QUIZ I: HINTS AND SOLUTIONS
1. Let A ∈ B(X, Y ). Then
∥Axn − Axm ∥Y ≤ ∥A∥B(X,Y ) ∥xn − xm ∥X .
The foregoing inequality implies that (Axn ) is a Cauchy sequence whenever (xn ) is so.
f (x)
2. Let x ∈ X. Since x0 ∈
/ N (f ), f (x0 ) ̸= 0. Let α := ∈ K and y = x − αx0 ∈ X. Note
f (x0 )
that
f (x)
f (y) = f (x) − αf (x0 ) = f (x) − f (x0 ) = 0.
f (x0 )
Thus, x = y + αx0 , where y ∈ N (f ), α ∈ K, showing the existence of such a representation.
To prove the uniqueness, let us assume
x = α1 x0 + y1 = α2 x0 + y2 ; α1 , α2 ∈ K, y1 , y2 ∈ N (f ).
Then
(α1 − α2 )x0 = y2 − y1 ∈ N (f ).
Since x0 ∈
/ N (f ), the previous equation implies α1 = α2 , which in turn asserts y1 = y2 .
3. Let X and Y be normed linear spaces with dim(X) = dim(Y ) = n < ∞. Let BX =
{x1 , x2 , . . . , xn } be a basis for X and let BY = {y1 , y2 , . . . , yn } be a basis for Y . Set F (xi ) = yi
for i = 1, 2, . . . , n. Let F : X → Y be defined as follows:
( n )
∑ ∑ n
F (x) = F ki x i = ki yi .
i=1 i=1
Then F is a bijective linear map. Since X and Y are finite dimensional, the linear maps
F : X → Y and F −1 : Y → X are bounded. That is, F is a homeomorphism from X onto Y .
4. First let us recall the fundamental theorem of calculus (see Theorem 0.1 below), which, in
certain sense, says that integration and differentiation are inverse operations.
Theorem 0.1. Let f be Riemann integrable on [a, b]. For a ≤ x ≤ b, set
∫ x
F (x) = f (t) dt.
a
Then F is continuous on [a, b]. Furthermore, if f is continuous at a point x0 of [a, b], then
F is differentiable at x0 , and F ′ (x0 ) = f (x0 ).
Note that in view of the preceding theorem, T : C[0, 1] → C[0, 1] given by
∫ t
x 7→ T x(= y), where y(t) = x(u) du.
0
is well-defined.
(i) Clearly, T is linear. For any t ∈ [0, 1]
∫ t ∫ t ∫ 1
|T x(t)| = x(u) du ≤ |x(u)| du ≤ ∥x∥∞ du = ∥x∥∞ .
0 0 0
Consequently,
∥T x∥∞ := sup |T x(t)| ≤ ∥x∥∞ .
t∈[0,1]
Ergo, T is a bounded linear operator and ∥T ∥ ≤ 1.
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1. Let A ∈ B(X, Y ). Then
∥Axn − Axm ∥Y ≤ ∥A∥B(X,Y ) ∥xn − xm ∥X .
The foregoing inequality implies that (Axn ) is a Cauchy sequence whenever (xn ) is so.
f (x)
2. Let x ∈ X. Since x0 ∈
/ N (f ), f (x0 ) ̸= 0. Let α := ∈ K and y = x − αx0 ∈ X. Note
f (x0 )
that
f (x)
f (y) = f (x) − αf (x0 ) = f (x) − f (x0 ) = 0.
f (x0 )
Thus, x = y + αx0 , where y ∈ N (f ), α ∈ K, showing the existence of such a representation.
To prove the uniqueness, let us assume
x = α1 x0 + y1 = α2 x0 + y2 ; α1 , α2 ∈ K, y1 , y2 ∈ N (f ).
Then
(α1 − α2 )x0 = y2 − y1 ∈ N (f ).
Since x0 ∈
/ N (f ), the previous equation implies α1 = α2 , which in turn asserts y1 = y2 .
3. Let X and Y be normed linear spaces with dim(X) = dim(Y ) = n < ∞. Let BX =
{x1 , x2 , . . . , xn } be a basis for X and let BY = {y1 , y2 , . . . , yn } be a basis for Y . Set F (xi ) = yi
for i = 1, 2, . . . , n. Let F : X → Y be defined as follows:
( n )
∑ ∑ n
F (x) = F ki x i = ki yi .
i=1 i=1
Then F is a bijective linear map. Since X and Y are finite dimensional, the linear maps
F : X → Y and F −1 : Y → X are bounded. That is, F is a homeomorphism from X onto Y .
4. First let us recall the fundamental theorem of calculus (see Theorem 0.1 below), which, in
certain sense, says that integration and differentiation are inverse operations.
Theorem 0.1. Let f be Riemann integrable on [a, b]. For a ≤ x ≤ b, set
∫ x
F (x) = f (t) dt.
a
Then F is continuous on [a, b]. Furthermore, if f is continuous at a point x0 of [a, b], then
F is differentiable at x0 , and F ′ (x0 ) = f (x0 ).
Note that in view of the preceding theorem, T : C[0, 1] → C[0, 1] given by
∫ t
x 7→ T x(= y), where y(t) = x(u) du.
0
is well-defined.
(i) Clearly, T is linear. For any t ∈ [0, 1]
∫ t ∫ t ∫ 1
|T x(t)| = x(u) du ≤ |x(u)| du ≤ ∥x∥∞ du = ∥x∥∞ .
0 0 0
Consequently,
∥T x∥∞ := sup |T x(t)| ≤ ∥x∥∞ .
t∈[0,1]
Ergo, T is a bounded linear operator and ∥T ∥ ≤ 1.
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