Bài tập toán cao cấp chương II
Trang 1Ch ’u ’ong 2 Khˆ ong gian ¯ di.nh chu ’ˆan
10 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) =Rb
a
K(t, s)x(s)ds, Bx(t) =Rb
a
H(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`a Rb
a
H(t, u)K(t, s)ds.
Gi ’ai
(BA)x(t) = B(Ax(t)) = Rb
a
H(t, u)Ax(u)du
=Rb
a
H(t, u)
µRb
a
K(u, s)x(s)ds
¶
du =Rb
a
µRb
a
H(t, u)K(u, s)x(s)ds
¶
du.
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´ıch
phˆan ta ¯d ’u ’o.c
BAx(t) =Rb
a
µb R
a
H(t, u)K(u, s)x(s)du
¶
ds =Rb
a
(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.ch Rb
a
H(t, u)K(t, s)ds.
Ch ’u ’ong 4 C´ ac nguyˆ en l´ı c ’o b ’an c’ua gi ’ai t´ıch h` am
5 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 G A = {(x, f (x)) : x ∈ M} ¯ d´ong trong M × Y th`ı f liˆen tu.c
trˆen M
Gi ’ai
Gi ’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.i
M, t´’uc l`a t `ˆon ta.i d˜ay {x n } ⊂ M sao cho x n → x nh ’ung f(x n ) 9 f (x).
Suy ra t `ˆon ta.i s ´ˆo ε0 > 0 sao cho ∀n, ∃n k > n (n k > n k−1 ) sao cho kf (x n k ) − f (x)k ≥
ε0
Ta c´o {f (x n k )} k ⊂ f (M) v`a f (M) nˆen t ` ˆon ta.i d˜ay con {f(x n kj )} j, v´’oi
f (x n kj ) → y ∈ f (M).
Khi ¯d´o
Trang 2(x n kj , f (x n kj )) → (x, y).
Do G f d´ong nˆen (x, y) ∈ G¯ f Suy ra y = f (x) Do ¯ d´o f (x n kj ) → f (x) (vˆo l´y).
Ch ’u ’ong 5 Khˆ ong gian Hilbert
4 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 hAx, xi ≥ αhx, xi, 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 kA −1 k ≤ 1
α
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 αkxk2 = αhx, xi ≤ hAx, xi ≤ kAxk.kxk ⇒ αkxk ≤ kAxk Do ¯ d´o A −1 :
ImA → X liˆen tu.c v`a kA −1 k ≤ 1
α
* 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 = hAz, zi ≥ αhz, zi Do ¯ d´o hz, zi = 0 T`’u ¯d´o
z = 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´’ung minh r`˘ang n ´ˆeu v´’oi m ˜ˆoi u ∈ X, phi ´ˆem h`am
x 7→ hAx, ui, 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 {(x n , Ax n )} ⊂ G(A) v`a lim
n→∞ (x n , Ax n ) = (x, y) Khi ¯d´o lim
∀u ∈ X M˘a.t kh´ac, do phi ´ˆem h`am c ’ua ¯d `ˆe b`ai liˆen tu.c nˆen
lim
Trang 3∀u ∈ X T`’u (i) v`a (ii), ta suy ra
hy, ui = hAx, ui ⇔ hy − Ax, ui = 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 {e n } n l`a mˆo.t c ’o s’’o c’ua khˆong gian Hilbert X v`a
P n x =
n
X
k=1
hx, e k ie k , x ∈X, n = 1, 2,
l`a d˜ay ph´ep chii ´ˆeu tr ’u.c giao Ch´’ung minh r`˘ang d˜ay {P n } n hˆo.i tu ¯di ’ˆem ¯d´ˆen to´an t ’’u ¯d `ˆong
nh ´ˆat I nh ’ung khˆong hˆo.i tu ¯d `ˆeu ¯d ´ˆen I.
Gi ’ai
P n l`a ph´ep chi ´ˆeu tr ’u.c giao lˆen khˆong gian con tuy ´ˆen t´ınh L{e1, , e n } V`ı {e n } l`a
mˆo.t c ’o s’’o c’ua X nˆen v´’oi mo.i x ∈ X ta c´o
x =
∞
X
n=1
hx, e n ie n Khi ¯d´o
lim
n→∞ P n x = lim
n→∞
n
X
k=1
hx, e k ie k =
∞
X
n=1
hx, e n ie n = x = Ix, ∀x ∈ X.
Vˆa.y d˜ay {P n } hˆo.i tu ¯d´ ˆen I.
Gi ’a s ’’u d˜ay {P n } hˆo.i tu ¯d `ˆeu ¯d ´ˆen I Khi ¯d´o, lim
n→∞ kP n − Ik = 0 Do ¯ d´o, kP n0− Ik < 1
v´’oi n0 d ’u l´’¯ on L ´ˆay x = e n0+1 th`ı
k(P n0 − I)e n0+1k ≤ kP n0 − Ik.ke n1+1k < 1.
M˘a.t kh´ac, ta c´o
k(P n0 − I)e n0+1k = kP n0e n0+1− e n0+1k = ke n0+1k = 1 (vˆo l´y)