Basic Analysis – Gently Done Topological Vector Spaces

94 Basic Analysis
and we deduce that �A� ≤ 1+�A m �. Thus A ∈ B(X,Y).
We must now show that, indeed, A n → A with respect to the norm in B(X,Y).
Let ε > 0 be given. Then there is N ∈ N such that
�A n x−A m x� ≤ �A n −A m ��x� ≤ ε�x�
for any m,n > N and for any x ∈ X. Taking the limit n → ∞, as before, we
obtain
�Ax−A m x� ≤ ε�x�
for any m > N and any x ∈ X. Taking the supremum over x ∈ X with �x� ≤ 1
yields �A−A m � ≤ ε for all m > N. In other words, A m → A in B(X,Y) and the
proof is complete.
Remark 8.14 If S and T belong to B(X), then ST : X → X is defined by
STx = S(Tx), for any x ∈ X. Clearly ST is a linear operator. Also, �STx� ≤
�S��Tx� ≤ �S��T��x�, which implies that ST is bounded and �ST� ≤ �S��T�.
Thus B(X) is an example of an algebra with unit (the unit is the bounded linear
operator 1lx = x, x ∈ X). If X is complete, then so is B(X). In this case B(X) is
an example of a Banach algebra.
Examples 8.15
1. Let A = (a ij ) be any n × n complex matrix. The map x ↦→ Ax, x ∈ C n , is
a linear operator on C n . Clearly, this map is continuous (where C n is equipped
with the usual Euclidean norm), and so therefore it is bounded. By slight abuse
of notation, let us also denote by A this map, x ↦→ Ax.
To find �A�, we note that the matrix A ∗ A is self adjoint and positive, and so there
exists a unitary matrix V such that VA ∗ AV −1 is diagonal:
VA ∗ AV −1 =
⎛
λ1 0 ... 0
⎞
⎜ 0 λ2 ... 0 ⎟
⎜
⎝ .
.
.
. .
.
..
. ⎟
. ⎠
.
0 0 ... λn where each λ i ≥ 0, and we may assume that λ 1 ≥ λ 2 ≥ ... ≥ λ n . Now, we have
�A� 2 = sup{�Ax� : �x� = 1} 2
= sup{�Ax� 2 : �x� = 1}
= sup{(A ∗ Ax,x) : �x� = 1}
= sup{(VA ∗ AV −1 x,x) : �x� = 1}
= sup �� n
k=1 λ k |x k |2 : � n
k=1 |x k |2 = 1 �
= λ 1 .
Department of Mathematics King’s College, London