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Functional Analysis Exercises Solutions – Sheet 12 Prof. Dr. Reiner Lauterbach Summer term 2016 Jan Henrik Sylvester Functional analysis exercises Solutions – Sheet 12 Problem 45. 1. Show the resolvent equation: for T ∈ L(X) and λ, µ ∈ PT , we have R(T, λ) − R(T, µ) = (λ − µ)R(T, λ)R(T, µ) and conclude that the resolvent is either nowhere compact or compact for every µ ∈ PT . 2. Show that for a compact operator on an infinite dimensional B-space 0 is always in the spectrum. Which part of the spectrum can contain 0? Give examples. Solution. 1. We have R(T, λ) − R(T, µ) = R(T, λ)((T − µ1) − (T − λ1))R(T, µ) = (λ − µ)R(T, λ)R(T, µ). If R(T, λ) is compact, the same applies for the right side of the resolvent equation, hence for the left side, too, and R(T, µ) is compact, too. 2. We have already show in Exercise 44 that 0 is part of the spectrum of every compact operator on an infinite dimensional B-space. E For the zero operator, we have 0 ∈ Σ0 . 2 2 1 T : ` → ` : {xn}n∈ 7→ xn is compact. T − 01 = T is injective N n n∈N C with dense image, but not surjective. Thus, we have 0 ∈ ΣT . sh+ ◦ T is compact, injective and the image is not dense. We have 0 ∈ R Σsh+ ◦T . Problem 46. Let X, Y be B-spaces and L : X → Y a linear mapping. Show that L is norm continuous, iff L is continuous with respect to the weak topologies on X and Y . 00 0 (Hint for the only-if part: Consider the functionals Lx ∈ Y with Lx(y ) = 0 X y (Lx) for x ∈ B1 (0) and apply the uniform boundedness principle / the Banach-Steinhaus theorem.) Solution. Let L be norm continuous, U ⊆ Y be an open basis set given by n 0 0 0 0 functionals y1, . , yn ∈ Y and the inequalities |yi(y)| < ε. Then we have −1 0 L (U) = {x ∈ X | |yi(Lx)| < ε ∀ i} 0 0 = {x ∈ X | |L (yi)(x)| < ε ∀ i} . 0 0 For the functionals L (yi), these are also the basis set of the weak topology on X, hence L is weakly continuous. For the other direction, let L be weakly continuous. Then y0 ◦ L : X → K is a weak functional for every y0 ∈ Y 0 and by Exercise 43 also a weak functional. 00 0 0 Consider the functionals Lx ∈ Y , Lx(y ) = y (Lx) for kxkX ≤ 1. Then we have 0 0 0 0 |Lx(y )| = |y (Lx)| ≤ ky ◦ LkX0 · kxkX < ky ◦ LkX0 1 Prof. Dr. Reiner Lauterbach Summer term 2016 Jan Henrik Sylvester Thus 0 X |Lx(y )| < ∞ ∀ x ∈ B1 (0). By the uniform boundedness principle, we have sup kLxkY 00 < ∞. X x∈B1 (0) X 0 0 0 For x ∈ B1 (0), let yx be a functional with kyxkY 0 = 1 and yx(Lx) = kLxkY . Then we have ∞ > sup kLxkY 00 kxkX ≤1 0 = sup sup |Lx(y )| 0 kxkX ≤1 ky kY 0 =1 = sup sup |y0(Lx)| 0 kxkX ≤1 ky kY 0 =1 0 ≥ sup |yx(Lx)| kxkX ≤1 = sup kLxkY . kxkX ≤1 Hence, L is norm continuous. Problem 47. 1. Let X be a B-space and L ∈ L(X, Y ). Show that if IM L has a finite dimensional algebraic complement, then IM L is already closed (hence, the requirement of Fredholm operators to have a closed image is not necessary). 2. Proof by counterexample that the proposition in the previous part of the exercise is not generally true, if it is not about the image of a continuous operator. That is: Find a finite codimensional subspace that is not closed. Solution. 1. WLOG, L is injective, otherwise we restrict it to the complement of its kernel. Let W ⊆ Y be a finite dimensional subspace with IM L ⊕ W = Y (algebraically). W is finite dimensional, hence there is a closed complement W ⊥ ⊆ Y of W . Let P : Y → W ⊥ be a continuous projector onto W ⊥. Then P ◦ L : X → W ⊥ is an isomorphism: Clearly, it is continuous and injective. Consider any y ∈ W ⊥. Since Y = IM L⊕W , we have y = Lx + w for some w ∈ W . Then we get P (y − Lx) = 0 = y − P Lx and hence y ∈ IM P ◦L. By the open mapping theorem, P ◦L is a toplinear ⊥ isomorphism. Let G : W → X be its inverse and xn ∈ X a sequence with ⊥ Lxn → y ∈ Y . Then we have P ◦ L(xn) → P y ∈ W . Set x = G ◦ P (y). Consider Lxn = L◦G◦P ◦Lxn. The left side converges to y and the right side to Lx, hence y = Lx and IM L is closed. 0 2. In C ([0, 1]; R), consider the subspace U of the polynomials. Let {pi}i∈I be a basis of U. Complete that set to a basis B of C0([0, 1]; R). Let f be an element of the basis that is not a polynomial. The span of B \{f} is a subspace of codimension 1, which contains all polynomials and is therefore dense and not closed. Problem 48. 1. Show that the spectrum of a bounded linear operator T ∈ L(X) is in the ball B (0) ⊂ . kT kL(X) C 2 Prof. Dr. Reiner Lauterbach Summer term 2016 Jan Henrik Sylvester 2. The spectrum of a bounded linear operator is not empty. 3. For sh± : `p → `p, calculate the spectrum (divided in point spectrum, continuous spectrum, and residual spectrum). Solution. 1. If |λ| > kT kL(X), then the sequence ∞ X λ−nT n n=0 converges and represents the operator (1 − λ−1T )−1. Multiplying by −λ−1, we obtain (T − λ1)−1. 2. The resolvent is holomorphic on the resolvent set. If the spectrum ΣT = ∅, then the resolvent is holomorphic on the whole space C. Since T is bounded, we write the resolvent as R(T, λ) = (T − λ1)−1 = −λ−1(1 − λ−1T )−1. For λ ∈ B (0), this function is bounded, for λ > kT k it is kT kL(X) L(X) a perturbation of the identity with norm smaller than 1 and arbitrarily small for large values of |λ|. Thus, R(T, λ) is bounded and hence it is constant by Liouville’s theorem. This is obviously not the case and by this contradiction we got ΣT 6= ∅. 3. We begin with the observation that both shift operators have norm 1. For 1 ± 1 |λ| > 1, we obtain form the Neumann sequence that the operator λ sh − is invertable. λ cannot be in any part of the spectrum. Now we calculated the point spectrum: If sh+ x − λx = 0, then x = 0. Thus, sh+ −λ1 is injective for any λ. If sh− x − λx = 0, then we have n−1 p xn = λ x1 for any x1 ∈ C. This sequence is in ` for 1 ≤ p < ∞, iff |λ| < 1. Thus, the point spectrum of sh− is given by the unit ball p B1(0) ⊆ C. For p = ∞, the sequence is in ` for |λ| = 1, too, hence in that case, the point spectrum is B1(0) ⊆ C and the other parts of the spectrum are empty. To calculate the other parts of the spectrum in the case 1 ≤ p < ∞, we observe that an operator has a dense image, iff the dual operator is injective (Theorem 2.4.5.6). The dual operator of sh− is sh+ and vice verse (canonically identifying the dual spaces). Then sh− −λ1 has a dense image + for every λ ∈ C, the residual spectrum is empty. Moreover, sh −λ1 has a dense image for |λ| ≥ 1 and the residual spectrum is the open unit ball for 1 < p < ∞ and the closed unit ball for p = 1. ± We still need to consider λ ∈ S1 in the case 1 ≤ p < ∞. There, sh are both injective and have a dense image. We will show that sh+ −λ1 is not surjective for these values of λ. By the closed range theorem, sh− −λ1 is not surjective and we obtain that the unit sphere is the continuous spectrum of both operators. 3 Prof. Dr. Reiner Lauterbach Summer term 2016 Jan Henrik Sylvester We need to solve + iϕ iϕ iϕ sh x − e x = (−e x1, x1 − e x2,... ) = y for x for any y ∈ `p, ϕ ∈ [0, 2π]: iϕ e xk = −yk + xk−1 and by induction iϕ −iϕ −i2ϕ −i(k−1)ϕ e xk = −yk − e yk−1 − e yk−2 − · · · − e y1. The sequence on the left side is obviously in `p. Then it is sufficient to chose y such that the right side diverges. This is the case for e−i(k−1)ϕ y = , k k since then we have k X 1 −y − e−iϕy − e−i2ϕy − · · · − e−i(k−1)ϕy = −e−ikϕ · . k k−1 k−2 1 j j=1 In the case p = ∞, we still need the other part of the spectrum for sh+. For |λ| ≤ 1, we will see that the image is not dense. Solving (sh+ −λ1)x = y Pk −(k−i+1) as above, we get xk = i=1 −λ yi. Thus, for y = {1}n∈N, the sequence xk would not be bounded and the same applies for any near by y. Hence, the closed unit ball is the residual spectrum. Exercises are due on July 5, 2016. Space of the week Name: HP ( 1 ) C 2π p f : B1 (0) → C 1 R iθ p Definition: lim f re dθ < ∞ holomorphic r→1 2π 0 1 2π p 1 R iθ p Norm: kfkHp = lim 2π f re dθ r→1 0 Dual space: Dual space of: Reflexive: Criterion for com- pactness: Criterion for weak convergence: Additional aspects: 4.
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