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Bài tập toán cao cấp chương II
1Ch’u’ong 2. Khˆong gian ¯di.nh chu’ˆan10. Cho A, B l`a hai to´an t’’u t´ıch phˆan trong C[a,b]v´’oi ha.ch l`ˆan l’u’o.t l`a K(t, s), H(t, s)Ax(t) =baK(t, s)x(s)ds, Bx(t) =baH(t, u)x(u)du.Ch´’ung minh B ◦ A c˜ung l`a to´an t’’u t´ıch phˆan v`a ha.ch l`abaH(t, u)K(t, s)ds.Gi’ai(BA)x(t) = B(Ax(t)) =baH(t, u)Ax(u)du=baH(t, u)baK(u, s)x(s)dsdu =babaH(t, u)K(u, s)x(s)dsdu.V`ı c´ac h`am H(t, u), K(u, s) liˆen tu.c trˆen {a ≤ t, s ≤ b} v`a h`am x(s) liˆen tu.c trˆen[a, b] nˆen h`am H(t, u)K(u, s)x(s) liˆen tu.c trˆen {a ≤ u, s ≤ b}. Thay ¯d’ˆoi th´’u t’u.l´ˆay t´ıchphˆan ta ¯d’u’o.cBAx(t) =babaH(t, u)K(u, s)x(s)duds =ba(H(t, u)K(u, s)du) x(s)ds.Vˆa.y B ◦ A l`a to´an t’’u t´ıch phˆan v`a ha.chbaH(t, u)K(t, s)ds.Ch’u’ong 4. C´ac nguyˆen l´ı c’o b’an c’ua gi’ai t´ıch h`am5. Cho X, Y l`a c´ac khˆong gian ¯di.nh chu’ˆan, M ⊂ X, v`a f : M → Y sao cho f(M)compact. Ch´’ung minh n´ˆeu GA= {(x, f(x)) : x ∈ M} ¯d´ong trong M × Y th`ı f liˆen tu.ctrˆen M.Gi’aiGi’a s’’u f khˆong liˆen tu.c trˆen M. T`’u ¯d´o t`ˆon ta.i x ∈ M sao cho f khˆong liˆen tu.c ta.iM, t´’uc l`a t`ˆon ta.i d˜ay {xn} ⊂ M sao cho xn→ x nh’ung f(xn) f(x).Suy ra t`ˆon ta.i s´ˆo ε0> 0 sao cho ∀n, ∃nk> n (nk> nk−1) sao cho f(xnk) − f(x) ≥ε0.Ta c´o {f(xnk)}k⊂ f(M) v`a f (M) nˆen t`ˆon ta.i d˜ay con {f(xnkj)}j, v´’oif(xnkj) → y ∈ f(M).Khi ¯d´o 2(xnkj, f(xnkj)) → (x, y).Do Gf¯d´ong nˆen (x, y) ∈ Gf. Suy ra y = f(x). Do ¯d´o f(xnkj) → f(x) (vˆo l´y).Ch’u’ong 5. Khˆong gian Hilbert4. Cho X l`a khˆong gian Hilbert th’u.c v`a A : X → X l`a to´an t’’u tuy´ˆen t´ınh liˆen tu.c.To´an t’’u A go.i l`a x´ac ¯di.nh d’u’ong n´ˆeu ∀x ∈ X ta c´o Ax, x ≥ αx, x, trong ¯d´o α > 0.Ch´’ung minh n´ˆeu A x´ac ¯di.nh d’u’ong th`ı A l`a song ´anh v`a A−1 ≤1α.Gi’ai* A l`a ¯d’on ´anh.* A l`a to`an ´anh ⇔ ImA = X.A : X → ImA l`a song ´anh.∀x ta c´o αx2= αx, x ≤ Ax, x ≤ Ax.x ⇒ αx ≤ Ax. Do ¯d´o A−1:ImA → X liˆen tu.c v`a A−1 ≤1α.* ImA l`a khˆong gian con ¯d´ong c’ua X.* X = ImA ⊕ (ImA)⊥. Ta ch´’ung minh (ImA)⊥= {0}.∀z ∈ (ImA)⊥th`ı Az ∈ ImA nˆen 0 = Az, z ≥ αz, z. Do ¯d´o z, z = 0. T`’u ¯d´oz = 0.Vˆa.y A l`a song ´anh.12. Gi’a s’’u X l`a khˆong gian Hilbert, A : X → X l`a mˆo.t to´an t’’u tuy´ˆen t´ınh. Ch´’ungminh r`˘ang n´ˆeu v´’oi m˜ˆoi u ∈ X, phi´ˆem h`amx → Ax, u, x ∈ X¯d`ˆeu liˆen tu.c th`ı A liˆen tu.c.Gi’ai* Ta ch´’ung minh A l`a to´an t’’u ¯d´ong ⇔ G(A) = {(x, Ax) : x ∈ X} l`a khˆong gian con¯d´ong c’ua X × X.Gi’a s’’u {(xn, Axn)} ⊂ G(A) v`a limn→∞(xn, Axn) = (x, y). Khi ¯d´olimn→∞Axn, u = y, u (i)∀u ∈ X. M˘a.t kh´ac, do phi´ˆem h`am c’ua ¯d`ˆe b`ai liˆen tu.c nˆenlimn→∞Axn, u = Ax, u (ii) 3∀u ∈ X. T`’u (i) v`a (ii), ta suy ray, u = Ax, u ⇔ y − Ax, u = 0, ∀u ∈ X⇔ y − Ax = 0 ⇔ y = Ax.Do ¯d´o (x, y) = (x, Ax) ∈ G(A).* V`ı X l`a khˆong gian Banach nˆen A liˆen tu.c.13. Gi’a s’’u {en}nl`a mˆo.t c’o s’’o c’ua khˆong gian Hilbert X v`aPnx =nk=1x, ekek, x ∈X, n = 1, 2, . . .l`a d˜ay ph´ep chii´ˆeu tr’u.c giao. Ch´’ung minh r`˘ang d˜ay {Pn}nhˆo.i tu.¯di’ˆem ¯d´ˆen to´an t’’u ¯d`ˆongnh´ˆat I nh’ung khˆong hˆo.i tu.¯d`ˆeu ¯d´ˆen I.Gi’aiPnl`a ph´ep chi´ˆeu tr’u.c giao lˆen khˆong gian con tuy´ˆen t´ınh L{e1, . . . , en}. V`ı {en} l`amˆo.t c’o s’’o c’ua X nˆen v´’oi mo.i x ∈ X ta c´ox =∞n=1x, enen.Khi ¯d´olimn→∞Pnx = limn→∞nk=1x, ekek=∞n=1x, enen= x = Ix, ∀x ∈ X.Vˆa.y d˜ay {Pn} hˆo.i tu.¯d´ˆen I.Gi’a s’’u d˜ay {Pn} hˆo.i tu.¯d`ˆeu ¯d´ˆen I. Khi ¯d´o, limn→∞Pn− I = 0. Do ¯d´o, Pn0− I < 1v´’oi n0¯d’u l´’on. L´ˆay x = en0+1th`ı(Pn0− I)en0+1 ≤ Pn0− I.en1+1 < 1.M˘a.t kh´ac, ta c´o(Pn0− I)en0+1 = Pn0en0+1− en0+1 = en0+1 = 1 (vˆo l´y) . . h`am c’ua ¯d`ˆe b`ai liˆen tu.c nˆenlimn→∞Axn, u = Ax, u (ii) 3∀u ∈ X. T`’u (i) v`a (ii) , ta suy ray, u = Ax, u ⇔ y − Ax, u = 0, ∀u ∈ X⇔ y −. gian Hilbert X v`aPnx =nk=1x, ekek, x ∈X, n = 1, 2, . . .l`a d˜ay ph´ep chii´ˆeu tr’u.c giao. Ch´’ung minh r`˘ang d˜ay {Pn}nhˆo.i tu.¯di’ˆem ¯d´ˆen to´an