Linear Functional Analysis Prof. M. A. Sofi Department of Mathematics University of Kashmir Srinagar-190006 1 Bounded linear operators In this section, we shall characterize continuity of linear operators acting between normed spaces. It turns out that a linear operator is continuous on a normed linear space as soon as it is continuous at the origin or for that matter, at any point of the domain of its definition. Theorem 1.1. Given normed spaces X and Y and a linear map T : X ! Y; then T is continuous on X if and only if 9 c > 0 such that kT (x)k ≤ ckxk; 8 x 2 X: (1) Proof. Assume that T is continuous. In particular, T is continuous at the origin. By the definition of continuity, since T (0) = 0; there exits a neigh- bourhood U at the origin in X such that T (U) ⊆ B(Y ): But then 9r > 0 such that Sr(0) ⊂ U: This gives T (Sr(0)) ⊆ T (U) ⊆ B(Y ): (2) x Let x(6= 0) 2 X; for otherwise, (1) is trivially satisfied. Then kxk 2 Sr(0); so rx that by (2), T 2kxk 2 B(Y ): In other words, 2 kT (x)k ≤ ckxk where c = r which gives (1). Conversely, assume that (1) is true. To show that T is continuous on X; let 1 n n x 2 X and assume that xn −! x in X: Then xn − x −! 0 in X: Thus, 8 > 0; 9 N 3: 8 n > N; kT (xn) − T (x)k = kT (xn − x)k ≤ ckxn − xk ≤ , n which holds for all n > N: In other words, T (xn) −! T (x) in Y and therefore, T is continuous at x 2 X: Since x 2 X was arbitrarily chosen, it follows that T is continuous on X: Definition 1.2. Given a continuous linear operator T : X ! Y; we define kT k = sup kT (x)k: (3) kxk≤1 Exercise 1.3. Show that the above quantity is finite. Theorem 1.4. Given normed linear spaces X and Y; the space L(X; Y ) of continuous linear maps from X into Y can be made into a normed space by means of formula (3): kT k = sup kT (x)k: kxk≤1 Further, L(X; Y ) is a Banach space if Y is a Banach space. Proof. We leave the details of proof of the first statement of the theorem as an exercise. To show that L(X; Y ) is complete, let fTng be a Cauchy sequence in L(X; Y ): Thus, 8 > 0; 9 N 3: kTm − Tnk ≤ , 8 m; n > N: In particular, we have kTm(x) − Tn(x)k ≤ kxk; 8 x 2 X; m; n > N: (4) This shows that fTn(x)g is Cauchy in Y and therefore, converges to y = T (x) 2 Y as Y is assumed to be complete. Claim: T 2 L(X; Y ) and Tn ! T with respect to (3). Writing T (x) = lim Tn(x); x 2 X; it is easily seen that T is linear. Further, n!1 using the fact k:k defines a continuous function on X(verify!), we see that kT (x)k = k lim Tn(x)k n!1 = lim kTn(x)k: (5) n!1 2 Since Cauchy sequence are bounded in a metric space, 9 c > 0 such that kTnk ≤ c; 8 n ≥ 1 ) kTn(x)k ≤ ckxk 8 n ≥ 1; 8 x 2 X: Combining with (5), the above estimate gives: kT (x)k ≤ lim ckxk = ckxk; 8 x 2 X n!1 which means that T is continuous, by virtue of Theorem 1.1. Finally, letting m ! 1 in (4) gives: kTn(x) − T (x)k ≤ kxk; x 2 X; 8 x 2 X; 8 n > N: Taking kxk ≤ 1; gives: kTn − T k = sup kTn(x) − T (x)k ≤ , x 2 X; 8 x 2 X; 8 n > N; kxk≤1 n which shows that Tn −! T in L(X; Y ) with respect to the norm defined by (3). Remark 1.5. We shall prove later that converse of Theorem 1.4 also holds that is L(X; Y ) is complete if and only if Y is complete. Exercise 1.6. Show that for T 2 L(X; Y ); one has kT k = sup kT (x)k = sup kT (x)k: kxk=1 kxk<1 Solution. Denote the above quantities on the right by A and B; respectively. x Fix x(6= 0) 2 X such that kxk ≤ 1: Then for y = kxk ; kyk = 1; so that kT (y)k ≤ A: Equivalently, kT (x)k ≤ Akxk: Taking supremum over kxk ≤ 1 gives kT k = sup kT (x) ≤ A ≤ kT k kxk≤1 which gives kT k = A: To show that kT k = B; choose x 2 X; kxk ≤ 1 and > 0 arbitrarily. x Then for y = kxk+ ; kyk < 1; so that kT (y)k ≤ B: Equivalently, kT (x)k ≤ B(kxk + ) ≤ B(1 + ): Taking supremum over kxk ≤ 1 gives: kT k = sup kT (x)k ≤ B(1 + ): kxk≤1 Since > 0 was arbitrarily chosen, letting ! 0 gives kT k ≤ B ≤ kT k; i.e., kT k = B: 3 Definition 1.7. For a normed linear space X; the space X∗ = L(X; K) is called the (topological) dual of X: Elements of X∗ shall be referred to as the bounded(continuous) linear functionals on X: Example 1.8. (a) The linear map T : lp ! lp(1 ≤ p ≤ 1) defined by T (x1; x2; ··· ; xn; ··· ) := (x2; x3; ··· ; xn; ··· ) is a bounded linear operator on lp: Indeed, 1 !1=p 1 !1=p X p X p kT (¯x)k = jxnj ≤ jxnj = kx¯k; 8 x¯ 2 lp: n=2 n=1 Example 1.8. (b) T : C[0; 1] ! C[0; 1] defined by x Z T (f) := f(t)dt; x 2 [0; 1] 0 is a bounded linear operator on C[0; 1](verify!). Example 1.8. (c) T : l1 ! l1 defined by x 1 T (¯x) := n n2 n=1 gives a well-defined linear map such that 1 1 ! x 1 X n X kT (¯x)k = 2 ≤ 2 supjxnj n n n≥1 n=1 n=1 π2 = kx¯k; 8 x¯ 2 l ; 6 1 π2 which show that T is continuous and kT k = 6 : 2 Finite Dimensional Banach Spaces In this section, we shall see that as opposed to the infinite dimensional Banach spaces where it is possible for a linear map to be discontinuous, every linear map on a finite dimensional Banach space is always continuous. To show that it is indeed the case, we shall prove a series of result valid for finite dimensional spaces which will yield the stated results as a simple consequence. 4 Definition 2.1. The norms k:k1 and k:k2 on a vector space X are defined to be equivalent if the topologies generated by the metrics induced by each of them are coincident. Proposition 2.2. The norms k:k1 and k:k2 are equivalent on X if and only if 9 c1; c2 > 0 such that c1kxk1 ≤ kxk2 ≤ c2kxk1; 8 x 2 X: (6) Proof. Necessity: Assume that k:k1 and k:k2 are equivalent and consider the identity map i :(X; k:k1) ! (X; k:k2) i(x) = x: The given condition yields that i is continuous as a linear and therefore, by Theorem 1.1, 9 c2 > 0 such that kxk2 = ki(x)k2 ≤ ckxk1; 8 x 2 X: This gives the right side of the desired inequality. Interchanging the roles of kxk1 and kxk2 gives the other inequality. Conversely, to show that the norms k:k1 and k:k2 are equivalent, it suffices to show that i given above is continuous. Indeed, let xn ! 0 in (X; k:k1) and fix > 0 arbitrarily. Then (6) gives: kxnk2 ≤ c2kxnk1 ≤ c2 = , 8 n > N: c2 Here N has been chosen so that kxnk1 < /c2; 8 n > N: This gives xn ! 0 in (X; k:k2) which shows that i is continuous. In a similar manner, we can −1 show that i :(X; k:k2) ! (X; k:k1) is continuous. Since i = i it follows that is a homeomorphism yielding thereby that the topologies induced by the given norms are coincident. n n n o Theorem 2.3. B (l1) = x¯ 2 : max jxij ≤ 1 is compact. R 1≤i≤n n Proof. Since we are dealing with metric spaces, it suffices to show that B (l1) (m) 1 n is sequentially compact. To this end, let (x )m=1 be a sequence in B (l1) : There exist M > 0 such that (m) (m) jxi j ≤ kx k ≤ M; 8 m > 1; 8 1 ≤ i ≤ n 5 (m) (m) 1 (m) 1 where x = (xi )i=1: Thus, for each 1 ≤ i ≤ n; the sequence (xi )m=1 is a bounded sequence of real numbers and by the Bolzano-Weierstrass theorem, 1 1 (m) 1 (mk)1 (mk) k (x1 )m=1 has a convergent subsequence x1 k=1: Let x1 −! x1 2 R: 1 (mk)1 Now, x2 k=1 being bounded has a subsequence which converges to x2; 1 1 say. Passing to the corresponding subsequence of (mk)k=1 yields a single subsequence (mk) of N such that (mk) k (mk) k x1 −! x1; x2 −! x2: Repeating this procedure for i = 1; 2; ··· ; n yields a single subsequence, denoted again by (mk) such that (mk) k xi −! xi; 1 ≤ i ≤ n: (mk) (mk) ) kx − x¯k1 = max jx − xij ! 0 1≤i≤n n n as k ! 1 wherex ¯ = (xi)i=1 2 B (l1) : In other words, the subsequence (mk) 1 n (x )k=1 converges tox ¯ in B (l1) and this completes the proof.