HOMEWORK 1 1. P 229. # Then ; then we have. goes to 1 uniformly as n goes to infinity. Therefore. e x2 /n dx = = sin x.

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1 HOMEWORK 1 SHUANGLIN SHAO 1. P 229. # Proof. (). Let f (x) x The x 66 + x3 f x 33 s goes to ifiity. We estimte the differece, f (x) x 33 5 x x 66 5, for ll x [1, 3], which goes to zero s goes to ifiity. Thus for y ɛ >, there exists N N such tht for y N, f (x) x 33 5 < ɛ. This proves tht f coverges to x 33 uiformly o [1, 3] s goes to ifiity. 3 x x 3 dx x 33 dx x (b). For x 2, e x2 e x 1; the we hve 2 1 e x2 1 e 4 1. The sequece {e 4 1} goes to zero s goes to ifiity. By similr proof i prt(), e x2 goes to 1 uiformly s goes to ifiity. 2 e x2 / dx 2 e dx 2. (c). For x 3, si x + x + 1 x + 1 si x si x + x x + 1 si 3, 1

2 if 1. The sequece 3 goes to zero s goes to ifiity. By the proof i prt (), si x + x + 1 coverges to zero uiformly o [, 3]. Thus 3 si x 3 + x + 1dx 14 x + 1dx P229. # Proof. If f coverges to f uiformly o E, the by the uiform Cuchy criterio, for y 1 > ɛ >, there exists N N such tht for y m, N, Let m N + 1. The f m (x) f (x) ɛ. f (x) f N+1 (x) + ɛ f N+1 (x) + 1. Ech f is bouded for 1 N + 1. Thus {f } is uiformly bouded o E. O the other hd, to show tht f is bouded, we use the uiform covergece of f to f. We omit the detils. 3. P23. # Proof. The sequece {f } coverges to f uiformly o E: for y ɛ >, there exists N N such tht for y N, f (x) f(x) < ɛ/3. For f N d for the sme ɛ >, there exists δ > such tht for x y < δ, f N (x) f N (y) ɛ/3. f(x) f(y) f(x) f N (x) + f N (x) f N (y) + f N (y) f(y) f(x) f N (x) + f N (x) f N (y) + f N (y) f(y) < 3 ɛ 3 ɛ. Thus f is uiformly cotiuous o E. 2

3 4. P23. # Proof. Let f (x) (1 + x ) e x. For x, f () coverges to 1 s goes to ifiity. Ech f is differetible o (, b). Next we show tht f (x) coverges uiformly o (, b): f (x) (1 + x ) 1 e x (1 + x ) e x (1 + x ) 1 (1 1 x )e x x (1 + x ) 1 e x x b, sice (1 + x ) is icresig d coverges to e x. f (x) coverges uiformly o (, b). By Theorem 7.12, f coverges uiformly o (, b). Thus (1 + x ) e x dx 1dx b. 5. P23. # Proof. The sequece f coverges to f uiformly o E: for y ɛ >, there exists N N such tht for > N, f (x) f(x) < ɛ/2, i.e., ɛ/2 + f(x) < f (x) < f(x) + ɛ/2. We split the terms s follows. f 1 (x) + f 2 (x) + + f (x) f(x) This implies f 1(x) + + f N 1 (x) g (x) + f N(x) + + f (x) f(x). g (x) N f(x)+nɛ 2 ɛ/2 + f N(x) + + f (x) f(x) f 1 (x) + f 2 (x) + + f (x) f(x) g (x) N f(x) Nɛ 2 +ɛ/2. Sice ech f is bouded d f coverges uiformly to f(x), f is bouded. Moreover, g (x) N Nɛ f(x) + 2 g (x) N 3 f(x) Nɛ 2.

4 for the sme ɛ >, there exists N 1 N such tht for N 1, ɛ 2 < g (x) N Nɛ f(x) + 2 < ɛ/2, ɛ/2 < g (x) N Nɛ f(x) 2 < ɛ/2. We tke mx{n, N 1 }. For N, we hve ɛ < f 1 (x) + f 2 (x) + + f (x) f(x) < ɛ. f 1 (x) + f 2 (x) + + f (x) f(x). 6. P23. # Proof. The sequece f coverges to f uiformly o E d ech f is itegrble d hece is bouded; thus f is bouded by M >, i.e., f < M for some M >. O the other hd, we hve b 1, f (x) f(x), uiformly o E, f (x)dx The lst oe is implied by the secod oe. Note tht b is icresig. So for y ɛ >, there exists N N such tht for, m N b 1 < ɛ 3M, f m (x) f(x) < ɛ, for ll x E, 3 f (x)dx f m (x)dx < ɛ 3. Tkig m N. The f (x)dx f(x)dx f (x)dx f N (x)dx + f N (x)dx f(x)dx (f (x) f N (x))dx + (f N (x) f(x))dx + f(x)dx b f (x) f N (x) dx + f N (x) f(x) dx + M(1 b ) < ɛ 3 3 ɛ. 4

5 To coclude, f (x)dx This completes the proof. 7. P235. # Proof. (). Let x be poit i the bouded itervl i R. The for sufficietly lrge k, si x x. By the compriso test, the series k1 si x coverges bsolutely d hece coverges t x. O the other hd, let f k (x) si x, f k (x) 1 cos x. f k (x) 1 for ll x i the bouded itervl. By the Weierstrss test, f k (x) coverges uiformly. Thus k1 f k(x) coverges. (b). For ll x i this closed subitervl of (, ), e kx e k, where is the left edpoit of this itervl. We c >. Ideed, the itervl is cotied i (, ) d so. If, the itervl is ope o the left edpoit. Thus >. by the Weierstrss test, k1 e kx coverges uiformly. 8. P236. # Proof. Sice cos kx 1, the series π 2 1 f(x)dx k1 k1 k1 k1 k1 π 1 2 cos kxdx 1 π 2 ( 1) k (2k + 1) 3. 5 cos kx coverges uiformly o R. π/2 1 k (si kx) dx cos kxdx

6 9. P236. # Proof. Let M > be the boud of g 1 o E d the prtil sum S (x) of k1 f k(x). Let S(x) be the uiform it of the series k1 f k(x). The series f k1 f k(x) coverges to S(x) uiformly o E: for y ɛ >, there exists N N such tht > m > N, S (x) S m 1 (x) ɛ M, d S m 1 (x) S(x) < ɛ 3M, for ll x E. The f k (x)g k (x) km 1 S (x)g (x) S m 1 (x)h m (x) + S k (x) (g k (x) g k+1 (x)) km 1 (S (x) S m 1 (x))g (x) + S m 1 (x)(g (x) g m (x)) + S k (x) (g k (x) g k+1 (x)) km km km (S (x) S m 1 (x))g (x) + (S m 1 (x) S(x))(g (x) g m (x)) ( S)(g k (x) g k+1 (x)) + S k (x) (g k (x) g k+1 (x)) (S (x) S m 1 (x))g (x) + (S m 1 (x) S(x))(g (x) g m (x)) 1 + (S k (x) S(x)) (g k (x) g k+1 (x)) km < M S (x) S m 1 (x) + ɛ 3M M + ɛ 3M M < M ɛ 3M + ɛ 3M M + ɛ 3M M ɛ, for ll x E. By the Uiform Cuchy Criterio, k1 f k(x)g k (x) coverges uiformly o E. 1. P236. # Proof. Let g (x) ( k1 f k(x)) 1. Sice fk (x) f k+1 (x), g (x) (f (x)) 1 1 f (x) 2f (x) 6

7 for lrge eough becuse 1 1. Sice f coverges to f uiformly o [, b], g coverges uiformly o [, b]. g (x)dx exists d equls g (x)dx. Sice 1 N, So g (x) (1 + ɛ)f (x) d hece 1, for y ɛ >, there exists N N such tht for Tkig its o both sides, 1 < 1 + ɛ. g (x)dx (1 + ɛ) g (x)dx (1 + ɛ) Sice ɛ > is rbitrry, g (x)dx Sice f k+1 f k (x) o [, b], ( k1 f k (x) ) 1 ( N 1 k1 f k (x) + ( fn(x) kn f (x)dx f(x)dx kn ) 1 ( N) 1 fn (x). tkig, g (x)dx Tkig N, g (x)dx f k (x) f N (x)dx. ) 1 g (x)dx 7

8 Deprtmet of Mthemtics, KU, Lwrece, KS 6645 E-mil ddress: 8