Basic Analysis – Gently Done Topological Vector Spaces

Now let s ∈ K and T ∈ B(X,Y). Then
�sT� = sup{�sTx� : �x� ≤ 1}
= sup{|s|�Tx� : �x� ≤ 1}
= |s| sup{�Tx� : �x� ≤ 1}
= |s|�T�.
Finally, we see from the above, that for any S,T ∈ B(X,Y),
and the proof is complete.
�S +T� = sup{�Sx+Tx� : �x� ≤ 1}
≤ sup{(�S�+�T�)�x� : �x� ≤ 1}
= �S�+�T�
8: Banach Spaces 93
Proposition 8.13 Suppose that X is a normed space and Y is a Banach space.
Then B(X,Y) is a Banach space.
Proof All that needs to be shown is that B(X,Y) is complete. To this end, let
(A n ) be a Cauchy sequence in B(X,Y); then
�A n −A m � = sup{�A n x−A m x�/�x� : x �= 0} → 0,
as n,m → ∞. It follows that for any given x ∈ X, �A n x − A m x� → 0, as
n,m → ∞, i.e., (A n x) is a Cauchy sequence in the Banach space Y. Hence there
is some y ∈ Y such that A n x → y in Y. Set Ax = y. We have
A(sx+x ′ ) = lim n A n (sx+x ′ )
= lim n (sA n x+A n x ′ )
= slim n A n x+lim n A n x ′
= sAx+Ax ′ ,
for any x,x ′ ∈ X and s ∈ K. It follows that A : X → Y is a linear operator. Next
we shall check that A is bounded. To see this, we observe first that for sufficiently
large m,n ∈ N, and any x ∈ X,
�A n x−A m x� ≤ �A n −A m ��x� ≤ �x�.
Taking the limit n → ∞ gives the inequality �Ax−A m x� ≤ �x�. Hence, for any
sufficiently large m,
�Ax� ≤ �Ax−A m x�+�A m x�
≤ �x�+�A m ��x�