472 CHAPTER 17. BANACH SPACES

Theorem 17.6.9 If T is a Fredholm operator, then T X is closed in Y .

Proof: Recall that Y = T X ⊕E where E is a closed subspace of Y . In fact, E is finitedimensional, but it is only needed that E is closed. Let T0 ∈L (X×E,T X⊕E) be givenby

T0 (x,e)≡ T x+ e

Let the norm on X×E be

∥(x,e)∥X×E ≡max{∥x∥X ,∥e∥E}

Thus T0 (x,e) = 0 implies both T x = 0 and e = 0. Thus ker(T0) = ker(T )×{0}. Also,T0 (X×E) is closed in Y because in fact it is all of Y , T X⊕E. By Proposition 17.6.8, thereexists δ > 0 such that

∥T0 (x,e)∥Y ≥ δ dist((x,e) ,ker(T0))

= δ dist((x,e) ,ker(T )×{0})≥ δ dist(x,ker(T ))

Then∥T x∥Y ≡ ∥T0 (x,0)∥Y ≥ δ dist(x,ker(T ))

and by Proposition 17.6.8, T X is closed.Actually, the above proves the following corollary.

Corollary 17.6.10 If T X ⊕E is closed in Y and E is a closed subspace of Y , then T X isclosed. Here T ∈L (X ,Y ).

Note that it appears that dim(ker(T ))< ∞ was not really needed.Let B be a Hamel basis for T X and consider A ≡ {x : T x ∈B} . Then this is a lin-

early independent set of vectors in X . Suppose now that ker(T ) = span(z1, · · · ,zn) where{z1, · · · ,zn} is linearly independent so here the assumption that ker(T ) has finite dimen-sions is being used. Then if x ∈ X ,T x ∈ T X and so there are finitely many vectors xi ∈Asuch that

T x = ∑i

ciT xi.

Hence

T

(x−∑

icixi

)= 0

so

x−∑i

cixi =n

∑j=1

a jz j

Hence X = span(A ) + ker(T ) . In fact, {A ,{z1, · · · ,zn}} is linearly independent as iseasily seen and so this is a basis for X . Hence

X = span(A )⊕ker(T )≡ X1⊕ker(T )