MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous rel-vlued function on I), nd let L 1 (I) denote the completion of C(I; R) with respect to. 1. (1) Show tht the Riemnn integrl : C(I; R) R, f b f() d, etends to bounded liner mp : L 1 (I) R (which would be clled the Lebesgue integrl). () Suppose f L 1 (I), i.e. f is the equivlence clss of Cuchy sequence {f n } of continuous functions. We sy tht f if there eists Cuchy sequence {f n } representing f such tht f n or ll n. Show tht if f nd g then f + g, nd if c R, c then cf. Show lso tht if f then f. (Thus, it mkes sense to sy whether n element of L 1 (I) is non-negtive on I. We lso sy f g if f g, so it mkes sense to sy tht n element of L 1 (I) is greter thn nother element of L 1 (I).) (3) Show tht if f nd f then f. Use this to conclude tht is prtil order on L 1 (I), i.e. f g nd g h implies f h, nd f g nd g f implies f g. (4) Suppose tht {f (k) } k1 is sequence in L1 (I) nd f (k) f in L 1 (I). Show tht if f (k) for ll k then f. (Hint: Consider f (k) C(I; R) such tht f (k) nd f (k) is close to f (k).) (5) Suppose I. Show tht there is no continuous liner mp E : L 1 (I) R such tht E (f) f() for ll f C(I; R). Tht is, evluting continuous functions t cnnot be etended in resonble mnner to L 1 (I), i.e. elements of L 1 (I) do not hve vlues t ny given point. (Hint: If such mp eisted, nd f L 1 (I) were represented by Cuchy sequence {f n }, wht would E (f) be? Now find different Cuchy sequences representing the sme f.) Solution. (1) Recll the B.L.T. theorem: Suppose T is bounded liner trnsformtion from normed vector spce (V 1,. 1 ) to complete liner spce (V,. ). Then T cn be uniquely etended to bounded liner trnsformtion (with the sme bound), T, from the completion of V 1 to (V,. ). Note tht (C(I; R),. 1 ) is normed vector spce with completion L 1 (I) nd R is complete vector spce. We only need to show tht the Riemnn integrl : C(I; R) R is bounded liner trnsformtion, then we cn pply the B.L.T. theorem bove to etend it to bounded liner mp on L 1 (R). Linerity of the Riemnn integrl is well-known result from elementry nlysis. For
boundedness, simply note tht f f f 1. So f is bounded liner trnsformtion with 1 (ctully equlity holds). () Suppose f nd g. By definition, there re Cuchy sequences {f n } nd {g n } representing f nd g respectively with f n, g n for ll n. Then {f n + g n } is non-negtive Cuchy sequence representing f + g. Thus, f + g. On the other hnd, if c R nd c, then {cf n } is non-negtive Cuchy sequence representing cf, hence cf. Now, ssume tht f nd let {f n } be non-negtive Cuchy sequence representing f. By definition of the Lebesgue integrl in prt (1), we hve f lim f n. Since ech f n is non-negtive, f n. This implies tht f. (3) Assume tht f nd f. Then there eist two non-negtive Cuchy sequences {f n } nd {f n} for f nd f respectively. Note tht { f n} is nonpositive Cuchy sequence representing f, therefore, lim f n +f n 1. Since both f n nd f n re non-negtive, f n + f n 1 f n 1 + f n 1. Hence lim f n 1 nd therefore {f n } lso represents. We conclude tht f. Net, we show tht is prtil order on L 1 (I). For trnsitivity, suppose f g nd g h. Then f g nd g h. Since ddition preserves non-negtivity (see ()), f h (f g) + (g h), i.e. f h. For ntisymmetry, ssume f g nd g f, then f g nd g f. Apply the result we hve just proved, we hve f g, in other words, f g. (4) Let {f (k) } k1 be sequence in L1 (I) with f (k) f in L 1 (I). Suppose further tht f (k) for ll k. We wnt to prove tht f. For ech k, let {f n (k) } be non-negtive Cuchy sequence in C(I; R) representing f (k). Note tht f n (k) f (k) in L 1 (I) s n goes to infinity, we cn choose n(k) such tht f (k) n(k) f (k) 1 1. k Therefore, {f (k) n(k) } is non-negtive Cuchy sequence representing f nd f. (5) Suppose I. Assume the contrry tht there eists continuous liner mp E : L 1 (I) R such tht E (f) f() for ll f C(I; R). Suppose f L 1 (I) is represented by Cuchy sequence {f n }. By continuity of E, E (f) lim E (f n ) lim f n (). Consider the Cuchy sequence of functions f n C(I; R) defined by { n( ) + 1, [, + 1 f n () ] n., otherwise Then it is not hrd to see tht f n converges to in L 1 (I) but f n () 1 for ll n. Therefore, lim f n () 1, which is contrdiction. Problem. Suppose tht V is n inner product spce, D is liner subspce, nd A : D V is liner opertor (not necessrily continuous). We sy tht A is symmetric if (Au, v) (u, Av) for ll u, v D. We sy tht non-zero vector v D is n eigenvector of A with eigenvlue λ C if Av λv.
(1) Show tht if A : D V is symmetric, then ll eigenvlues of A re rel, nd ll eigenvectors corresponding to different eigenvlues re orthogonl, i.e. Av λv, Aw µw, λ µ implies (v, w). () Let A 1 d on D {f i d C1 ([, π]) : f() f(π)}, with V C([, π]). Show tht A is symmetric, nd the functions e in re orthogonl to ech other on [, π]. (3) Let V C([, b]), nd let D be subspce of C ([, b]). Under wht conditions on D is A, given by Af f, symmetric? (Hint: Clculte (Af, g) (f, Ag).) (4) Show tht the functions sin n, n 1 integer, re orthogonl to ech other on [, π]. (5) Show tht the functions sin(n + 1 ), n, 1,,... re orthogonl to ech other on [, π]. 3 Solution. Note tht we dopt the convention tht n inner product is conjugte liner in the second slot. (1) Suppose A : D V is symmetric nd λ is n eigenvlue of A with n eigenvector v D, i.e. Av λv nd v. By symmetry of A, we hve λ(v, v) (λv, v) (Av, v) (v, Av) (v, λv) λ(v, v). Since (v, v), dividing through gives λ λ. Hence λ is rel. All eigenvlues of symmetric opertor is rel. Suppose λ, µ re two distinct eigenvlues of A nd Av λv, Aw µw. We hve to show tht v nd w re orthogonl, i.e. (v, w). We hve lredy proved tht λ nd µ re rel. Note tht by symmetry gin, λ(v, w) (λv, w) (Av, w) (v, Aw) (v, µw) µ(v, w) µ(v, w). Therefore, (λ µ)(v, w) nd λ µ, hence (v, w). () Note tht for ny f, g V, we define (f, g) f()g() d to be the inner product on V. Note tht D is liner subspce of V nd for ll f, g D, (Af, g) ( 1 df i d, g) 1 df i d gd 1 i fg π 1 f dg d (f, Ag), i d where the first term drops out becuse f, g D. Thus, A is symmetric. Since A(e in ) 1 d i d (ein ) ne in, e in is n eigenvector corresponding to the eigenvlue n. By the result of prt (1), e in re orthogonl to ech other on [, π].
4 (3) Following the hint: (Af, g) (f, Ag) ( f, g) (f, g ) (f, g) + (f, g ) b f gd + b b b f g f g b fg d f g d + fg b fg f g b Hence, in order to gurntee tht A is symmetric, we need f()g () f ()g() f(b)g (b) f (b)g(b) for ll f, g D. (4) Note tht sin n re smooth nd A(sin n) (sin n) n sin n. Let f() sin n nd g() sin m, where n, m re positive integers. Then f()g () f ()g() f(π)g (π) f (π)g(π) since f() g() f(π) g(π). Therefore, sin n re eigenvectors of the symmetric opertor A with respect to the eigenvlue n. By prt (1), they re orthogonl on [, π]. (5) Similrly sin(n + 1) re smooth nd A(sin(n + 1 (n + 1 ) ) sin(n + 1 ). Let f() sin(n + 1) nd g() sin(m + 1). Then f () (n + 1) cos(n + 1) nd g () (m + 1) cos(m + 1). Moreover, f()g () f ()g() since f() g(), nd f(π)g (π) f (π)g(π) since f (π) g (π). Therefore, they re orthogonl to ech other on [, π] by prt (1). Problem 3. (cf. Tylor 3.1.5, 3.1.6) Suppose f is π-periodic C 1 function on R, i.e. f C 1 p([, π]). Let c n (f, e in ), b n (f, e in ), where (f, g) (π) 1 f() g() d. (1) Prove tht b n < nd conclude tht n c n <. () Prove tht c n <. (3) Prove tht M n M c ne in is uniformly convergent s M. (4) Let S N (f)(θ) N c ne inθ. Prove tht (S N f)(θ) 1 π The π-periodic function D N () sin(n + 1 ) sin(/) is known s the Dirichlet kernel. f(θ + ) sin(n + 1 ) sin(/) e in d.
(5) Using tht 1 π D π N () d 1, show tht for f Cp([, 1 π]) ( ) 1 f(θ) S N f(θ) sin(n + ), g θ, 5 g θ () f(θ+) f(θ) sin(/), g θ C ([, π]). (6) Using tht sin(n + 1 ), N, 1,,... is n orthonorml set in L ([, π]), show tht S N f(θ) f(θ), nd conclude tht the Fourier series of f C 1 p([, π]) converges uniformly to f. Solution. (1) We know tht e in / π is n orthonorml set in L ([, π]), nd we know tht f C ([, π]), so f L ([, π]), so we hve tht > f (e in / π, f ) b n. Moreover, b n 1 π e in f ()d in π e in f()d inc n, so > b n inc n n c n. () Write c n 1 n nc n, then Cuchy-Schwrz gives: ( ) 1/ 1 ( ) 1/ cn n c n n <. (3) C ([, π]) is Bnch spce, so it is enough to show tht c n e in / π is bsolutely summble, but cn e in / π c n e in / π 1 π cn <, so the sequence is summble nd thus converges in C, i.e. converges uniformly. (4) First observe tht e in e in n e in ei(n+1) e in e i 1 ei(n+ 1 ) e i(n+ 1 ) sin(n + 1). e i/ e i/ sin(/)
6 Then, using the bove, nd the substitution + θ, S N (f)(θ) c n e inθ / π ( 1 π 1 π 1 π ) e in f()d e inθ / π π e in( θ) f()d f() sin(n + 1 )( θ) d sin( θ)/ f(θ + ) sin(n + 1) d. sin / (5) First we show tht the function g θ C ([, π]). This is clerly C, ecept possibly t the endpoints. Using the reltion tht lim 1, we see tht g θ () lim f(θ + ) f(θ) sin(/) lim f( + θ) f(θ) sin sin(/) f (θ), nd then we use the periodicity of f, nd the fct tht sin( + π) sin to see g θ (π) lim π f( + θ) f(θ) sin(/) lim f( + π + θ) f(θ) sin(/ + π) f( + θ) f(θ) lim f (θ), / sin(/) so g C ([, π]). We hve tht f(θ) 1 π f(θ)d π N ()d, so f(θ) S N f(θ) 1 π 1 π (f(θ) f(θ + ))D N ()d f(θ) f(θ + ) sin(n + 1/) d sin(/) ( sin(n + 1/), gθ ), s desired. (6) Note tht the functions g θ C ([, π]), so g θ L ([, π]), so using the ssumption tht sin(n + 1/) is n orthonorml set (we cn proved this by
problem (), since they re eigenfunctions of Af f with boundry condition f() f(π) ), we hve g θ ( sin(n + 1 ), g θ), n so in prticulr, ( sin(n + 1 ), g θ), so f(θ) S N f(θ). We know, then, tht S N f converges to f pointwise, nd, from prt (3), tht S N converges to some limit h uniformly. Uniform convergence implies pointwise convergence, nd pointwise limits re unique, so f h, nd so S N converges to f uniformly. Problem 4. (cf. Tylor 3.1.8.) Suppose now tht f is piecewise C 1 nd π-periodic on R, i.e. there eist finite number of points j [, π), j 1,,..., n, such tht f( j ±) lim j ± f() nd lim j ± f () eist, but re not necessrily equl to ech other, nd wy from f is C 1. S { j + πk : j 1,..., n, k Z}, Show tht t ech / S, the Fourier series converges to f(), while t ech j, the Fourier series converges to 1 (f( j+) + f( j )). Hint: use tht sin(n + 1 ), N, 1,,... is n orthonorml set in L ([, π]) with inner product (f, g) L ([,π]) π 1 f() g() d. 7 Solution. Recll the useful Riemnn-Lebesgue Lemm: if f L 1 ([, π]), then c n (f, e in ) s n. A proof of Riemnn-Lebesgue lemm is s follows: note tht if f C 1 ([, π]), then we hve lredy proved in problem 3 (). Net, we know tht C 1 -functions re dense in L 1 ([, π]) (continuous functions re clerly dense in L 1 by completion, nd ny continuous function on [, π] cn indeed be pproimted by smooth functions, sy polynomils by Stone-Weierstrss theorem), so for ny ɛ >, there eists g C 1 such tht f g L 1 < ɛ/. Since g C 1, there eists N such tht (g, e in ) < ɛ/ for ll n N. Hence for ny n N, we hve (f, e in ) (f g, e in ) + (g, e in ) f g L 1 + (g, e in ) < ɛ. So the lemm is proved.
8 Let θ / S, then f is continuously differentible in neighborhood of θ. Therefore, g θ () f(+θ) f(θ) sin(/) is in L 1 ([, π]) s f L 1 nd g θ is continuous t. Apply Riemnn- Lebesgue lemm to g θ, we get 1 π g θ ()e in d π s n goes to infinity. By the sme clcultions we did in the previous problem, f(θ) S N f(θ) 1 π sin(n + 1 ) g θ()d. We wnt to show tht the right hnd side converges to s N. Observe tht sin(n + 1/) sin N cos / cos N sin /. Hence sin(n + 1 ) g θ()d sin N (cos g θ())d cos N (sin g θ())d. Apply Riemnn-Lebesgue lemm to the L 1 functions in brckets, we see tht they converge to s N. Therefore, the Fourier series converges to f(θ) when θ / S. Now, consider those singulr points j, by doing trnsltion if necessry, we cn ssume tht j (note tht f is π-periodic.) Note tht S N f() N (f, ein ) nd Using periodicity, f() cos nd (f, e in ) + (f, e in ) π (f, cos n) 1 π f() cos nd+ f() cos nd f() cos nd. (f()+f( )) cos nd. Hence S N f() only depends on the even prt of f: f even () f()+f( ). More precisely, S N f() 1 π f even ()D N ()d sin(n+1/) where D N is the Dirichlet kernel. Note tht we cn re-define f sin(/) even () f(+)+f( ) so tht it is continuous t. Consider S N f() f even () 1 (f even () f even ())D N () d π ( f even () f even (), ) sin(n + 1 sin ) L ([,π]) which converges to since sin(n + 1/), N, 1,,..., is n orthonorml bsis in L ([, π]) nd f even () f even () L ([, π]). sin
(Note tht f even () f even () lim f (+) f ( ) + eists, so the function bove is bounded round.) 9 Alterntive Solution: sin(n + 1/) is n orthonorml set in L ([, π]) since they re eigenfunctions of the opertor Af f with f() f (π) (see Problem ). If f is C 1 t, then the sme rgument s in the previous problem shows tht g θ is L, hence the Fourier series converges to f(). If f is singulr t, then S N f() 1 f(θ) π D N(θ)dθ + 1 π by periodicity of f nd chnge of vrible. Therefore, f(+) + f( ) S N f() ( f(θ) f(+) sin θ/, sin(n + 1 )θ f( θ) D N (θ)dθ ) ( f( θ) f( ) + sin θ/, sin(n + 1 ) )θ. Apply the rgument in the previous problem with g θ replced by the functions f(θ) f(+) sin θ/ respectively on [, π]. This yields our desired conclusion. nd f( θ) f( ) sin θ/