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Notes on Functional Analysis 5. Functional analysis 5.1. Normed spaces and linear maps. For this section, X will denote a vector space over F = R or C.(WewillassumeF = C unless stated otherwise.) Definition 5.1.1. A seminorm on X is a function : X [0, )whichis k·k ! 1 (homogeneous) λx = λ x • (subadditive) xk+ yk | |·x k+k y • k kk k k k We call a norm if in addition it is k·k (definite) x =0impliesx =0. • k k Recall that given a norm on a vector space X, d(x, y):= x y is a metric which k·k k − k induces the norm topology on X.Twonorms 1, 2 are called equivalent if there is a c>0suchthat k·k k·k 1 c− x x c x x X. k k1 k k2 k k1 8 2 Exercise 5.1.2. Show that all norms on Fn are equivalent. Deduce that a finite dimensional subspace of a normed space is closed. Note: You may assume that the unit ball of Fn is compact in the Euclidean topology. Exercise 5.1.3. Show that two norms 1, 2 on X are equivalent if and only if they induce the same topology. k·k k·k Definition 5.1.4. A Banach space is a normed vector space which is complete in the induced metric topology. Examples 5.1.5. (1) If X is an LCH topological space, then C0(X)andCb(X)areBanachspaces. (2) If (X, ,µ)isameasurespace, 1(X, ,µ)isaBanachspace. 1 M L M (3) ` := (xn) F1 xn < { ⇢ | | | 1} Definition 5.1.6. SupposeP (X, )isanormedspaceand(xn) (X, )isasequence. k·k N ⇢ k·k We say x converges to x X if x x as N .Wesay x converges n 2 n ! !1 n absolutely if xn < . P k k 1 P P Proposition 5.1.7. The following are equivalent for a normed space (X, ). P k·k (1) X is Banach, and (2) Every absolutely convergent sequence converges. Proof. (1) (2): Suppose X is Banach and x < .Let">0, and pick N>0suchthat ) k nk 1 x <".Thenforallm n>N, n>N k nk ≥ P m n m m P x x = x x x <". i − i i k ik k ik n+1 n+1 n>N X X X X X k (2) (1): Suppose (xn)isCauchy,andchoose n1 <n2 < such that xm xn < 2− ) ··· k − k whenever m, n > nk.Definey0 := 0 (think of this as xn0 by convention), and inductively define yk := xnk xnk 1 for all k N.Then − − 2 k y x + 2− = x +1< . k kkk n1 k k n1 k 1 k 1 ≥ X X99 Hence x := lim x = y exists in X.Since(x )isCauchy,x x. nk k n n ! ⇤ Proposition 5.1.8. PSuppose X, Y are normed spaces and T : X Y is linear. The following are equivalent: ! (1) T is uniformly continuous (with respect to the norm topologies), (2) T is continuous, (3) T is continuous at 0X , and (4) T is bounded, i.e., there exists a c>0 such that Tx c x for all x X. k k k k 2 Proof. (1) (2) (3): Trivial. ) ) (3) (4): Suppose T is continuous at 0 .ThenthereisaneighborhoodU of 0 such that ) X X TU y Y y 1 .SinceU is open, there is a δ>0suchthat x X x δ U. Thus⇢{x 2 δ|kimpliesk }Tx 1. Then for all x =0 { 2 |k k }⇢ k k k k 6 x Tx 1 δ δ = δ 1= Tx δ− x . · x ) · x )kk k k k k k k 1 (4) (1): Let ">0. If x x <c− ",then ) k 1 − 2k Tx Tx = T (x x ) c x x <". k 1 − 2k k 1 − 2 k k 1 − 2k ⇤ Exercise 5.1.9. Suppose X is a normed space and Y X is a subspace. Define Q : X X/Y by Qx = x + Y .Define ⇢ ! Qx =inf x y y Y . k kX/Y {k − kX | 2 } (1) Prove that is a well-defined seminorm. k·kX/Y (2) Show that if Y is closed, then X/Y is a norm. (3) Show that in the case of (2) above,k·k Q : X X/Y is continuous and open. Optional: is Q continuous or open only in! the case of (1)? (4) Show that if X is Banach, so is X/Y . Exercise 5.1.10. (1) Show that for any two finite dimensional normed spaces F1 and F2,alllinearmaps T : F F are continuous. 1 ! 2 Optional: Show that for any two finite dimensional vector spaces F1 and F2 endowed with their vector space topologies from Exercise 5.1.2, all linear maps T : F1 F2 are continuous. ! (2) Let X, F be normed spaces with F finite dimensional, and let T : X F be a linear map. Prove that the following are equivalent: ! (a) T is bounded (there is an c>0 such that T (B1(0X )) Bc(0F )), and (b) ker(T )isclosed. ✓ Hint: One way to do (b) implies (a) uses Exercise 5.1.9 part (3) and part (1) of this problem. Definition 5.1.11. Suppose X, Y are normed spaces. Let (X Y ):= bounded linear T : X Y . L ! { 100 ! } Define the operator norm on (X Y )by L ! T := sup Tx x 1 k k {k k|k k } =sup Tx x =1 {k k|k k } Tx =sup k k x =0 x k k6 ⇢ k k =inf c>0 Tx c x for all x X , { |k k k k 2 } Observe that if S (Y Z)andT (X Y ), then ST (X Z)and 2L ! 2L ! 2L ! STx S Tx S T x x X. k kk k·k kk k·k k·k k82 So ST S T . k kk k·k k Proposition 5.1.12. If Y is Banach, then so is (X Y ). L ! Proof. If (Tn)isCauchy,thensois(Tnx)forallx X.SetTx := lim Tnx for x X.One verifies that T is linear, T is bounded, and T T2. 2 n ! ⇤ Corollary 5.1.13. If X is complete, then (X):= (X X) is a Banach algebra (an algebra with a complete submultiplicative norm).L L ! Exercise 5.1.14 (Folland 5.1, #7). Suppose X is a Banach space and T (X). Let I (X)betheidentitymap.§ 2L 2L (1) Show that if I T < 1, then T is invertible. k − k n 1 Hint: Show that n 0(I T ) converges in (X) to T − . ≥ − L 1 1 (2) Show that if T (X) is invertible and S T < T − − ,thenS is invertible. (3) Deduce that the2P setL of invertible operatorsk GL− (Xk ) k (Xk)isopen. ⇢L n2 n2 c Exercise 5.1.15. Consider the measure space (Mn(C) ⇠= C ,λ ). Show that GLn(C) n2 ⇢ Mn(C)isλ -null. Exercise 5.1.16 (Folland 5.2, #19). Let X be an infinite dimensional normed space. § (1) Construct a sequence (xn)suchthat xn =1foralln and xm xn 1/2forall m = n. k k k − k (2) Deduce6 X is not locally compact. 5.2. Dual spaces. Definition 5.2.1. Let X be a (normed) vector space. A linear map X F is called a ! (linear) functional. The dual space of X is X⇤ := Hom(X F). Here, Hom means: ! linear maps if X is a vector space, and • bounded linear maps if X is a normed space. • Exercise 5.2.2. Suppose ','1,...,'n are linear functionals on a vector space X.Prove that the following are equivalent. n (1) ' = ↵k'k for some ↵1,...,↵n F. k=1 2 (2) There is an ↵>0suchthatforallx X, '(x) ↵ maxk=1,...,n 'k(x) . n P 2 | | | | (3) k=1 ker('k) ker('). ⇢ 101 T Exercise 5.2.3. Let X be a locally compact Hausdor↵space and suppose ' : C0(X) C is a linear functional such that '(f) 0wheneverf 0. Prove that ' is bounded. ! Hint: Argue by contradiction that '≥(f) 0 f 1 is≥ bounded using Proposition 5.1.7. { | } Proposition 5.2.4. Suppose X is a complex vector space. (1) If ' : X C is C-linear, then Re('):X R is R-linear, and for all x X, ! ! 2 '(x)=Re(')(x) i Re(')(ix). − (2) If f : X R is R-linear, then ! '(x):=f(x) if(ix) − defines a C-linear functional. (3) Suppose X is normed and ' : X C is C-linear. In Case (1), ' < implies! Re(') ' • In Case (2), k Re(k ')1< impliesk 'kkRe(k ') . Thus• ' = Re(k') . k 1 k kk k k k k k Proof. (1) Just observe Im('(x)) = Re(i'(x)) = Re(')(ix). − − (2) It is clear that ' is R-linear. We now check '(ix)=f(ix) if(i2x)=f(ix) if( x)=if(x)+f(ix)=i(f(x) if(ix)) = i'(x). − − − − (3, Case 1) Since Re(')(x) '(x) for all x X, Re(') ' . (3, Case 2) If '(x|) =0,then|| | 2 k kk k 6 '(x) = sgn('(x))'(x)='(sgn('(x)) x)=Re(')(sgn('(x)) x). | | · · Hence '(x) Re(') x ,whichimplies ' Re(') . | |k k·k k k kk k ⇤ Exercise 5.2.5. Consider the following sequence spaces. 1 ` := (xn) C1 xn < x 1 := xn ⇢ | | 1 k k | | n o c0 := (xn) C1 xXn 0asn x :=X sup xn { ⇢ | ! ! 1}k k1 | | c := (xn) C1 lim xn exists x := sup xn ⇢ n k k1 | | n !1 o `1 := (xn) C1 sup xn < x := sup xn { ⇢ | | | 1}k k1 | | (1) Show that every space above is a Banach space.
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