Mathematics 522 Bounded Linear Operators and the Definition of Derivatives Definition. Let V , W be normed vector spaces (both over R or over C). A linear transformation or linear operator T : V ! W is bounded if there is a constant C such that (1) kT xkW ≤ CkxkV for all x 2 V . Remark: We use the linearity of T and the homogeneity of the norm in W to see that x T (x) kT (x)k T = = W kxkV W kxkV W kxkV we see that T is bounded, satisfying (1), if and only if sup kT (x)kW ≤ C: kxkV =1 Theorem. Let V , W be normed vector spaces and let T : V ! W be a linear transformation. The following statements are equivalent. (i) T is a bounded linear transformation. (ii) T is continuous everwhere in V . (iii) T is continuous at 0 in V . Proof. (i) =) (ii). Let C as in the definition of bounded linear transfor- mation. By linearity of T we have kT (v) − T (~v)kW = kT (v − v~)kW ≤ Ckv − v~kV which implies (ii). (ii) =) (iii) is trivial. (iii) =) (i): If T is continuous at 0 there exists δ > 0 such that for all v 2 V with kvk < δ we have kT vk < 1. Now let x 2 V and x 6= 0. Then x x δ = δ=2 and thus T (δ ) < 1: 2kxkV V kxkV W But by the linearity of T and the homogeneity of the norm we get x T (x) δ 1 ≥ T (δ ) = δ = kT xkW kxkV W 2kxkV W 2kxkV and therefore kT xkW ≤ CkxkV with C = 2/δ. Notation: If T : V ! W is linear one often writes T x for T (x). Definition. We denote by L(V; W ) the set of all bounded linear trans- formations T : V ! W . 1 2 L(V; W ) form a vector space. S + T is the transformation with (S + T )(x) = S(x) + T (x) and cT is the operator x 7! cT (x). On L(V; W ) we define the operator norm (depending on the norms on V and W ) by kT vkW kT kL(V;W ) ≡ kT kop = sup : v6=0 kvkV We can think of kT kL(V;W ) as the best constant for which (1) holds. Note that kT xkW ≤ kT kL(V;W )kxkV : Using the homogeneity of the W -norm we also can write kT kL(V;W ) = sup kT xkW : kxkV =1 We use the k · kop notation if the choice of V , W and the norms are clear n m from the context. In the textbook, Rudin considers V = R , W = R with the standard Euclidean norms and simply writes kT k for the operator norm. Lemma. Let V and W be normed spaces. If V is finite dimensional then all linear transformations from V to W are bounded. Pn Proof. Let v1; : : : ; vn be a basis of V . Then for v = j=1 αjvj we have n n n X X X kT vkW = αjT vj ≤ jαjjkT vjkW ≤ kT vjkW max jαkj: W k=1;:::;n j=1 j=1 j=1 The expression maxk=1;:::;n jαkj defines a norm on V . Since all norms on V are equivalent, there is a constant C1 such that n X max jαjj ≤ C αjvj j=1;:::;n V j=1 for all choices of α1; : : : ; αn. Thus we get kT vkW ≤ CkvkV for all v 2 V , Pn where the constant C is given by C = C1 j=1 kT vjk: n m Pn 2 1=2 Lemma. On R , R use the Euclidean norms kxk2 = ( j=1 jxjj ) , Pm 2 1=2 kyk2 = ( i=1 jyij ) . Let A be an m × n matrix and consider the linear n m operator T : R ! R defined by T (x) = Ax. Let m n X X 2 1=2 kAkHS := ( jaijj ) : i=1 j=1 Then kT kop ≤ kAkHS: 2 Pm Pn 2 Proof. kAxk2 = i=1 j j=1 aijxjj . By the Cauchy-Schwarz inequality Pn 2 Pn 2 2 j j=1 aijxjj ≤ j=1 jaijj kxk2 and hence kAxk2 ≤ kAkHSkxk2 for all n x 2 R . Thus kT kop ≤ kAkHS. 3 Remark. We often write A for both the matrix and the linear operator x ! Ax. The expression kAkHS is a norm on the space of m × n matrices called the Hilbert-Schmidt norm of A. In many cases it is substantially larger then the operator norm (and so the estimate in the Lemma is rather inefficient). To illustrate this let D an n × n diagonal matrix (with entries λ1,...,λn on the diagonal and zeroes elsewhere. Then it is easy to verify the inequality kDxk2 ≤ max jλjjkxk2 j=1;:::;n and this is seen to be sharp by testing on the unit vectors x = ej. Thus kDk = max jλ j while kDk = (Pn jλ j2)1=2. In particular for op j=1;:::;n j HS j=1 j p the identity operator I we have kIkop = 1 but kIkHS = n. Normed vector spaces and compleness. Let V be a normed space. Then V is a metric space with respect to the metric d(x; y) = kx−yk. Recall 1 that a sequence fxngn=1 is a Cauchy-sequence in V if for every " > 0 there is K such that kxn − xmkV < " for m; n ≥ K. Definition. A Banach space is a normed space which is complete (i.e. every Cauchy sequence converges). Lemma: A finite dimensional normed space over R or C is complete. Proof. Let v1; : : : ; vN be a basis of V and for x 2 V let β1(x); : : : ; βN (x) be the coordinates of f with respect to the basis v1; : : : ; vN . We need to show that fxng has a limit in V . For every n we can write N X xn = βi(xn)vi: i=1 The expression kxk1 := max jβi(x)j i=1;:::;N defines a norm on V which is equivalent to the given norm. In particular we have jβi(xn) − βi(xm)j ≤ Ckxn − xmkV ; i = 1;:::;N: This shows that for each i = 1;:::;N the numbers βi(xn) form a Cauchy- PN sequence of scalars and thus converge to a scalar βi. Define x = i=1 βivi (so that βi = βi(x)). Then N n X X kxn − xkV = k βi(xn) − βi vikV ≤ jβi(xn) − βijkvikV i=1 i=1 which converges to 0 as n ! 1. Hence xn ! x in V . 4 Theorem: Let V and W be normed spaces and assume that W is com- plete. Then the space L(V; W ) of bounded linear operators from V to W is a Banach space. Proof. We need to show that L(V; W ) is complete. Let Tn be a Cauchy sequence in L(V; W ) (with respect to the norm kT k = sup kT xk ). op kxkV =1 W Then for each x 2 V , Tnx is Cauchy in W and by assumption Tnx converge to some vector T x. Check, using limiting arguments, that T : V ! W is linear. Since Tn is Cauchy the sequence TN is bounded in L(V; W ) and thus there exists M with kTnkop ≤ M. We need to show that T is a bounded operator. We have for all n and all x 2 V kT xkW ≤ kT x − TnxkW + kTnxkW ≤ kT x − TnxkW + kTnkopkxkW : Letting n ! 1 we see that kT xkW ≤ MjxkV . Hence T 2 L(V; W ) with kT kop ≤ M. Finally we need to show that Tn ! T in L(V; W ). Given " > 0 we choose N such that kTn − Tmkop < "=2 for m; n ≥ N. We prove the following Claim: For each x 2 V , x 6= 0, and n ≥ N we have kTnx − T xkW < "kxkV : Let n ≥ N. Since Tmx ! T x in W we can choose m(x) ≥ N such that kTm(x)x − T xkW ≤ "=4kxkV . We estimate we estimate (with m ≥ N) kT x − TnxkW ≤ kT x − Tm(x)x + Tm(x)x − Tn(x)kW ≤ kT x − Tm(x)xkW + kTm(x)x − TnxkW " ≤ kxk + kT − T k kxk 4 V m(x) n op V " " ≤ kxk + kxk < "kxk : 4 V 2 V V Hence the claim is established and thus kTn − T kop < " for n ≥ N. This shows Tn ! T with respect to the operator norm. Definition. (i) Let V be a normed vector space over R. Bounded linear transformations from V to R are called bounded linear functionals. The ∗ space L(V; R) with the operator norm is called the dual space to V , or V . (ii) Similarly if V be a normed vector space over C we call the bounded linear transformations from V to R bounded linear functionals and refer to ∗ the space L(V; C) with the operator norm as the dual space V . Since R and C are complete the above theorem about completeness of L(V; W ) immediately yields the Corollary. Dual spaces of normed vector spaces are Banach spaces. 5 n n Exercise: The space L(R ; R) of linear functionals ` : R ! R can be n identified with R, since for any such ` there is a unique vector y 2 R such Pn n n that `(x) = j=1 xjyj for all x 2 R . We consider the p-norm on R , Pn p 1=p kxkp = ( j=1 jxjj ) , 1 ≤ p < 1. Also set kxk1 = maxj=1;:::;n jxjj. The norm on R is just the absolute value j · j. (i) Prove Pn j=1 xjyj sup = kyk1 kxk16=0 kxk1 and Pn j=1 xjyj sup = kyk1: kxk16=0 kxk1 (ii) Prove that if 1 < p < 1 then Pn xjyj j=1 0 p k`kop ≡ sup = kykp0 where p = : kxkp6=0 kxkp p − 1 Hint: H¨older'sinequality should be used here.