Prof. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf

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1 Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some order. Recll (see pge 7 in Knpp). Let {c n } be sequence in R (or even R ). Then the sequence is monotone decresing sequence in R nd {sup c k } k n lim c n lim sup c n lim n n sup n k n c k R inf sup n N k n Remrk. Did you notice the mistkes in the solution hints in the bck of Knpp s book? Proof. Let { n } nd {b n } be two sequences in R. Let (lim n n, lim n b n ) not be equl to (, ) nor (, ); thus, the right side of () is not n indeterminte form of nor +. Fix n N. Since for ny r N such tht r n we hve tht sup k n r + b r sup k + sup b k k n k n ( k + b k ) sup k n c k. k + sup b k. (2) k n Cse. Both lim n n nd lim n b n re finite. Tking the limit s n in (2) gives tht [ ] lim ( n+b n ) lim sup ( k +b k ) (2) lim sup k + sup b k lim sup k + lim sup b k lim n+ lim b n. n n k n n k n k n n k n n k n n n So () holds. Cse 2. lim n n nd/or lim n b n. WLOG, lim n n. Then, by ssumpution, lim n b n. So the right side of () is. So () holds. Cse 3. lim n n nd/or lim n b n. WLOG, lim n n. So, by ssumpution, lim n b n +. Since lim n n, lim n n. Since lim n b n, the sequence {b n } n is bounded bove so there exists M b R so tht b n < M b for ech n N. So M R there exists N N so tht if n N then n < M M b nd so n + b n < M. Thus lim n ( n + b n ) lim n ( n + b n ). So () holds. Remrk. If you hd troubles with this problem, then red over my lim sup nd lim inf of sequences hndout posted on my Mth 555 homepge nd pges of Trench s book, which is posted on my Mth 73/74 homepge. Pge of 2

2 Homework 2 Knpp, Chpter, Section, # 7, p. 78. Use the Bolzno-Weierstrß theorem to prove Dini s theorem. Suppose tht f n : [, b] R is continuous nd tht f f 2 f 3. Suppose lso tht f(x) lim n f n (x) is continuous nd is nowhere +. Then {f n } converges uniformly to f for x b. Recll (see pge 9 of Knpp) Bolzno-Weierstrße Theorem. Every bounded sequence in R hs convergent subsequence with limit in R. Proof. Strt by mking some reductions. Define the sequence of functions g n ( ) : (f f n )( ): [, b] R then observe tht g n is continuous for ech n, {g n } is (pointwise) non-incresing, nd {g n } converges pointwise to on [, b], ll of which follow from the ssumptions on f n nd f. It suffices to show tht g n converges uniformly to on [, b], i.e. lim mx g n(x). n x [,b] For ech n N, since g n is continuous on compct set [, b], it ssumes it s mximum so we cn find x n [, b] nd M n R so tht mx g n(x) g n (x n ) M n. x [,b] It suffices to show tht lim n M n. Assume tht lim n M n. We wnt to find contrdiction. Note tht {M n } is non-incresing non-negtive sequence since if n k then M n g n (x n ) g k (x n ) g k (x k ) M k. So {M n } n decreses to some M >. By pssing to subsequence, we my ssume by the Bolzno- Weierstrss Theorem tht {x n } n converges, sy lim n x n x. Fix k N. For n k, note tht g k (x n ) g n (x n ) M n M nd so Since g k is continuous g k (x n ) M. lim g k(x n ) g k ( lim x n) g k (x ). n n Thus g k (x ) M. This contdicts the fct tht {g k } k converge pointwise to zero. Thus it is indeed true tht {f n } converge to f uniformly on [, b] Pge 2 of 2

3 Knpp, Chpter, Section, # 6, p. 79. Homework 3 Verify the following clcultions { of Fourier series. + for < x < π () The function f(x) for π < x < hs f(x) 4 π (b) The function f(x) e iαx on (, 2π) hs f(x) e iπα sin(πα) π provided α is not n integer. sin((2n )x). 2n e inx n + α, n Proof. Prt (). Note tht f is Riemnn integrble on [ π, π] nd tht f cn be regrded s periodic of period 2π on R. Note lso tht f, regrded s such, is n odd function. We know, then, tht the Fourier series of f exists, nd it will be comprised only of the sine terms. Tht is, there exist {b n } such tht f(x) b n sin(nx) where b n π We proceed to clculte the {b n }. b n π π π f(x) sin(nx) dx π π f(x) sin(nx) dx for n. π sin(nx) dx + π π( ) () sin(nx) dx π π ( cos(nx)) n + π π π ( cos(nx)) n ( ) ( ) cos() cos( nπ) + cos() cos(nπ) πn πn 2 ( ) cos(nπ). πn Cse : If n is odd, then cos(nπ). Thus, b n 2 ( ) 4 ( ) πn πn. Cse 2: If n is even, then cos(nπ). In this cse, b n 2 ( ) ). πn It follows tht f(x) k K 4 πk sin(kx), where K is the set of odd positive integers; equivlently, f(x) 4 π sin((2n )x). 2n Pge 3 of 2

4 Prt (b). Assume α is not n integer. Note tht f is Riemnn integrble on (, 2π) nd tht f cn be regrded s periodic of period 2π on R, so we know tht the Fourier series of f exists. We will find the complex form Fourier series by clculting the {c n } such tht f(x) n c n e inx where c n 2π f(x)e inx dx for n Z. 2π From the preceding formul (nd using the fct tht, by ssumption, n + α ), c n 2π 2π 2π 2π e iαx e inx dx e i(n+α)x dx i 2π 2π(n + α) e i(n+α)x i (e i(n+α)2π e i(n+α)()) 2π(n + α) i ( e i2πn e i2πα ) 2π(n + α) Note tht since n Z, e i2πn (e iπ ) 2n. Therefore, f(x) i ( e i2πα ) 2π(n + α) ( ) i (e iπα ( e iπα e iπα)) π(n + α) 2 ( e iπα ie iπα ie iπα ) Hello, complex form of sine! π(n + α) 2 e iπα π(n + α) sin(πα). n e iπα sin(πα) e inx e iπα sin(πα) π(n + α) π n Remrk. The complex form of sine is s follows. If z C, then e inx n + α. sin z eiz e iz 2i. Multiplying the RHS by i i nd using tht i2 gives sin z ie iz ie iz 2. Pge 4 of 2

5 Homework 4 Knpp, Chpter, Section, # 7, p. 79. Combining Prsevl s theorem nd the preceding problem, prove the following identities. () (2n ) 2 π2 8, nd (b) n n + α 2 π 2 sin 2 (πα). (Of course, in (b) we re ssuming α Z so tht we do not divide by zero.) Recll: Prsevl s theorem (Thm..6 in Knpp.) Let f : C be periodic of period 2π nd Riemnn integrble on [ π, π]. If f(x) n c ne inx, then lim N π f(x) s N (f; x) 2 dx 2π π nd π f(x) 2 dx 2π π n c n 2. Proof. In.6 (), we re given the function { + for < x < π f(x) for π < x <, which is integrble on [ π, π] nd hs period 2π when extended to ll of R. We lso showed tht f(x) b n sin((2n )x), with b n 4/(π(2n )). Prsevl s theorem then tells us, mong other things, tht π f(x) 2 dx b n 2. (3) π π It is not so hrd to figure out tht f(x) 2 {+ for ll x R}, so the left-hnd side of Eqution (3) becomes π () dx (2π) 2. π π π And b n 2 6/(π 2 (2n ) 2 ), so the right-hnd side of Eqution (3) becomes Putting everything together, we see tht 6 π 2 (2n ) 2 6 π 2 (2n ) π 2 (2n ) 2 π 2 8 (2n ) 2. In.6 (b), we re given the function f(x) e iαx defined nd integrble on (, 2π) nd of period 2π when extended to ll of R. And we showed tht f(x) e iπα sin(πα)/π n einx /(n + α), Pge 5 of 2

6 ssuming α is not n integer. theorem sttes tht For such function nd its complex Fourier series, Prsevl s 2π f(x) 2 dx 2π c n 2. (4) Here, f(x) 2 e iαx 2 e iαx (e iαx ) e iαx (e iαx ), so the left-hnd-side of Eqution (4) becomes 2π () dx (2π). 2π 2π Additionlly, c n 2 e iπα sin(πα) π(n + α) so the right-hnd side of Eqution (4) becomes 2 e iπα 2 sin 2 (πα) π 2 n + α 2 sin2 (πα) π 2 n + α 2, sin 2 (πα) π 2 n + α 2 sin2 (πα) π 2 n + α 2. Therefore! The left-hnd side nd right-hnd side together mke sin2 (πα) π 2 n + α 2 π 2 sin 2 (πα) n + α 2. Note it is importnt here tht α is not n integer, or we re dividing by sin 2 (πα). Trench, 8., # 3, p. 53. Suppose (A, ρ) is metric spce, nd define Show (A, ρ ) is metric spce. Homework 5 ρ (u, v) ρ(u, v) + ρ(u, v). Proof. By definition, (A, ρ ) is metric spce provided it stisfies (i), (ii), nd (iii) below. (i) ρ (u, v) for ll u, v A, nd ρ (u, v) if nd only if u v. Since ρ is metric, ρ(u, v). Then it is lso true tht + ρ(u, v). This implies tht for ll u, v A, ρ ρ(u,v) +ρ(u,v) ; ρ is lso defined, since the denomintor is non-zero. Furthermore, ρ (u, v) ρ(u, v) ρ(u, v), + ρ(u, v) nd since ρ is itself metric, ρ(u, v) if nd only if u v. So ρ (u, v) if nd only if u v. Pge 6 of 2

7 (ii) ρ (u, v) ρ (v, u) for ll u, v A. This follows immeditely from the symmetry of ρ. (iii) ρ (u, v) ρ (u, w) + ρ (w, v) for ll u, v, w A. Note tht ρ (u, v) ρ(u, v) + ρ(u, v) + ρ(u, v). By the tringle inequlity, ρ(u, v) ρ(u, w) + ρ(w, v). Then we hve ρ (u, v) + ρ(u, v) + ρ(u, w) + ρ(w, v) ρ(u, w) + ρ(w, v) + ρ(u, w) + ρ(w, v) ρ(u, w) + ρ(u, w) + ρ(w, v) + ρ(w, v) + ρ(u, w) + ρ(w, v) ρ(u, w) ρ(w, v) + + ρ(u, w) + ρ(w, v) ρ (u, w) + ρ (w, v). We conclude ρ is indeed metric on A. Trench, 8., # 4, p. 53. Let (A, ρ) be metric spce, nd let Homework 6 σ(u, v) ρ(u, v) + ρ(u, v). Let X A. Show tht X is open in (A, ρ) if nd only if X is open in (A, σ). Nottion. For A nd ɛ >, let N ρ ɛ ( ) : { A: ρ (, ) < ɛ} N σ ɛ ( ) : { A: σ (, ) < ɛ}. Proof of Homework 6. LTGBG. Note tht by the previous problem, (A, σ) is metric spce. Clerly the problem is true if X is the empty set so let X. Let X is open in (A, ρ). WTS tht X is open in (A, σ). Fix x X. Since X is open in (A, ρ), there is n ɛ > so tht Nɛ ρ (x ) X. (5) Pge 7 of 2

8 Note tht since if x N σ ɛ (x ) then σ(x, x ) < ɛ +ɛ +ɛ < nd so N σ ɛ (x ) Nɛ ρ (x ) (6) +ɛ ρ(x, x ) σ(x, x ɛ ) σ(x, x ) < +ɛ ɛ +ɛ ɛ. So N σ ɛ (x ) X by (5) nd (6). Thus X is open in (A, σ). +ɛ Let X is open in (A, σ). WTS tht X is open in (A, ρ). Fix x X. Since X is open in (A, σ), there is n ɛ > so tht Nɛ σ (x ) X. (7) Note tht since if x N ρ ɛ (x ) then ρ(x, x ) < ɛ nd so N ρ ɛ (x ) N σ ɛ (x ) (8) σ(x, x ) ρ(x, x ) + ρ(x, x ) < ρ(x, x ) < ɛ. So N ρ ɛ (x ) X by (7) nd (8). Thus X is open in (A, ρ). Remrk: Why is this problem useful? Well note tht σ(a) [, ) lthough ρ(a) cn be [, ). So often it s esier to work with σ rther thn ρ. But the open subsets (nd thus concepts involving convegence, continuity, etc.) of (A, ρ) nd (A, σ) coincide. Homework 7 Trench, 8., # 3, p Let (A, ρ) be metric spce. Let S nd S 2 be subsets of A. Prove: () (S S 2 ) S S 2 (b) S S 2 (S S 2 ). Recll. The following re equivlent for B A.. b is n interior point of B, in nottion, b B. 2. There is n open subset U of A such tht b U B. 3. There is ɛ > so tht N ɛ (b) B. Proof. LTGBG. in () If (S S 2 ) then the clim is trivilly true so ssume tht (S S 2 ). Fix x (S S 2 ). So there exists n ɛ > such tht N ɛ (x) S S 2. Thus N ɛ (x) S nd so x S. Also N ɛ(x) S 2 nd so x S 2. Thus x S S 2. So (S S 2 ) S S 2. Pge 8 of 2

9 in() If S S 2 then the clim is trivlly true so ssume tht S S 2. Fix x S S 2. So x S thus there exist ɛ > such tht N ɛ (x) S. Also, x S 2 thus there exist ɛ 2 > such tht N ɛ2 (x) S 2. Let ɛ : min{ɛ, ɛ 2 }. Then N ɛ (x) N ɛ (x) N ɛ2 (x) S S 2. So x (S S 2 ). So (S S 2 ) S S 2. (b) If S S 2 then the clim is trivlly true so ssume tht S S 2. Fix x S S 2. Thus x S or x S 2. WLOG, ssume x S. So there exists ɛ > such tht N ɛ (x) S. Since S S S 2, we hve N ɛ (x) S S 2. Thus x (S S 2 ). So S S 2 (S S 2 ). Homework 8 Trench, 8., # 25, p Consider the set C[, b] of continuous rel functions on the intervl [, b]. () Show tht is norm on C[, b]. f f(x) dx (9) (b) Show tht the sequence {f n } defined by is Cuchy sequence in (C[, b], ). f n (x) (c) Show tht (C[, b], ) is not complete. ( ) x n b Proof. LTGBG. It is understood, since f C[, b], tht the integrl in (9) is the Riemnn integrl. Recll tht if function f : [, b] R is continuous (or if the set of discontinuities of f is countble, or just tht the set of discontinuities of f is set of Lebesgue mesure zero), then f is Riemnn integrble. () By definition of norm, it suffices to show tht the following clims hold. Clim.: f for ll f C[, b]. Fix f C[, b]. By monotonicity of the integrl, since f(x) for ech x [, b], So Clim. holds. f : f(x) dx dx. Pge 9 of 2

10 Clim.2: If f then f. Let f, i.e., f(x) for ll x [, b]. Then f : f(x) dx dx. So Clim.2 holds. Clim.3: If f then f. We shll grue by counterpositive. Thus ssume tht f C[, b] is not the zero function. We wnt to show tht f >. Since f C[, b] is not the zero function, there is c [, b] so tht f(c). WLOG, f(c) >. thinking lnd: got to be creful in wording to cover cse c or c b. Since f is continuous nd f(c) : 2ɛ >, there is nondegenerte (i.e., hs more thn one point) intervl [, b ] so tht c [, b ] [, b] nd Thus f : f(x) ɛ x [, b ]. f(x) dx So Clim.3 holds. Clim 2. cf c f for ech c R nd f C[, b]. Fix c R nd f C[, b]. Then cf cf(x) dx f(x) dx ɛ (b ) >. c f(x) dx c f(x) dx c f. So Clim 2 holds. Clim 3. f + g f + g for ll f, g C[, b]. Fix f, g C[, b]. By the tringle inequlity on (R, ), we hve tht f(x) + g(x) f(x) + g(x) for ech x [, b]. Thus f + g : Thus Clim 3 holds. So is indeed norm on C[, b]. f(x) + g(x) dx f(x) dx + (b) LTGBG. First let s compute for n n N. f n ( ) x n b dx b ( f(x) + g(x) ) dx g(x) dx f + g. ( ) x n dx b (x )n+ (n + )(b ) n b b n +. () Fix n ɛ >. Pick N N so big tht 2(b ) N+ < ɛ. Then for ny n, m N, f n f m ineq. f n + f m () b n + + b m + b N + + b 2(b ) N + N + < ɛ. Thus {f n } is Cuchy. Pge of 2

11 Comment on (c) Note tht the sequence {f n } from prt (b) is Cuchy sequence in (C[, b], ). Consider the function f : [, b] R given by f(x) for ech x [, b]. Clerly f C[, b]. Also f n f f n b n + n. Thus {f n } converges to f in the -norm. (c) Let s first set some nottion. For f C[, b] let In prt (), we showed tht f L [,b] : f(x) dx. ( ) C[, b], L [,b] is normed (vector) spce. By diltion nd trnsltion, it suffices to find Cuchy sequence {f n } (C[, in ], L [,] tht does not converge, in the L [,]-norm, to function in C[, ]. For n N, define f n C[, ] s follows. Let f n (x) for x [, 2 2n ] nd f n(x) for x [ 2, ] nd f n(x) is liner for x [ 2 2n, 2 ]. Specificlly, drw yourself picture x 2 2n f(x) 2nx n + 2 2n < x < 2 2 x. Let n, k N. As simple picture shows, f n f n+k L [,] 2 2 2n f n (x) f n+k (x) dx re of right with bse 2n 2 2 2n f n (x) dx nd height 4n n,k. Thus {f n } is Cuchy in the L [,] -norm. Assume tht {f n } converges in the L [,]-norm to some function f C[, ]. Thus f n f L [,] : f n (x) f(x) dx n. ) WTF. Clim. f(x) for ech x [ 2, ]. Note tht (f ( ) ) [/2,] ( ) C[ 2 (C[/2,, ]. Since ], L [/2,] to show tht (f ( ) ) [/2,] ( ) L. But [/2,] ) is norm spce, it suffices (f ( ) ) [/2,] ( ) L [/2,] n N /2 f(x) f n (x) dx f(x) f n (x) dx n. So Clim holds. Clim 2. f(x) for ech x [, 2 ). Fix N N with N > nd let I N : [, 2 2N ]. It suffices to show tht f(x) for ech x I N. Pge of 2

12 Note tht f IN C(I N ). Since f IN L (I N ). But ( ) C(I N ), L (I N ) is norm spce, it suffices to show tht f IN L (I N ) 2 2n f(x) dx n N 2 2n f(x) f n (x) dx n. f(x) f n (x) dx So Clim 2 holds. Clim 3 We got ourselves contrdiction. So this f is continuous function on [, ] but it lso stisfies Clim nd Clim 2. Hey, this cnnot be! Thus prt (c) holds. Pge 2 of 2