Kenneth Kuttler

17.6. OPERATORS WITH CLOSED RANGE 471

Then0 = Rp

µ x = Rpµ y+Rp

µ z

Here p = k (µ). This is where it is important that µ ∈ Λ. However, N(Rk

)and Rk

λX

are invariant under Rpµ since it is clear that Rλ and Rµ commute. Thus Rp

µ y = −Rpµ z and

Rpµ y ∈ Rk

λX , −Rp

µ z ∈ N(Rk

)and these are equal. Hence they are both 0. Now it was just

shown that Rµ is one to one on N(Rk

)and so z = 0. Hence x = y ∈ Rk

λX .

Note that in the last step, we can’t conclude that y = 0 because we only know that Rµ

is one to one on Rkλ

X if µ is sufficiently close to λ . The above is about compact mappingsfrom a single space to itself. However, there are also mappings which have closed rangewhich map from one space to another. The Fredholm operators have this property that theirimage is closed. These are discussed next.

Suppose T ∈L (X ,Y ) . Then T X is a subspace of Y and so it has a Hamel basis B.Extending B to a Hamel basis for Y yields C . Then Y = span(B)⊕ span(C \B) . ThusY = T X⊕E. For more on this, see [55].

Definition 17.6.7 Let T ∈L (X ,Y ) . Then this is a Fredholm operator means

1. dim(ker(T ))< ∞

2. dim(E)< ∞ where Y = T X⊕E

Proposition 17.6.8 Let T ∈L (X ,Y ) . Then T X is closed if and only if there exists δ > 0such that

∥T x∥ ≥ δ dist(x,ker(T )) .

Proof: First suppose T X is closed. Let T̂ : X/ker(T )→ Y be defined as T̂ ([x])≡ T x.Then by Theorem 18.7.2, T̂ is one to one and continuous and X/ker(T ) is a Banach space,∥∥T̂∥∥ ≤ ∥T∥. Also T̂ has the same range as T . Thus T X is the same as T̂ (X/ker(T )) and

T̂ ∈L (X/ker(T ) ,Y ) . By the open mapping theorem, T̂ is continuous and has continuousinverse. Recall

∥[x]∥ ≡ inf{∥x+ z∥ : z ∈ kerT}= dist(x,ker(T ))

Then

dist(x,ker(T )) = ∥[x]∥=∥∥T̂−1T̂ [x]

∥∥≤ ∥∥T̂−1∥∥∥∥T̂ [x]∥∥= ∥∥T̂−1∥∥∥T x∥

and so,∥T x∥ ≥ δ dist(x,ker(T ))

where δ = 1/∥∥T̂−1

∥∥ .Next suppose the inequality holds. Why will T X be closed? Say {T xn} is a sequence

in T X converging to y. Then by the inequality,

∥T xn−T xm∥ ≥ δ dist(xn− xm,ker(T )) = δ ∥[xn]− [xm]∥X/ker(T )

showing that {[xn]} is a Cauchy sequence in X/ker(T ). Therefore, since this is a Banachspace, there exists [x] such that [xn]→ [x] in X/ker(T ) and so T̂ ([xn])→ T̂ ([x]) in Y. Butthis is the same as saying that T (xn)→ T (x). It follows that y = T x and so T X is indeedclosed.

17.6. OPERATORS WITH CLOSED RANGE 471Then0= Rix = Rhy +RizHere p =k(). This is where it is important that u € A. However, N (Ri ) and RUXare invariant under Rj, since it is clear that Ry and Ry commute. Thus Rjy = —Rjiz andRiy € R|X, —Riiz € N (Rj) and these are equal. Hence they are both 0. Now it was justshown that R,, is one to one on N(R‘) and soz =0. Hencex=y ERiX. WfNote that in the last step, we can’t conclude that y = 0 because we only know that Ryis one to one on RX if UW is sufficiently close to A. The above is about compact mappingsfrom a single space to itself. However, there are also mappings which have closed rangewhich map from one space to another. The Fredholm operators have this property that theirimage is closed. These are discussed next.Suppose T € & (X,Y). Then TX is a subspace of Y and so it has a Hamel basis &.Extending & to a Hamel basis for Y yields @. Then Y = span(#) @ span(@ \ F). ThusY =TX OE. For more on this, see [55].Definition 17.6.7 Let T € & (X,Y). Then this is a Fredholm operator means1. dim (ker(T)) <2. dim(E) < where Y =TX ®EProposition 17.6.8 Let T € & (X,Y). Then TX is closed if and only if there exists 5 > 0such that||Tx|| > 6 dist (x, ker (T)).Proof: First suppose TX is closed. Let 7 : X /ker(T) —+ Y be defined as 7 ([x]) = Tx.Then by Theorem 18.7.2, 7 is one to one and continuous and X / ker (T) is a Banach space,||7'|| < |Z. Also T has the same range as T. Thus TX is the same as T (X/ker(7)) andT €¢ &(X/ker(T) ,Y). By the open mapping theorem, 7 is continuous and has continuousinverse. Recall|| [x] || = inf {|| + z|] : z € ker T} = dist (x, ker (T))Thendist (x, ker (7) = ||[x] || =?'F [x||| <r'l|T {x\|| =|T'|\\I7-\Iand so,||Tx|| > 6 dist (x, ker (T))where 6 = 1/ i |Next suppose the inequality holds. Why will TX be closed? Say {Tx,} is a sequencein TX converging to y. Then by the inequality,|| 7Xn — TXm|| 2 6 dist (% —xXm,ker(T)) = 6 || [xn] — [xml Ilx/ker(7)showing that {[x,]} is a Cauchy sequence in X/ker(T). Therefore, since this is a Banachspace, there exists [x] such that [x,] — [x] in X/ker(T) and so T ([x,]) > 7 ([x]) in Y. Butthis is the same as saying that T (x,) — T (x). It follows that y = Tx and so TX is indeedclosed.