Basic Analysis – Gently Done Topological Vector Spaces

It follows that �A� = λ 1 , the largest eigenvalue of A ∗ A.
8: Banach Spaces 95
2. Let K : [0,1]×[0,1] → C be a given continuous function on the unit square.
For f ∈ C([0,1]), set
� 1
(Tf)(s) = K(s,t)f(t)dt.
0
Evidently, T is a linear operator T : C([0,1]) → C([0,1]). Furthermore, if we set
M = sup{|K(s,t)| : (s,t) ∈ [0,1]×[0,1]}, we see that
� 1
|Tf(s)| ≤ |K(s,t)||f(t)|dt
0
� 1
≤ M |f(t)|dt.
0
Thus, �Tf� 1 ≤ M�f� 1 , so that T is a bounded linear operator on the space
(C([0,1]), �·� 1 ).
3. With T defined as above, it is straightforward to check that
and that
�Tf� ∞ ≤ M �f� 1
�Tf� 1 ≤ M �f� ∞
so we conclude that T is a bounded linear operator from (C([0,1]), � · � 1 ) to
(C([0,1]), �·� ∞ ) and also from (C([0,1]), �·� ∞ ) to (C([0,1]), �·� 1 ).
4. Take X = ℓ 1 , and, for any x = (x n ) ∈ X, define Tx to be the sequence
Tx = (x 2 ,x 3 ,x 4 ,...). Then Tx ∈ X and satisfies �Tx� 1 ≤ �x� 1 . Thus T is a
bounded linear operator from ℓ 1 → ℓ 1 , with �T� ≤ 1. In fact, �T� = 1 (take
x = (0,1,0,0,...)). T is called the left shift on ℓ 1 .
Similarly, one sees that T : ℓ ∞ → ℓ ∞ is a bounded linear operator, with �T� = 1.
5. Take X = ℓ 1 , and, for any x = (x n ) ∈ X, define Sx to be the sequence
Sx = (0,x 1 ,x 2 ,x 3 ,...). Clearly, �Sx� 1 = �x� 1 , and so S is a bounded linear
operator from ℓ 1 → ℓ 1 , with �S� = 1. S is called the right shift on ℓ 1 .
As above, S also defines a bounded linear operator from ℓ ∞ to ℓ ∞ , with norm 1.