all absolutely convergent series are convergent) L^p(R^n) is separable Real and complex cases. Ex 3.6.2-9 are relevant for standard applications. CHAPTER 4 Fourier series (only series) Construction of countable orthonormal bases in separable Hilbert spaces through Gramm-Schmidt. Fourierseries of periodic functions. Relation between the two, convergence in mean. Brief discussion of other types of convergence (examining Fourier series for f and F with F'=f). All exercises are about L^2. CHAPTER 5 Operators 5.1 matrices, integral operators (sum over index j replaced by integral over y) For linear maps: continuity in one point <=> continuity everywhere 5.2 bounded linear operators For linear maps: continuity in one point <=> continuity everywhere <=> bounded The norm estimate for T : X -> Y gives uniform Lipschitz continuity The best M is the norm of T Y Banach => B(X,Y) Banach (collection of all bounded linear T: X -> Y) 5.3 Banach algebra's and spectral theory. Y=X gives B(X)=B(X,X), not only a Banach space, but also an algebra, with unit element I. see my old notes fa.pdf, section 20. In algebra's, such as B(X), we can talk about invertibility. Basic tool for invertibility (I-T)^(-1)=I+T+T^2+T^3.... just the geometric series, convergent if T is not too large, e.g. norm below 1. T invertible and S close to T gives S invertible. Get used to the role of z complex being played by lambda. For asci reasons I now use z instead of lambda. For T in B(R^n) or (B(C^n) (i.e. T a matrix) the spectrum of T is the set of complex eigenvalues. Now we want to have complex spaces X, all examples naturally come with real and complex version hand in hand, but there is an abstract complexifciation procedure (norm definition is not obvious. Definition. z belongs to spectrum of (T) <=> z-T is NOT invertible in B(X) (everything complex now) The inverse defines a B(X)-valued function on the complement of the spectrum. This complement is called the resolvent set of T. The inverse of z-T is called the resolvent function (depends on A and z, for now A is fixed) Theorem. T in B(X) => sigma(T) closed (easy), bounded (easy) and NON-EMPTY (non-trivial) Proof: resolvent is complex analytic, goes to zero as z goes to infty. Liouville's theorem says that this is impossible for globally defined analytic functions. Wow! Small detail: B(X)-valued functions instead of C-valued functions. Book cheats a bit, details are postponed. Resolvent has power series expansion in 1/z. Convergent for |z| > ||T||. (In fact for |z| larger than spectral radius r(T)= lim ||T^n||^(1/n), not proved here) NB Real part of the spectrum may be empty. 5.3 continued, compact operators. In general sigma(T) consist of eigenvalues and many more other points. Special case, T is a compact operator T is compact <=> the image of every bounded sequence has a convergent subsequence => 0 is in sigma(T), all other point in sigma(T) are isolated eigenvalues, no accumulation points except possibly zero. K(X) = { compact bounded linear operators T: X -> X } K(X) is a closed ideal in B(X) 5.4 Invariant subspaces General question. Can we decompose T : X -> X into smaller parts? (think of a 4x4 matrix with zeros in the 2x2 squares top-right and bottom-left). Related question. Does T have an invariant subspace? (think of a 4x4 matrix with zeros in the 2x2 square bottom-left). Answer. In general no (counterexample hard), but yes if TK=KT for some compact bounded linear operator (non-zero of course, proof is hard/tricky). 5.4 back to concrete/easy cases, in the rest of the chapter X=H is a Hilbert space, T in B(H) is selfadjoint. Why complex? In Thm 5.18 we see that the spectrum is real. In fact the easiest theorem to prove follows from Lemma. Let H be a real Hilbert space, T: H -> H selfadjoint compact. The |(Tx,x)| has a maximum on the closed unit ball B AND every maximizer is an eigenvalue. Proof. Step 1. Maximum exists in view of compactness of T (a bit tricky, not hard). Step 2. Maximizers are eigenvectors (easy, not different from proof for 2x2 matrices). From Lemma a formulation and proof of spectral theorem follow easily. MORE TO FOLLOW