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Compactness, Fredholm Alternative and Spectrum I

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Compactness, Fredholm Alternative and Spectrum I Compactness, Fredholm Alternative and Spectrum I. Compactness Definition. We say that E ⊆ X is compact if from eavery open covering of E it is possible to extract a finite subcovering of E. Definition. A subset E of a normed space X is sequentially compact if for every sequence {xk}⊂E there exists a subsequence {xk}, converging in X. Theorem 1. Let X be a normed space and E ⊂ X. Then E is compact iff it is sequentially compact. • While a compact set is always closed and bounded, the following example exhibits a closed and bounded set which is not compact in `2. Example. Consider the real Hilbert space ∞ 2 ∞ 2 ` = {x = {xk}k=1 : X xk < ∞,xk ∈ R} k=1 endowed with ∞ ∞ 2 2 hx, yi = X xkyk, and kxk = X xk. k=1 k=1 1 2 Let E = {ek}k≥1, where e = {1, 0, 0, ···}, e = {0, 1, 0, ···}, etc. 2 Observe that E constitute an orthonormal basis in ` . Thus, E is closed√ and 2 bounded in ` . Howeve, E is not sequentially compact. Indeed, kei −ejk = 2, k if j =6 k, and therefore no subsequence of {e }k≥1 can be convergent. • In fact, only in finite-dimensional spaces each closed and bounded set is compact. Theorem 1. Let B be a Banach space. B is finite-dimensional iff the unit ball {x : kxk≤1} is compact. II. Weak Convergence and Compactness. Definition. Let H be a Hilbert space with inner product h·, ·i and norm k·k.A sequence {xk}⊂H converges weakly to x ∈ H and we write xk *x,if ∗ hF, xki∗ →hF, xi∗, ∀F ∈ H . • The convergence in norm is called strong convergence. • From Riesz’s representation theorem, it follows that {xk}⊂H converges weakly to H iff hxk,yi→hx, yi, ∀y ∈ H. • The weak limit is unique, since xk → x and xk → z implv hx − z,yi, ∀y ∈ H, whence x = z. Typeset by AMS-TEX 1 2 • Strong convergence implies weak convergence, since Schwarz’s inequality gives khxk,yik ≤ kxk − xkkyk. – The two notions of convergence are equivalent in the finite-dimensiona spaces. – It is not so in infinite dimensions, as the following example shows. 2 Example. Let H = L (0, 2π). The sequence vk = cos kx, k ≥ 1, is weakly converhent to zero. In fact, ∀f ∈ L2(0, 2π), the Riemann-Lbesgue theorem on the Fourier coefficients of f implies that 2π Z 0 hf,vki = f(x) cos kx dx → 0, as k →∞. However √ kukkL2(0,2π) = π and therefore {vk}k≥1 does not converge strongly. Example. Let {ϕ}i=1,2,··· denote an orthonormal system in the Hilbert space H, and we observe √ kϕi − ϕj k = 2, ∀i, j ∈ N,i=6 j. Consequently {ϕi}i=1,2,··· does not contains a strongly convergent subsequence. According to Bessel’s inequality, ∞ X |hϕ, fi|2 ≤kfk2 < ∞, ∀f ∈H, i=1 and we infer lim hϕi,fi =0=h0,fi∀f ∈H. i→∞ We therefore obtain ϕ*0asi →∞. Note that k0k < 1 = lim inf kϕk. k→∞ Principle of Uniform Boundedness. On the Hilbert space H let the sequence of bounded functionals An : H → C with n ∈ N be given, such that each element f ∈ H possesses a constant cf ∈ [0, +∞) with the property (1) |Anf|≤cf ,n=1, 2, ··· . Then we have a constant α ∈ [0, ∞) satisfying kAnk≤α, ∀n ∈ N. Proof. Step 1. Choose f0 ∈ H, ε>0 and a constant c ≥ 0. Let A : H → C denote a bounded linear functional such that (2) |Af|≤c ∀f ∈ H with kf − f0k≤ε. Then we have 2c kAk≤ ; ε k k≤ ∃ k − k≤ 1 − indeed, if x 1, f with f f0 ε such thar x = ε (f f0), and we infer 1 1 1 2c |Ax| = Af − Af0 ≤ (|Af| + |Af0|) ≤ . ε ε ε ε 3 Step 2. If the statement (2) does not hold true, Step 1 together with the continuity of the functional {An} enables us to construct a sequence of balls Σn := {f ∈ H : kf − fnk≤εn},n∈ N satisfying Σ1 ⊃ Σ2 ⊃··· with εn & 0asn →∞and an index-sequence 1 <n1 < n2 < ··· such that |Anj x|≥j, ∀x ∈ Σj ,j∈ N, which yields a contradiction to (1). • We have observed that the norm in a Hilbert space is strongly continuous. With respect to weak convergence, the norm is only lower semincontinuous, as poperty (b) in the following theorem shows. Theorem 2. Let {xk}⊂H such that xk weakly converges to x. Then (a) {xk} is bounded, (b) kxk≤lim infk→∞ kxkk. Proof. (a) follows from the principle of uniform boundedness. For (b), it is enough to observe that 2 kxk ≤ lim hxk,xi≤kxk lim inf kxkk. k→∞ k→∞ • The usefulness of weak convegence is revealed by the following compactness result. Theorem 3. Every bounded sequence in a Hilbert space H contains a subsequence which is weakly convergent to an element x ∈ H. Proof. Step 1. The sequence {xj } is bounded and thus we have a constant c ∈ [0, +∞) such that kxj k≤c, j =1, 2, ··· . On account of |hx1,xj i| ≤ ckx1k, ∀j ∈ N, (1) (1) we find a subsequence {xj }⊂{xj } such that limj→∞hx1,xj i exists. - Noting that (1) |hx2,xj i| ≤ ckx2k, ∀j ∈ N, (2) (1) (2) we select a further subsequence {xj }⊂{xj } such that limj→∞hx2,xj i exists. - The continuation of this process gives a chain of subsequence (1) (2) (k) {xj }⊃{xj }⊃{xj }⊃···⊃{xj } such that (k) lim hxi,x i exists for i =1, ···k. j→∞ j (k) - The Cantor’s diagonal procedure then gives us the sequence xk such that (k) lim hxi,x i exists for all i ∈ N. k→∞ k 4 Step 2. Denote the linear subspace of all fnite linear combinations of xj by M, namely N(x) M = {x : x = X αixi,αi ∈ C,N(x) ∈ N}. i=1 Then lim hx, x(k)i exists for all x ∈M. k→∞ k Making the transition to the linear closed subspace M⊂M⊂H, lim hy, x(k)i exists for all y ∈ M. k→∞ k Step 3. Due the projection theorem, each element y ∈ H can be represented in the form ⊥ y = y1 + y2,, with y1 ∈ M, y1 ∈M . This implies the existence of (k) (k) (k) lim hy, xk i = lim hy1 + y2,xk i = lim hy1,xk i, ∀y ∈ H. k→∞ k→∞ k→∞ Step 4. From Step 3, we may define a linear functional A in H by setting Ay = lim hy, x(k)i, ∀y ∈ H. k→∞ k Consider the bounded linear functional (i) Ai(y)=hxi ,yi,y∈ H. The principle of uniform boundedness gives us a constant C ∈ [0, ∞) such that (n) kxn k = kAnk≤C, ∀n ∈ N. This and Theorem 1(b) imply kAk≤C. Thus A is a bounded linear functional and Riesz representtion theorem then gives us an element x ∈ H satisfying A(y)=hx, yi, ∀y ∈ H. We then obtain (k) lim hy, x i = lim Ak(y)=A(y)=hx, yi, ∀y ∈ H, k→∞ k k→∞ (k) which means that the sequence {xk } converges weakly. III. Compact Operators • Let V1 and V2 be normed linear spaces. Every operator in L(V1,V2) transforms bounded sets in V1 into bounded sets in V2. The subclass of operators that transform bounded sets into pre-compact sets is particularly important. 5 Definition. A subset Σ of a normed linear space V is called precompact, if each sequence {ym}⊂Σ contains a stronly convegent subsequence {ymk }, which means lim kym − yn k =0. k,`→∞ k ` Definition. Let V1 and V2 be normed linear spaces and L ∈L(V1,V2). L is said to be compact if for every bounded subset E of V1, the image L(E) is pre-compact in V2. This means that each sequence {xm} in V1 with kxmkV1 ≤ r, for some r ∈ R and all n ∈ N, contains a subsequence {xmk } such that {Lxmk }⊂V2 converges strongly. Proposition 4. A compact operator K : V1 → V2 is bounded. Proof. If K : V1 → V2 were not bounded, there would exist a sequence {xn}⊂V1 with kxmk = 1, for all n ∈ N and kKxmk2 →∞as n →∞. Therefore, we cannot select a convergent subsequence from {Kxn} in V2. • An equivalent characterization of compact operators between Hilbert spaces may be given in terms of weak convergence. Indeed, an operator is compact iff it converts weak convergence to strong convergence. Precisely, we have the following. Proposition 5. Let H1 and H2 be Hilbert spaces and L ∈L(H1,H2). L is compact iff for every sequence {xk}⊂H1, xk *x in H1 weakly implies Lxk → Lx in H2 strongy. Proof. (⇒) Let {xm}⊂H1 be a sequence with kxmk≤1, m ∈ N. According to Theorem 3, we have a subsequence {xmk }⊂{xm} which converges weakly to x in H1 as k →∞. Then, by assumption Kxmk → Kx strongly,k→∞. Consequently, the operator K : H1 → H2 is compact. (⇐) Now let K be compact and {xm} denote a sequence weakly converges to x.We then have to prove Kxm → Kx strongly as m →∞in H2. If the latter statement were false, there would exist a number d>0 and a subse- quence {xmk } of {xm} satisfying (3) kKxmk − Kxk≥d>0, ∀n ∈ N. – On the other hand, by Theorem 2(b) the sequence {xm} is bounded in H1. Since the operator is compact, we have a further subsequence {xmk }⊂{xmk } with Kxmk → y strongly, as n →∞. e e – Moreover, {xm} converges weakly to Kx, which implies y = Kx. Hence kKxmk − Kxk→0asmk →∞, in contradiction with (3). Example.
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