ASSIGNMENT II
MTL 411 FUNCTIONAL ANALYSIS
1. For x = (x(1), x(2), . . . ) ∈ l∞ , let T : l∞ → l∞ be defined by
( )
x(2) x(3)
T x = x(1), , ,...
2 3
Prove that
(i) T is a bounded linear operator
(ii) T is injective
(iii) Range of T is not a closed subspace of l∞ .
2. If T : X → Y is a linear operator such that there exists c > 0 and 0 ̸= x0 ∈ X satisfying
∥T x∥ ≤ c∥x∥ ∀ x ∈ X; ∥T x0 ∥ = c∥x0 ∥,
then show that T ∈ B(X, Y ) and ∥T ∥ = c.
3. For x = (x(1), x(2), . . . ) ∈ l2 , consider the right shift operator S : l2 → l2 defined by
Sx = (0, x(1), x(2), . . . )
and the left shift operator T : l2 → l2 defined by
T x = (x(2), x(3), . . . )
Prove that
(i) S is a abounded linear operator and ∥S∥ = 1.
(ii) S is injective.
(iii) S is, in fact, an isometry.
(iv) S is not surjective.
(v) T is a bounded linear operator and ∥T ∥ = 1
(vi) T is not injective.
(vii) T is not an isometry.
(viii) T is surjective.
(ix) T S = I and ST ̸= I. That is, neither S nor T is invertible, however, S has a left inverse
and T has a right inverse.
Note that item (ix) illustrates the fact that the Banach algebra B(X) is not in general
commutative.
4. Show with an example that for T ∈ B(X, Y ) and S ∈ B(Y, Z), the equality in the submulti-
plicativity of the norms
∥S ◦ T ∥ ≤ ∥S∥∥T ∥
may not hold.
5. Let X and Y be normed linear spaces and A : X → Y be a linear map. Show that A is an
isometry if and only if ∥Ax∥ = ∥x∥ for all x ∈ X. Deduce that a linear isometry A is injective
and ∥A∥ = 1.
6. Show with an example that an injective bounded linear map A with ∥A∥ = 1 need not be an
isometry.
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MTL 411 FUNCTIONAL ANALYSIS
1. For x = (x(1), x(2), . . . ) ∈ l∞ , let T : l∞ → l∞ be defined by
( )
x(2) x(3)
T x = x(1), , ,...
2 3
Prove that
(i) T is a bounded linear operator
(ii) T is injective
(iii) Range of T is not a closed subspace of l∞ .
2. If T : X → Y is a linear operator such that there exists c > 0 and 0 ̸= x0 ∈ X satisfying
∥T x∥ ≤ c∥x∥ ∀ x ∈ X; ∥T x0 ∥ = c∥x0 ∥,
then show that T ∈ B(X, Y ) and ∥T ∥ = c.
3. For x = (x(1), x(2), . . . ) ∈ l2 , consider the right shift operator S : l2 → l2 defined by
Sx = (0, x(1), x(2), . . . )
and the left shift operator T : l2 → l2 defined by
T x = (x(2), x(3), . . . )
Prove that
(i) S is a abounded linear operator and ∥S∥ = 1.
(ii) S is injective.
(iii) S is, in fact, an isometry.
(iv) S is not surjective.
(v) T is a bounded linear operator and ∥T ∥ = 1
(vi) T is not injective.
(vii) T is not an isometry.
(viii) T is surjective.
(ix) T S = I and ST ̸= I. That is, neither S nor T is invertible, however, S has a left inverse
and T has a right inverse.
Note that item (ix) illustrates the fact that the Banach algebra B(X) is not in general
commutative.
4. Show with an example that for T ∈ B(X, Y ) and S ∈ B(Y, Z), the equality in the submulti-
plicativity of the norms
∥S ◦ T ∥ ≤ ∥S∥∥T ∥
may not hold.
5. Let X and Y be normed linear spaces and A : X → Y be a linear map. Show that A is an
isometry if and only if ∥Ax∥ = ∥x∥ for all x ∈ X. Deduce that a linear isometry A is injective
and ∥A∥ = 1.
6. Show with an example that an injective bounded linear map A with ∥A∥ = 1 need not be an
isometry.
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