Problem 1: [27 Marks]
1-1. Let X =/=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.
[3+2]
1-1-2. A Borel a-algebra on X.
[3]
1-1-3. A measurable space on X.
[1]
1-1-4. A measure on X.
[3]
1-1-5. A measure space on X.
[1]
1-2. Let EC JRa non-empty set. Show that F = {0,E, Ee, 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
~c.
Problem 2: [35 Marks]
Let (X, 11· 11)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'f'(n)}such that
[6]
1
Vk EN, llx'f'(k)- x'f'(k-1)11:S 2k-l
and show that
[6]
n
X'f'(n)= I)x'f'(k) - X'f'(k-1))+ X'f'(O),for any n 2: l.
k=l
2-2-3. Deduce from question 2-2-2 that the sequence {X<p(n)c}onverges.
[6]
2-2-4. Conclude that {xn} converges and therefore Xis 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,i),where X = C1[0, 1] and llflloo,1= sup lf(x)I + sup IJ'(x)I and also consider
xE[O,l]
xE[O,l]
(Y, II· lloo),where Y = C[0, 1].
3-1-1. Show that T = d~: X--+ Y is a bounded linear operator.
[7]
3-1-2. Show that T = ddx: D(T) £; Y--+ Y is an unbounded linear operator, where D(T) = C1[0, 1]. [10]
(Hint: use un(x) = sin(mrx)).
t, <c,o, } (t, r 3-2. We recall that /l,2 or /l,2 sometimes denoted /l,2 (N0 ) is the space of sequences defined by
l
e'- {X - (Xn)neo' lxnI'
No- Nu {O}' and llxll,,-
Ix.I'
Show that the following operators are linear and continuous and compute their norms.
3-2-1. T1: /l,2 --+/l,2 : T1 ((xn)n2'.0)= (Xn+1)n2'.0·
[9]
f 3-2-2. T2: L2 ([0, 1])--+ C: T2 (!) = 01x2 f(x)dx, where:
[12]
{!: oo} L2([0, 1]) =
f [0, 1]--+ JR:
1
0
lf(x)l2dx
<
(f and IIJIIL2=
1
0
lf(x)l2dx)
1
2
.
God bless you !!!