The Riesz–Fredholm Theory

Compact Operators. Spectral Decomposition of Self-Adjoint Compact Operators

6.2 The Riesz–Fredholm Theory

clearlyT_εsatisfies (3).

This kind of approximation is very useful, for example, to deduce Schauder’s fixed-point theorem from Brouwer’s fixed-point theorem (see, e.g., K. Deimling [1], A. Granas–J. Dugundji [1], J. Franklin [1], and Exercise 6.26). A similar construction, combined with the Schauder fixed-point theorem, has also been used in a surprising way by Lomonosov to prove the existence of nontrivial invariant subspaces for a large class of linear operators (see, e.g., C. Pearcy [1], N. Akhiezer–I. Glazman [1], A. Granas–J. Dugundji [1], and Problem 42). Another linear result that has a simple proof based on the Schauder fixed-point theorem is the Krein–Rutman theorem (see Theorem 6.13 and Problem 41).

Proposition 6.3.LetE,F, andGbe three Banach spaces. LetT ∈ L(E, F )and S∈K(F, G)[resp. T ∈K(E, F )andS∈L(F, G)]. ThenS◦T ∈K(E,G).

The proof is obvious.

Theorem 6.4 (Schauder).IfT ∈K(E,F ), thenT∈K(F,E). And conversely.

Proof. We have to show thatT(B_F)has compact closure inE. Let (v_n)be a sequence inB_F. We claim that(T(v_n))has a convergent subsequence. SetK = T (B_E); this is a compact metric space. Consider the setH⊂C(K)defined by

H= {ϕ_n:x∈K−→ v_n, x;n=1,2, . . .}.

The assumptions of Ascoli–Arzelà’s theorem (Theorem 4.25) are satisfied. Thus, there is a subsequence, denoted byϕ_nk, that converges uniformly onK to some continuous functionϕ∈C(K). In particular, we have

sup

u∈BE

v_nk, T u −ϕ(T u)−→0

k→∞ . Thus

sup

u∈B_E

v_nk, T u − v_n, T u−→0

k,→∞, i.e.,Tv_nk−Tv_nE −→0

k,→∞. ConsequentlyTv_nk converges inE.

Conversely, assume T ∈ K(F, E). We already know, from the first part, thatT ∈ K(E,F). In particular,T(B_E)has compact closure inF. But T (B_E)=T(B_E)andF is closed inF. ThereforeT (B_E)has compact closure inF.

Remark2.LetEandFbe two Banach spaces and letT ∈K(E,F ). If(u_n)converges weaklytouinE, then(T u_n)convergesstronglytoT u. The converse is also true if Eis reflexive (see Exercise 6.7).

6.2 The Riesz–Fredholm Theory

We start with some useful preliminary results.

Lemma 6.1 (Riesz’s lemma).LetEbe ann.v.s.and letM⊂Ebe a closed linear space such thatM =E. Then

∀ε >0 ∃u∈Esuch thatu =1and dist(u, M)≥1−ε.

Proof. Letv∈Ewithv /∈M. SinceMis closed, then d =dist(v, M) >0.

Choose anym0∈Msuch that

d ≤ v−m0 ≤d/(1−ε).

Then

u= v−m0

v−m0

satisfies the required properties. Indeed, for everym∈M, we have u−m =

v−m₀ v−m₀−m

≥ d

v−m₀ ≥1−ε, sincem₀+ v−m₀m∈M.

Remark3.IfMis finite-dimensional (or more generally ifMis reflexive) one can chooseε=0 in Lemma 6.1. But this is not true in general (see Exercise 1.17).

•Theorem 6.5 (Riesz). Let E be an n.v.s. with B_E compact. Then E is finite- dimensional.

Proof. Assume, by contradiction, that E is infinite-dimensional. Then there is a sequence (E_n) of finite-dimensional subspaces of E such that E_n−1 ⊂ E_n and E_n−1 = E_n. By Lemma 6.1 there is a sequence (u_n) withu_n ∈ E_n such that u_n =1 and dist(un,E_n−1)≥ 1/2. In particular,u_n−u_m ≥ 1/2 form < n.

Thus(u_n)has no convergent subsequence, which contradicts the assumption thatB_E is compact.

•Theorem 6.6 (Fredholm alternative).LetT ∈K(E). Then (a)N (I−T )is finite-dimensional,

(b)R(I−T )is closed, and more preciselyR(I−T )=N (I −T)^⊥, (c)N (I−T )= {0} ⇔R(I−T )=E,

(d) dimN (I −T )=dimN (I −T).

Remark4.The Fredholm alternative deals with the solvability of the equation u−T u=f. It says that

• eitherfor everyf ∈Ethe equationu−T u=f has a unique solution,

• orthe homogeneous equationu−T u=0 admitsnlinearly independent solutions, and in this case, the inhomogeneous equationu−T u=f is solvable if and only iff satisfiesnorthogonality conditions, i.e.,

6.2 The Riesz–Fredholm Theory 161

f ∈N (I−T)^⊥.

Remark5.Property (c) is familiar in finite-dimensional spaces. If dimE < ∞, a linear operator fromE into itself is injective(= one-to-one) if and only if it is surjective(=onto). However,in infinite-dimensional spaces a bounded operator may be injective without being surjective and conversely, for example the right shift (resp. the left shift) in² (see Remark 6). Therefore, assertion (c) is a remarkable property of the operators of the formI−T withT ∈K(E).

Proof.

(a) LetE₁=N (I−T ). ThenB_E1 ⊂T (B_E)and thusB_E1is compact. By Theorem 6.5,E1must be finite-dimensional.

(b) Letfn =un−T un→f. We have to show thatf ∈R(I−T ). Setdn=dist(un, N (I −T )). SinceN (I−T )is finite-dimensional, there existsvn∈N (I −T ) such thatdn= un−vn. We have

(4) f_n=(u_n−v_n)−T (u_n−v_n).

We claim thatu_n−v_nremains bounded. Suppose not; then there is a subse- quence such thatu_nk−v_nk → ∞. Setw_n=(u_n−v_n)/u_n−v_n. From (4) we see thatw_nk −T w_nk → 0. Choosing a further subsequence (still denoted byw_nk for simplicity), we may assume thatT w_nk → z. Thus w_nk → zand z∈N (I −T ), so that dist(wnk,N (I −T ))→0. On the other hand,

dist(wn, N (I−T ))= dist(un, N (I−T )) u_n−v_n =1 (sincevn∈N (I −T )); a contradiction.

Thusun−vnremains bounded, and sinceT is a compact operator, we may extract a subsequence such thatT (u_nk−v_nk)converges to some limit. From (4) it follows thatu_nk−v_nk →f+. Lettingg=f+, we haveg−T g=f, i.e.,f ∈R(I−T ). This completes the proof of the fact that the operator(I−T ) has closed range. We may therefore apply Theorem 2.19 and deduce that

R(I−T )=N (I −T)^⊥, R(I−T)=N (I−T )^⊥. (c) We first prove the implication⇒. Assume, by contradiction, that

E₁=R(I−T ) =E.

ThenE1is a Banach space andT (E1)⊂ E1. ThusT_|E1 ∈ K(E1)andE2 = (I−T )(E1)is a closed subspace ofE1. Moreover,E2 =E1(since(I−T )is injective). LettingEn=(I−T )ⁿ(E), we obtain a (strictly) decreasing sequence of closed subspaces. Using Riesz’s lemma we may construct a sequence(un) such thatun∈En,un =1 and dist(un, En+1)≥1/2. We have

T u_n−T u_m= −(u_n−T u_n)+(u_m−T u_m)+(u_n−u_m).

Note that ifn > m, thenE_n+1⊂E_n⊂E_m+1⊂E_mand therefore

−(u_n−T u_n)+(u_m−T u_m)+u_n∈E_m+1.

It follows thatT u_n−T u_m ≥dist(um, E_m+1)≥1/2. This is impossible, since T is a compact operator. Hence we have proved thatR(I−T )=E.

Conversely, assume thatR(I−T )=E. By Corollary 2.18 we know thatN (I− T)=R(I−T )^⊥= {0}. SinceT∈K(E), we may apply the preceding step to infer thatR(I −T)= E. Using Corollary 2.18 once more, we conclude thatN (I −T )=R(I−T)^⊥= {0}.

(d) Setd = dimN (I −T ) andd = dimN (I −T). We will first prove that d ≤ d. Suppose not, thatd < d. SinceN (I −T )is finite-dimensional, it admits a complement inE (see Section 2.4, Example 1). Thus there exists a continuous projectionP fromEontoN (I−T ). On the other hand,R(I−T )= N (I−T)^⊥has finite codimensiond(see Section 2.4, Example 2) and thus it has a complement (inE), denoted byF, of dimensiond. Sinced < d, there is a linear map : N (I −T ) → F that isinjectiveandnot surjective. Set S=T +◦P. ThenS∈K(E), since◦P has finite rank.

We claim thatN (I−S)= {0}. Indeed, if

0=u−Su=(u−T u)−(◦P u), then

u−T u=0 and ◦P u=0, i.e.,u∈N (I−T )andu=0. Therefore,u=0.

Applying (c) to the operatorS, we obtain thatR(I−S)=E. This is absurd, since there exists somef ∈ F withf /∈R(), and so the equationu−Su=f has no solution.

Hence we have proved thatd≤d. Applying this fact toT, we obtain dimN (I−T)≤dimN (I−T)≤dimN (I−T ).

ButN (I −T)⊃N (I−T )and therefored=d.