FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...

16 CHRISTOPHER HEIL
The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space.
Theorem 2.18. If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banach
space.
Proof. Assume that {An}n∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy in
operator norm. For any given f ∈ X, we have
Amf − Anf ≤ Am − An f,
so we conclude that {Anf}n∈N is a Cauchy sequence of vectors in Y . Since Y is complete,
this sequence must converge, say Afn → g ∈ Y . Define Af = g. This gives us a candidate
limit operator A.
Exercise: Show that A defined in this way is a linear operator.
To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we
must have C = sup An < ∞. If f ∈ X, then since Anf → Af we have
Af = lim Anf ≤ sup
n→∞
n∈N
Anf ≤ sup An f = C f.
n∈N
Hence A is bounded, and A ≤ C.
Finally, we must show that An → A in operator norm. Fix any ε > 0. Then there exists
an N such that
m, n > N =⇒ Am − An < ε
2 .
Choose any f ∈ X with f = 1. Then since Amf → Af, there exists an m > N such that
Af − Amf < ε
2 .
Hence for any n > N we have
Af −Anf ≤ Af −Amf+Amf −Anf ≤ Af −Amf+Am−An f < ε ε
+ = ε.
Taking the supremum over all unit vectors, we conclude that A − An ≤ ε for all n > N.
Thus An → A.
Corollary 2.19. If X is a normed space, then its dual space X ′ = B(X, F) is a Banach
space.
The next exercise deals with the problem of extending an operator defined only a dense
subspace to the entire space.
Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed
space X, and let Z be a Banach space.
(a) Suppose that L: Y → Z is a bounded linear operator. Show that there exists a
unique bounded linear operator ˜ L: X → Z whose restriction to Y is L. Prove that
˜ L = L.
(b) Prove that if L: Y → range(L) is a topological isomorphism, then ˜ L: X → range(L)
is also.
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