# ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.

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1 ACM 05: Applied Rel nd Functionl Anlysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov Problem. Riemnn-Lebesgue Theorem. Theorem (G.F.B. Riemnn, H.L. Lebesgue). If f is n integrble function on (, + ), then lim + 0 f(x) cos dx = 0 Proof. We first note tht if the intervl of integrtion were finite [, b] with f(x) on it, then / f(x) cos dx = cos dx = cos(y)dy = O() 0 s 0, since the integrl of the cosine is bounded. We shll reduce the generl cse to this simple cse. First off, due to the integrbility of f on (, + ) nd the following estimtes, f(x) cos ( x ) dx + + b / + f(x) cos f(x) cos ( x ) dx we cn lwys find finite intervl [, b], such tht + f(x) cos dx dx f(x) cos f(x) cos f (x) dx + dx < ε 3 + b dx f (x) dx (by bounding the integrls long the tils ). Furthermore, since f is integrble, there exists sequence of integrble simple functions f n (x), such tht implying tht for N lrge enough, f(x) dx = lim n f N (x) dx f n (x) dx f(x) dx < ε 3

2 We thus hve f(x) cos dx f N (x) cos dx dx f(x) f N (x) cos f(x) f N (x) dx < ε 3 Now, we cn lso find (see proof of Royden s Proposition 3.22 in Homework ) set A with µ([, b]\a) < δ, for some δ > 0, nd step function g(x), such tht g(x) = f N (x) on A with f N (x) dx g(x) dx < ε 3 But g(x) is step function, so there exists finite number of points x j, j = 0,..., m with = x 0 < x <... < x m = b, such tht g(x) const = c j on [x j, x j ]. Therefore g(x) cos dx = m x j j= x j c j cos dx = m ( ) x j c j sin 0 j= x j s 0. Since the bove mnipultions gurntee tht + f(x) cos dx g(x) cos dx < ε for n rbitrrily smll ε, we re done. Problem 2. Lx Equivlence Theorem: Solution Opertor. For the initil vlue problem du(t) dt = Lu(t), 0 t T u(0) = u 0, () nd the norm V in the spce V u(t), we denote S(t) the solution opertor: if u(t) is the solution of (), then u(t) = S(t)u 0, u 0 V 0. We hve lim t 0 (u(t + t) u(t) Lu(t)) t = 0 (2). From the continuous dependence property we hve sup S(t)(u 0 u 2 (u 0) C 0 0 u 2 V 0) nd 0 t T sup S(t)u 0 V C 0 u 0 V u 0 V 0 0 t T This implies tht S(t) C 0, so S(t) is bounded on V 0. Since V 0 is dense in V, the required extension of S(t) on the whole spce V exists nd is unique by the extension theorem for opertors. 2 V

3 b. (continuity of generlized solutions). Choose sequence {u 0,n } V 0 which converges to u 0 in V : u 0,n u 0 V 0 s n Fix t 0 [0, T ] nd let t [0, T ]. We write u(t) u(t 0 ) = S(t)u 0 S(t 0 )u 0 = S(t)(u 0 u 0,n ) + (S(t) S(t 0 ))u 0,n S(t 0 )(u 0 u 0,n ) In estimting the norm of this difference, we use the fct tht S(t) C 0 in the first nd the lst terms of the right-hnd side: u(t) u(t 0 ) V 2C 0 u 0,n u 0 V + (S(t) S(t 0 ))u 0,n V Given n ε > 0 we choose sufficiently lrge n, such tht 2C 0 u 0,n u 0 V < ε 2 For this n we use the definition of the solution (2) to come up with δ > 0, such tht for t t 0 < δ, (S(t) S(t 0 ))u 0,n V < ε 2 Then for t [0, T ] with t t 0 < δ,we hve u(t) u(t 0 ) V < ε, so tht u(t) is continuous. c. (semi-group property). The solution to the problem () is u(t) = S(t)u 0. We hve u(s) = = S(s)u 0 nd S(t)u(s) is the solution of the differentil eqution on [s, T ] with the initil condition u(s) imposed t time s. By the uniqueness of solution, i.e., S(t)u(s) = u(t + s) S(t)S(s)u 0 = S(t + s)u 0, nd therefore S(t + s) = S(t)S(s), s required. Problem 3. Lx Equivlence Theorem: Proof. Recll the definitions. Definition. A difference method is one-prmeter fmily of opertors C( t) : V V, nd there exists t 0 > 0, such tht t (0, t 0 ] C( t) c (C is uniformly bounded). The pproximte solution is u t (m t) = C( t) m, m =, 2,... Definition 2. A difference method is consistent if V c V, dense subspce of V, such tht u 0 V c for the corresponding solution of the initil vlue problem (), we hve u(t + t) C( t)u(t) lim = 0 t 0 t uniformly on [0, T ]. Definition 3. A difference method is convergent if for ny fixed t [0, T ] nd ny u 0 V we hve lim t i 0 (C( t i) m i S(t))u 0 = 0 3

4 where the two sequences: {m i }, of integers nd { t i }, of step sizes re such tht lim i m i t i = t. Definition 4. A difference method is stble if the opertors {C( t) m 0 < t t0, m t T } re uniformly bounded, i.e., there exists constnt M 0 > 0 such tht C( t) m V V M 0 m : m t T, t t 0 Theorem (P.D. Lx). Suppose the initil vlue problem () is well-posed. Then for consistent difference method, stbility is equivlent to cnvergence. Proof. (Stbility = Convergence). Consider the error C( t) m u 0 u(t) = m j= C( t) j [C( t)u((m j) t) u((m j) t)] + u(m t) u(t) We use the hint: first consider u 0 V c. Then since the method is stble, C( t) m u 0 u(t) M 0 m t sup C( t)u(t) u(t + t) t t + u(m t) u(t) The sup goes to 0 by consistency, while continuity implies u(m t) u(t) 0, so convergence is estblished in this cse. Now consider the generl cse when u V nd not necessrily in V c. Since V c is dense in V, we hve sequence {u 0,n } V 0 such tht u 0,n u 0 in V. Writing (recll the construction in Problem 2) C( t) m u 0 u(t) = C( t) m (u 0 u 0,n ) + [C( t) m S(t)]u 0,n S(t)(u 0 u 0,n ), we obtin, by tringle inequlity, C( t) m u 0 u(t) C( t) m (u 0 u 0,n ) + [C( t) m S(t)]u 0,n + S(t)(u 0 u 0,n ) Since the initil vlue problem () is well-posed nd the method is stble, C( t) m u 0 u(t) c u 0 u 0,n + [C( t) m S(t)]u 0,n Given ny ε > 0, we cn find sufficiently lrge n, so tht For this n, tke t sufficiently smll, so tht c u 0 u 0,n < ε 2 [C( t) m S(t)]u 0,n ε 2 t smll, m t t < t nd we hve convergence. (Convergence = Stbility). We gin use the hint nd invoke proof by contrdiction. Suppose the method is not stble. Then there re sequences { t k } nd {m k } such tht m k t k T nd lim C( t k) m k = k 4

5 Since t k t, we my ssume tht the sequence { t k } is convergent. (Techniclly, we only know tht, since { t k } is bounded, it hs convergent subsequence, but we cn ssume here, without loss of generlity, tht it is the whole sequence.) First ssume tht {m k } is bounded. Then sup k C( t k ) m k sup C( t k ) m k < k which is contrdiction. Thus necessrily m k nd t k 0 s k. By the convergence of the method, sup C( t k ) m k u 0 < k The principle of uniform boundedness then implies lim C( t k) m k < k u 0 V contrdicting the ssumption tht the method is not stble, so the theorem is proved completely. Problem 4. Het Eqution: Discretiztion. u t = u xx + f(t, x) in [0, π] [0, T ] u(0, t) = u(π, t) = 0 0 t T u(x, 0) = u 0 (x) 0 x π (3). Let V = C 0 [0, π] = {v C[0, π], v(0) = v(π) = 0} with the norm v C0 = mx 0 x π v(x) Since both continuity nd the vlues t the boundries re preserved under liner combintions, V is obviously liner spce. To prove completeness, consider Cuchy (fundmentl) sequence in V : for ny ε > 0 N = N(ε) such tht for ny n, m > N we hve v n (x) v m (x) = mx 0 x π v n(x) v m (x) < ε By the uniform convergence criterion, the condition bove implies tht the Cuchy sequence converges uniformly on [0, π] to function v(x). But uniform limit of continuous functions is continuous, so v C[0, π]. By continuity, since v n (0) = v n (π) = 0 n, it follows tht lso v(0) = v(π) = 0, so v V, nd V is complete liner normed spce, i.e., Bnch spce. Now tke V 0, the spce of liner combintions of sines. Obviously ny v V 0 is continuous nd vnishes t the boundries (0 nd π). Therefore, V 0 V. A well-known result from Fourier nlysis sys tht ny continuous function on [0, π], vnishing t the boundries, cn be represented s Fourier sine series, v(x) = k sin(kx) Therefore for ny function v V we cn find sequence {v n } V 0, v n (x) = k sin(kx) so tht v n v. Thus the closure of V 0, V 0 V, so V 0 is dense in V. 5

6 b. Define C( t, r)v(x) = ( 2r)v(x) + r(v(x + x) + v(x x)) Here h t = t, h x = x = t/r. If x ± x / [0, π], then we tke the odd 2π-periodic extension of v. Obviously, C( t) : V V is liner opertor. Moreover, so C( t)v V = ( 2r)v(x) + r(v(x + x) + v(x x)) V ( 2r)v V + rv V + rv V = ( 2r + 2r) v V nd the fmily {C( t)} is uniformly bounded. C( t) 2r + 2r, (4) For consistency nlysis, tke V c = V 0. For ny initil condition u 0 (x) V c we then obtin smooth solution s Fourier sine series u(t, x) = b k e k2t sin(kx) + F k (t) sin(kx) Here u 0 (x) = n b k sin(kx) nd F k (t) re due to the source term. (It cn be shown tht F k looks like F k (t) = t 0 f k (τ)e k2 (τ t) dτ where f(t, x) = n f k (t) sin(kx).) Now use Tylor series expnsion in (x, t) with reminder to write C( t)u(x, t) u(x, t + t) = ( 2r)u(x, t) + r(u(x + x, t) + u(x x, t)) u(x, t + t) = ( 2r)u(x, t) + r(2u(x, t) + u xx (x, t)( x) 2 ) + + r 4! ( x)4 (u xxxx (x + θ x, t) + u xxxx (x θ 2 x, t)) u(x, t) u t (x, t) t 2 u tt(x, t + θ 3 t)( t) 2 = t [ u t (x, t) + r ( x)2 u xx (x, t) + t +r ( x)2 ( x) 2 t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) ] 2 u tt(x, t + θ 3 t) t Since x = t/r, (r/ t)( x) 2 =, nd the first two terms combine to f(x, t) t. For consistency, the remining terms must go to zero fster thn t. We hve r ( x)2 ( x) 2 t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) 2 u tt(x, t + θ 3 t) t = [ x x = t t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) ] 2 u tt(x, t + θ 3 t) Now it is indeed cler tht the h x /h t = x/ t c is suffiecient to bound the term in the squre brckets, so tht (C( t)u(x, t) u(x, t + t) f(x, t)) t c t 0 uniformly, i.e., the difference method bsed on C( t) is consistent. 6

7 c. We prove tht r /2 is the necessry nd sufficient condition for stbility nd then invoke Lx s theorem to get the sme result for convergence. We hve estblished erlier (4) tht C( t) 2r +2r. It immeditely follows tht if r /2, then C( t), so tht C( t) m, i.e., uniformly bounded, so the method is stble. For the other direction (necessity), we prove tht for r > 0 (4) is ctully n equlity, i.e., tht mx C( t)v V = = ( 2r + 2r) v mx V is chieved. We just need to present the function v mx t which this mximum is ttined. For r /2, the mximum is obviously ttined on v mx. For 0 < r < /2, let x j = (j/π) x be the sptil discretiztion. Define piecewise liner function by { 0, j = 0 or j = n v mx (x j ) = ( ) j, otherwise with stright line segments connecting the vertices. (This is no other thn the fmilir swtooth shpe.) Obviously, the supremum norm of this function is, nd since its vlues on the lttice re lternting + s nd - s, the norm C( t)v mx = ( 2r)(±)+ +2r( ) = 4r = ( 2r + 2r) v V. Thus the inequlity in (4) is in fct n equlity. It is now esy to estblish the required result. If the method is stble, then C( t) m is uniformly bounded. Obviously, if C( t) >, then C( t) m, so the uniform bound cn exist only if C( t), i.e., 2r + 2r r /2. Thus r /2 is necessry nd sufficient for stbility of the difference method. Since the method is consistent, by the Lx equivlence theorem it is lso necessry nd sufficient for convergence. Problem 5. Geometric Series Theorem.. Geometric Series Theorem. Let V be Bnch spce nd L : V V with L <. Then the opertor I L is bijection on V, its inverse is bounded opertor nd (I L) L. (5) Proof. Define the following sequence of opertors on V, M n = n k=0 L k, n 0. For p, n+p n+p M n+p M n = L k n+p L k L k Since L <, we hve k=n+ k=n+ M n+p M n L n+ L k=n+ so sup M n+p M n 0 s n, nd {M n } is Cuchy sequence in L(V ), the spce of p liner opertors on V. But this spce is complete, so there exists n opertor M : V V, such tht M n M 0 s n. Now, (I L)M n = M n LM n = 7 L k k=0 L k+ = I L n+ k=0

8 Similrly, M n (I L) = I L n+, so (I L)M n = M n (I L) = I L n+. Letting n, we get (I L)M = M(I L) = I proving tht I L is bijection nd M = (I L) = lim L k = L n n k=0 n=0 Furthermore, M n k=0 Tking the limit s n, we obtin L k = L n+ L L M = (I L) L proving (5) nd concluding the proof of the geometric series theorem. We cn puse for second here to recognize tht wht we hve just obtined is just nother generliztion of the geometric series formul: if 0 < q <, then the sum of the geometric series + q + q q n +... = q We lso know from complex nlysis tht if z C is lso inside the unit disc (i.e., z < ), then the series z n n=0 converges uniformly to the function ( z), which is bounded on the unit disc. Now we lso know tht the sme holds for liner opertors cting on Bnch spce! b. The liner integrl eqution of the second kind, λu(x) k(x, y)u(y) dy = f(x) (6) with λ 0, k(x, y) C([, b] [, b]) nd f C[, b], cn be written in the form (λi K)u = f, where K is the integrl opertor generted by the kernel k(, ). Equivlently, By the geometric series theorem, if (I L)u = λ f, L = λ K L = λ K < then the inverse opertor (I L) exists nd (I L) L 8

9 Since K = mx x b k(x, y) dy we thus hve tht, s long s K < λ, (λi K) exists nd Therefore, if only (λi K) λ K mx k(x, y) dy < λ x b for ny f C[, b], there is unique solution u C[, b] to the integrl eqution of the second kind (6). Problem 6. Different Bnch Spces with the Sme Norm.. Consider the spce l of bounded rel sequences with the (i.e., supremum) norm. All xioms of the norm re trivilly verified. Also, liner combintions preserve boundedness, so l is liner normed spce. Consider Cuchy sequence {x n } l : ε > 0 N = N(ε) such tht n, m > N sup x n,k x m,k < ε k By the uniform convergence criterion x n x s n, so for n sufficiently lrge, sup x n,k x k < ε k so sup k x k sup k x n,k + ε for n rbitrry ε > 0, so x is lso bounded, proving tht l is complete, nd therefore Bnch spce. b. Following the sme line of rguments s bove, we estblish tht c is normed liner spce, nd if {x n } c is Cuchy sequence, then x n x with sup x n,k x k < ε k 2 for n rbitrry ε > 0 nd ll n > N (ε). Since {x n } c, it consists of converging sequences, so for ll k > N 2 (ε), we hve x n,k C < ε/2. Therefore, for ll k > mx{n, N 2 }, we hve x k C ε so x c nd c is Bnch spce. The exct sme rgument pplies to c 0. c. First consider c 0, the spce of sequences converging to 0, with the supremum norm. We show tht the spce of bounded liner functionls on c 0 is isomorphic to l, the spce of bsolutely summble sequences (i.e., those with x k < ), with the norm f = 9 f k

10 To prove this, first note tht for ny element f = (f, f 2,..., f n,...) l, the formul f(x) = x k f k (7) defines functionl f on the spce c 0, which is clerly liner. Moreover, so f(x) x f k f f (8) Consider the following vectors in c 0 : e = (, 0,..., 0,...), e 2 = (0,,..., 0,...), etc., nd let x (n) f k = f k e k (if f k = 0, then we set f k / f k = 0). Then x (n) c 0 nd x (n). Moreover, f(x (n) ) = f k f k f(e k ) = f k f k = f s n. This implies f f lim x (n) f. Together with (8), the opposite n inequlity estblished erlier, this implies f = f. Therefore the mpping crrying f into f is norm-preserving mpping from the spce of liner functionls on c 0 to l. It remins to prove tht every functionl hs unique representtion like (7), where f l. Let x = (x,..., x k,...) c 0. Then x = x k e k where the series on the right converges in c 0 to the element x since n x x k = sup x k 0 k>n s n. Since the functionl f is continuous, ( ) f(x) = f lim x k e k = lim n n f(x k e k ) = x k f(ek ) so f hs unique representtion of the form (7). We only need to estblish tht f(e k ) < (9) For ny n, let x (n) = f(e k ) f(e k ) e k 0

11 Then x (n) c 0 nd x (n), so f(e f(e k ) k ) = f(e k ) f(e k ) = f( x (n) ) f < Since n cn be mde rbitrrily lrge, this proves (9). Thus the spce of bounded liner functionls on c 0 is isomorphic to l : x c 0 f(x) = f k x k with f k = f(e k ) (e k defined bove) nd f k <. Now consider the more generl spce c of ll bounded convergent sequences. The obvious ide is to notice tht for ny x c, converging to x 0, x = x x 0 c 0, so we should be ble to use the result for functionls on c 0 for the cse of c. We mke this ide precise in the following wy. Consider the vectors e = (, 0, 0,..., 0,...), e 2 = (0,, 0,..., 0,...), etc., for ll k nd lso e 0 = (,,,...). Then for {e n } n 0 c, nd for ny liner functionl f on c we cn define f n = f(e n ), n = 0,,... We use these vlues to represent ll bounded liner functionls on c. We first prove tht the set x (n) = λ i e i, i=0 λ i Q is dense subset of c. Indeed, if = (, 2,..., k,...) c, then 0 : lim k = 0 <. k Then for ny ε > 0 we cn find sufficiently lrge n, so tht 0e 0 ( i 0 )e i < ε 2 i=0 since the sequence ( 0 ) k 0 s k. Now, since i R, there exist rtionl numbers λ i Q, such tht λ i ( i 0 ) < ε/2 i+, so tht nd therefore λ i e i ( i 0 )e i < ε 2 i= i= x (n) = 0e 0 λ i e i < ε i= so the x (n) s re indeed dense in c. Moreover, since λ i Q, {x (n) } is countble dense subset of c. For ny x {x (n) } nd ny liner functionl f we cn represent f(x) = f 0 λ 0 + λ i f i i= where f i = f(e i ). This functionl will be bounded s long s f i <. Since x (n) is dense in c, then by the Hhn-Bnch theorem we cn extend the liner functionl f to

12 bounded liner functionl on the whole spce c. Since this extension is unique, it hs to look like f(x) = f(x, x 2,..., x k,...) = f 0 x 0 + f i x i where, s bove, f i = f(e i ). In other words, we hve proved tht c = c 0 R, so tht the spce of bounded liner functionls on c is isomorphic to l R, just s we hd expected. References [] Royden H.L., Rel Anlysis, 3rd ed., Prentice-Hll, 988. [2] Kolmogorov A.N., Fomin S.V., Introductory Rel Anlysis, Dover, 975. [3] Atkinson K., Hn W., Theoreticl Numericl Anlysis: Functionl Anlysis Frmework, Springer, 200. [4] Zorich V.A., Mthemticl Anlysis (Clculus), 2nd ed., MCNMO, 998 (in Russin). i= 2

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