Problem 1: [27 Marks]
1-1. Let X -f 0. Give the definition of the following concepts:
1-1-1. A a-algebra on X and a a-algebra generated by a family C of subsets of X.
1-1-2. A Borel a-algebra on X.
[3+2]
[3]
1-1-3. A measurable space on X.
[l]
1-1-4. A measure on X.
[3]
1-1-5. A measure space on X.
[l]
1-2. Let EC JRa non-empty set. Show that F = {0, E, EC,JR}is the a-algebra of subsets of JRgenerated
by {E}.
[9]
1-3. Let X = {l, 2, 3,4} and consider C = {{1},{2,3}} C P(X). Determine a(C) the a-algebra generated
by C.
[5]
Problem 2: [35 Marks]
Let (X, II· II) be a normed space.
2-1. Assume that X is a Banach space.
Show that any absolutely summable series is summable.
[6]
2-2. Now we assume that X is a normed space in which any absolutely summable series is summable.
2-2-1. Let {xn} be a Cauchy sequence in X. Show that if {xn} has a convergent subsequence {xnk}, {xn}
converges to the same limit.
[6]
2-2-2. Show that we can construct a subsequence {X,p(n)} such that
[6]
1
Vk E N, llx,p(k)- X,p(k-1) II 2k-l
and show that
[6]
n
X,p(n) = I)x<p(k) - X,p(k-1))+ X,p(O), for any n 2: l.
k=l
2-2-3. Deduce from question 2-2-2 that the sequence {X,p(n)} converges.
[6]
2-2-4. Conclude that {Xn} converges and therefore X is a Banach space.
[3]
2-3. What is the general rule that you can establish from the main results obtained above.
[2]
Problem 3: [38 Marks]
3-1. Consider (X, II· lloo,1),where X = C1[0,l] and 111001,11= sup IJ(x)I + sup IJ'(x)I and also consider
xE[0,l]
xE[0,l]
! : (Y, II· lloo),where Y = C[O,l].
3-1-1. Show that T = X Y is a bounded linear operator.
[7]
3-1-2. Show that T = d~: D(T) £; Y
(Hint: use Un(x) = sin(mrx)).
Y is an unbounded linear operator, where D(T) = C1[0,l]. [10]
3-2. We recall that £2 or £2 sometimes denoted £2 (N0 ) is the space of sequences defined by
t, oo•} (t, t' {x (xn)neo, lx.12 <
No~ NU (0), and llxlll'
I
Ix.I')'
Show that the following operators are linear and continuous and compute their norms.
3-2-1. T1: £2 f2 : T1 ((xn)n;::o)= (xn+i)ne::O·
[9]
f 3-2-2. T2: £ 2([0,l]) C: T2(!) = 01x2J(x)dx, where:
[12]
{!: oo} L2([0,l]) =
J [0,l]
JR:
1
0
IJ(x)J2 dx
<
l
and
11!11=£2(J01
IJ(x)l2dx)
2
.
God bless you !!!