen fg canno be coninuous everywhere on [, ]. if Q Le f() = g() = { if Q Then (fg)() = for all real.. If G: [, ] R is coninuous hen, for every cr, he se S = { [, ]: G() = c} canno be counably infinie. Define G() = { sin if (, ] if =. Le I = (, ). If g: I R is coninuous hen g(i) is no a compac inerval. Le g() = sin(). Then g(i) = [-, ]. 6. Le F: R R be coninuous and le A, B R. Then F(A B) = F(A) F(B). Le F: R R be defined by F() = for all R. Le A = {, } and B = {, -}. Then F(A B) = F({}) = {}. Bu F(A) F(B) = {, }{, }={, }.

2 7. If wo funcions, f: R R and g: R R saisfy he condiions ha hen g f() eiss and equals. f() = and g() = Le f() = for all R. Le g( ) if if 8. There does no eis a funcion f: R R ha is coninuous only a =. Le f() = D() where D is he Dirichle funcion defined by: if Q D() = { if Q. If F: [, ] R is coninuous and {bn} is a sequence in [, ] for which {F(bn)} converges hen {bn} mus converge. Define F() = for all [, ]. Define b n = { if n Z+ is even if n Z + is odd If S and T are sequenially compac subses of R, hen S T ={S T} is sequenially compac. Le S = [, ] and Le T = {}. Then S T = (, ], which is cerainly no sequenially compac.. Le h: [, ] R be a funcion ha achieves a maimum value on he inerval [, ]. Then he funcion F() = (h()) also achieves a maimum value on he inerval [, ]. if < Define h() = { if = Noe ha h has a maimum value of on [, ] bu ha (h()) has no maimum on [, ].. There does no eis a funcion f: R R ha is coninuous everywhere ecep a a counably infinie se of poins. Le D() denoe he Dirichle funcion defined by:

3 if Q D() = { if Q Le g()= D() (sin ). Then g is coninuous only a poins of he form n where n Z.. There does no eis a funcion H: (, ) R ha is bijecive. Define H() = an π( ). Le f: (, ) R be coninuous and le {an} be a Cauchy sequence in (, ). Then {f(an)} is a Cauchy sequence. Le f() = /. Le n = /n. Par II [ ps each] Insrucions: Selec any of he following 7 problems. You may answer more han o obain era credi.. Le f: (, ) R (a) Le p (, ). Wrie he negaion of he saemen f: (, ) R is coninuous a = p. (Noe: f is no coninuous a p is no a sufficien answer. Use and in your answer.) Since he definiion of coninuiy of f a = p is: ε > δ > such ha f() f(p) < ε whenever (, ) and p < δ. The logical negaion of his senence is: ε > δ > (, ) such ha f() f(p) ε and p < δ. (b) Wrie he negaion of he saemen f: (, ) R is uniformly coninuous on (, ). (Noe: f is no uniformly coninuous on R is no a sufficien answer. Use and in your answer.) Since he definiion of uniform coninuiy of f on (, ) is: ε > δ > such ha f(a) f(b) < ε whenever a, b (,) and a b < δ. The logical negaion of his senence is: ε > δ > a, b (, ) such ha f(a) f(b) ε and a b < δ.. Le < a < b and le f be a posiive funcion defined on [a, b]. If f saisfies he Inermediae Value Propery on [a, b], prove ha saisfies he Inermediae Value Propery on [a, b]. f

4 The definiion of is: () = for all [a, b]. f f f() Now suppose is a number beween (a) and (b), inclusive. f f Then β lies beween f(a) and f(b), inclusive. Since f saisfies he IVP, c [a, b] such ha f(c) = β. Bu his means ha (c) = β. f Hence we have shown ha saisfies he IVP. f. Le f: (, ) R be funcion for which f() =. Prove ha f() sin =. Soluion: Le > be given. Then: f ( ) K such ha K f ( ). Thus K f ( )sin f ( )sin f ( ). Hence we obain f ( )sin. Le G: [, ] R be coninuous. We know from he Boundedness Theorem ha = sup G([, ]) eiss. Le {an} be a sequence of poins in [, ] saisfying (a) G(an) > /n for all n. Mus i follow ha {an} converge? Eplain! No, le G() = for all [, ]. Then = sup G([, ]) =. Define a n = ½ +(-½) n. Then G(a n ) = > /n for all n. And ye { a n } diverges. (b) Suppose ha an p. Prove ha G(p) =. Soluion: Suppose ha an p. Since [, ] is sequenially compac, i follows ha p[, ]. Now, since G is coninuous on [, ], i is sequenially coninuous. Thus G(an) G(p) (using he fac ha p[, ]). Since G(an) > /n for all n, we may invoke he Squeeze Theorem o obain ha G(an) eiss, and ha G(an) =. Now, since G(p) = G(an), we obain: G(p) =.

5 . Le f: [, ] [, ] be a coninuous funcion. Prove ha here eiss a poin p[a, b] such ha f(p) = p. Hin: Consider he funcion G() = f(). Soluion: Define h: [, ] R as follows: h() = f() for all [, ]. Now since f and he ideniy funcion are coninuous, h mus be coninuous by he lineariy heorem. Ne noe ha h() = f(), by definiion of f, and ha h() = f(). Now, if eiher h() = or h() =, hen we are done, for i follows ha eiher f() = or f() =. So le us assume ha h() < and h() >. Then, invoking Bolzano s Theorem, here eiss p[a, b] such ha h(p) =. And so, f(p) = p. (Noe: p is called a Fied Poin of f.) 6.. ) ( d g Le Prove ha g() eiss and find is value. Soluion: We claim ha ) ( g Observe ha, since : Thus: (**) d d d Now d d and so: as d Finally, applying he Squeeze Theorem o (**), we find ha. ) ( is his and eiss g 7. Le G: (a, b) R be a uniformly coninuous funcion and le {n} be a Cauchy sequence in (a, b). Prove ha {G(n)} is a Cauchy sequence.

6 Le ε > be given. Since G is uniformly coninuous, δ > such ha p, q (a, b) p q < δ G(p) G(q) < ε. Since {n} is a Cauchy sequence, n m < δ for m, n. Thus for a given ε > δ > such ha G( n ) G( m ) < ε for m, n. 6