Functional Analysis I Course Notes by Stefan Richter Transcribed and Annotated by Gregory Zitelli Bounded Linear Operators Let ' 2 L1(µ) and X = Y = Lp(µ). Then we can define a linear operator p p M' : L ! L by pointwise multiplication M'(f)(x) ! '(x)f(x). Furthermore, this operator is bounded, since Z Z p p p p p p kM'fkLp(µ) = j'fj dµ ≤ k'kL1(µ) jfj dµ = k'kL1(µ)kfkLp(µ) X X which imples that kM'k ≤ k'kL1(µ). In many situations, we actually have equality for operators defined in a similar way. Now let (X; M; µ) be a finite measure space. Then there is a natural inclusion Lp(µ) ⊆ Lq(µ) for p > q ≥ 1. This fact follows directly from Holder's inequality, by p 1 1 letting r = q > 1 and s + r = 1, Z Z 1=s Z 1=r q s qr 1=r jfj dµ ≤ 1 dµ jfj = µ(X) kfkLp(µ) X X X The inclusion i : Lp(µ) ! Lq(µ) is a bounded operator, with kik ≤ µ(X)1=sq. The differential properties of the Sobolev spaces are not preserved by a dif- ferential operator within an individual space itself, however we doe have that d k;p k−1;p dx : W [a; b] ! W [a; b] is a bounded linear operator, since k Z p X (j) p kfkW k;p = jf j dµ j=0 p k Z d X (j) p p f = jf j dµ ≤ kfkW k;p dx k−1;p W j=1 Functional Analysis I Part 1 1 Taking the derivative of functions in C [a; b] with the supremum norm kfkC1 = n jf(0)j+kfk1 is still a linear operator, but is certainly unbounded since kx kC1 = 1 d n 1 but dx x C1 = n. Note that C [a; b] is not a Banach space since it is not complete. Hamel Bases Recall that if X is a vector space, then a Hamel basis of X is a maximal linearly independent set. A Hamel basis B for a vector space always exists, such that for any y 2 X there exists unique elements b1; : : : bn 2 B and scalars α1; : : : ; αn 2 F Pn where y = i=1 αibi. If E ⊆ X is a linearly independent set, then there always exists a Hamel basis B for X such that E ⊆ B, by Zorn's lemma. This allows us to make a precise definition of finite and infinite-dimension vector spaces, where depending on the cardinality of the space's Hamel basis. Hamel basis can also be used to construct examples of unbounded functions on infinite dimensional Banach spaces. The typical construction follows. If B is a Hamel basis for X, and if we are given a specific scalar ab 2 F for each b 2 B, then we can define a linear transformation on X given by f : X ! F such that f(b) = ab. Pn Such a transformation can be extended linearly so that for y 2 X, y = i=1 αibi, Pn bi we have f(y) = i=1 αia . 1 Suppose that we have C[0; 1] under the supremum norm, and we let ffngn=1 ⊆ C[0; 1] be a countably infinite, linearly independent set of elements. By the pre- ceding remarks there is a Hamel basis B = ffng [ E for C[0; 1]. Then we can n define a function ' : B ! F by '(fn) = 2 kfnk and '(b) = 0 for b 2 E, which by the previous paragraph can be extended linearly to all of X. Now if we set PN 1 gN = j fj, where kgN k ≤ 1, then we have the interesting situation i=1 2 kfj k N X 1 j'(g )j = '(f ) = N N 2jkf k j j=1 j which demonstrates that ' is unbounded. Such a construction relies heavily on the axiom of choice due to the very existence of the Hamel basis. In the finite dimensional case (where the axiom of choice cannot be evoked), all linear operators are bounded. Proposition. Let X be a normed space, x0 2 X, ' : X ! F a linear functional with '(x0) 6= 0. Then X = ker ' + fαx0 : α 2 Fg and ' is bounded () ker ' is closed () ker ' is not dense in X. 2 Functional Analysis I Part 1 Proof. Let x 2 X, then '(x) '(x) x = x − x0 + x0 '(x0) '(x0) '(x) '(x) where x − x0 2 ker ' and x0 = αx0 for some appropriate α. '(x0) '(x0) Now suppose that ' is bounded. Let xn 2 ker ' with xn ! x, then '(xn) = 0 for all n which implies that '(x) = 0 and so x 2 ker '. Next, suppose that ker ' is closed. Since '(x0) 6= 0 we know that x0 2= ker ', and so ker ' is not dense in X. Lastly, suppose that ' is not bounded, and we want to show that ker ' is dense in X. It is enough to show that there exists zn 2 ker ' such that zn ! x0, since then we can use the decomposition of X into ker ' + fαx0 : α 2 Fg to prove denseness. Using the assumption that ' is unbounded, take xn with kxnk = 1 where j'(xn)j ! 1. Now write xn = yn + αnx0 with yn 2 ker ', so that j'(xn)j = jα jj'(x )j, which implies that jα j ! 1. Then the elements z = − 1 y approx- n 0 n n αn n imate x0. Finite-Dimensional Spaces Definition. Let k · k1 and k · k2 be two norms over a vector space X. We say that the norms are equivalent if there exist constants c; C > 0 such that ckxk1 ≤ kxk2 ≤ Ckxk1 for all x 2 X. Two norms are equivalent whenever their convergence sequences are equivalent, or similar when their Cauchy sequences are equivalent. This is also the same as saying that the two norms generate the same topologies on X. Theorem. If X is a finite dimensional normed space, then any two norms are equivalent. N Proof. Let k·k be some norm on X, and fejgj=1 a Hamel basis. Then for any x 2 X, there exist unique scalars 'j(x) 2 F (which can be thought of as linear functionals PN on X) such that x = j=1 'j(x)ej. Let kxk1 = max1≤j≤n j'j(x)j, which is easy to verify as a norm on X. Now we can simply use the triangle inequality on the original norm so that 3 Functional Analysis I Part 1 N N N X X X kxk = 'j(x)ej ≤ j'j(x)jkejk ≤ kxk1 kejk j=1 j=1 j=1 PN so that we can take C = j=1 kejk. Now suppose that the lower inequality is not true, so that there is a sequence of xn 2 X such that kxnk1 = 1 but kxnk ! 0. However, j'j(xn)j ≤ kxnk1 ≤ 1. Using PN a truncated diagonalization argument, there is a subsequence xnk = j=1 'j(xnk )ej PN which converges to j=1 αjej in the original norm. Since kxnk ! 0, we have that αj ! 0 for each j, and so 'j(xnk ) ! 0 which contradicts the fact that kxnk k = 1. Therefore, the lower inequality must hold, and so the original norm k·k is equivalent to k · k1. Since this can be done for any k · k, we can show the equivalence of any two norms on the space. Since all norms on a finite dimensional space are equivalent, there is a natural isomorphism between them and Cn under the Euclidean norm. Corollary. If X and Y are normed spaces, with X finite dimensional, then every linear operator T : X ! Y is bounded. Proof. Let T :(X; k · k1) ! Y , then N ! N N X X X kT xkY = T 'j(x)ej ≤ j'j(x)jkT ejk ≤ kxk1 kT ejk j=1 j=1 j=1 Riesz Representation Theorems Recall that the dual space X∗ to a vector space X is the collection of all bounded linear functionals from X into a field, typically either R or C. For Hilbert spaces, the two spaces X and X∗ are naturally isomorphic. Theorem (Riesz Representation). Let H be a Hilbert space. Then for every ' 2 H∗ there exists a unique element x0 2 H such that '(x) = hx; x0i. Similarly, every ∗ function '(x) = hx; x0i for x0 2 H defines an element of H with k'k∗ = kx0k. Consequently, there is a natural isomorphism between H and H∗. 4 Functional Analysis I Part 1 Proof. For the second part, clearly '(x) = hx; x0i is linear and continuous since j'(x)j = jhx; x0ij ≤ kxkkx0k. For the first part, ' is continuous means that ker ' = M is a closed subspace and M 6= H. Then there exists an element y0 2 H such that y0 6= 0 and y0 ?M. Then for x 2 H we can write '(x) '(x) x = x − y0 + y0 '(y0) '(y0) where the first difference is in M and the last term is orthogonal to M. Then '(x) '(x) 2 '(x) 2 hx; y0i = x − y0; y0 + ky0k = ky0k '(y0) '(y0) '(y0) '(y0) Therefore, '(x) = 2 hx; y0i. ky0k There are a variety of other kinds of representation theorems as well. If (X; M; µ) is a measure space, then the spaces Lp(µ) and Lq(µ) are isomorphically dual to one 1 1 another when 1 < p; q < 1 where p + q = 1. The proof follows by naturally pairing elements of g 2 Lq(µ) with the dual elements G 2 (Lp(µ))∗ by G(f) = R fg dµ.