Functional Analysis

ASSIGNMENT II - HINTS TO SELECTED QUESTIONS


(4) Let X = Y = Z = R2 and consider the following maps
T (x, y) = (x, 0); S(x, y) = (0, y).
Then S ◦ T = 0, however ∥S∥ = ∥T ∥ = 1.

(7) Note that
∫ ∞ ∫ b
T (x) = x(t) dt = x(t) dt < ∞.
−∞ a
Further, T is a well-defined linear map. However, T is unbounded. To see this, consider


 0 if t ∈ (−∞, −1) ∪ (n + 1, ∞)

1 + t if t ∈ [−1, 0]
xn (t) =

 1 if t ∈ [0, n]

n + 1 − t if t ∈ [n, n + 1].
We have ∥xn ∥∞ = 1 and T xn = n + 1.

(8) Linearity of f is obvious. By the Schwartz inequality we have
∞
∑
|f (x)| = a(i)x(i) ≤ ∥a∥2 ∥x∥2 .
i=1

Therefore f is bounded and ∥f ∥ ≤ ∥a∥2 . Choose x = a
∞
∑
|f (x)| = (a(i))2 = ∥a∥2 .
i=1
sin nπt
(9) Consider a sequence {xn }, where xn (t) := .
n
sin nπt 1
∥xn ∥∞ = sup = ,
t∈[0,1] n n
whereas
1 1
∥Ixn ∥ = ∥xn ∥ = + sup | cos nπt| = 1 + .
n t∈[0,1] n
( 1 )
( 1 to the) limit x(t) = 0 in C [0, 1], ∥.∥∞ , but (Ixn ) does not
This shows that (xn ) converges
converge to the limit Ix in C [0, 1], ∥.∥ .

(10) If f1 (x) = 0 for all x ∈ X, then f2 (x) = 0 for all x ∈ X. In this case, take any k ̸= 0.
f2 (x0 )
Suppose f1 (x0 ) ̸= 0 for some x0 ∈ X. Then f2 (x0 ) ̸= 0. Let k := ̸= 0. Suppose x ∈ X
f1 (x0 )
f1 (x)
be arbitrary. Set α = . Then
f1 (x0 )
f1 (x) = αf1 (x0 ) = f1 (αx0 ).
Consequently,
f1 (x − αx0 ) = 0,
and
x − αx0 ∈ N (f1 ) = N (f2 ).
Thus, f2 (x − αx0 ) = 0, that is,
f2 (x) = αf2 (x0 ) = αkf1 (x0 ) = kf1 (x).
1

, 2 ASSIGNMENT II - HINTS TO SELECTED QUESTIONS




(11) Recall that for F ∈ B(X, Y )
∥F ∥ := sup S, where S := {∥F (x)∥ : x ∈ X, ∥x∥ ≤ 1} .
If ∥F ∥ = 0, S contains only 0. Let {xn } be a sequence in X such that ∥xn ∥ = 1 and ∥F (xn )∥ = 0
for all n ∈ N. Then
∥F (xn )∥ → 0 = ∥F ∥.
Assume ∥F ∥ ̸= 0. We have
1
∥F ∥ − ∥F ∥ < ∥F ∥ ∀ n ≥ 1.
n
So there exists αn ∈ S such that
1
0 ≤ ∥F ∥ − ∥F ∥ < αn .
n
zn
Let αn = ∥F (zn )∥, where zn ∈ X with ∥zn ∥ ≤ 1. Since αn ̸= 0, we have zn ̸= 0. Set xn = ,
∥zn ∥
then ∥xn ∥ = 1. Thus, ∥F (xn )∥ ∈ S and
∥F (xn )∥ ≤ ∥F ∥.
Further,
1 αn
∥F (xn )∥ = ∥F (zn )∥ = ≥ αn .
∥zn ∥ ∥zn ∥
Consequently,
1
∥F ∥ − ∥F ∥ < αn ≤ ∥F (xn )∥ ≤ ∥F ∥.
n
That is,
∥F (xn )∥ → ∥F ∥.

(12) For x ∈ C[0, 1] and t ∈ [0, 1], we have
∫ 1
|(Sx)(t)| ≤ t |x(u)|du ≤ t∥x∥∞ ≤ ∥x∥∞ ,
0
and
|(T x)(t)| = t|x(t)| ≤ t∥x∥∞ ≤ ∥x∥∞ .
Therefore,
∥Sx∥∞ ≤ ∥x∥∞ ∥T x∥∞ ≤ ∥x∥∞ ,
that is, S and T are bounded and

∥S∥ ≤ 1, ∥T ∥ ≤ 1, ∥T S∥ ≤ 1.
Letting z = T x, we obtain
|z(t)| = t|x(t)| ≤ t∥x∥∞ .
Thus,
∫ 1 ∫ 1
1
|(ST )x(t)| = |Sz(t)| ≤ t |z(u)|du ≤ u∥x∥∞ du = ∥x∥∞ ,
0 0 2
and
1
∥ST x∥∞ ≤ ∥x∥∞ .
2
Hence, we deduce that
1
∥ST ∥ ≤
.
2
Taking x0 (t) = 1 for all t ∈ [0, 1], we obtain ∥x0 ∥∞ = 1. Further,
t
Sx0 (t) = t, T x0 (t) = t, (T S)(x0 )(t) = t2 , (ST )(x0 )(t) = .
2
Consequently,
1 1
∥Sx0 ∥∞ = ∥T x0 ∥∞ = ∥(T S)x0 ∥∞ = 1 = ∥x0 ∥∞ , ∥(ST )x0 ∥∞ = = ∥x0 ∥∞ .
2 2