For M.Sc (Math) Theorem 2.4. Given a linear operator T : X ! Y , for normed spaces X and Y , the following three statements are equivalent. (i) T is continuous at some point in X. (ii) T is continuous at every point in X. (iii) T is bounded on X. Proof. Clearly (ii) implies (i). We shall show that (i) implies (ii). Suppose that T is continuous at x0 2 X. Let x 2 X and let " > 0 be given. Then there is ± > 0 such that if kz ¡ x0k < ± then kTz ¡ Tx0k < ". But then we have kTx0 ¡ Txk = kTx0 ¡ Tx + Tx0 ¡ Tx0k = kT(x0 ¡ x + x0) ¡ Tx0k ; by the linearity of T, < " whenever k(x0 ¡ x + x0) ¡ x0k < ±, i.e., kx0 ¡ xk < ±. This shows that T is continuous at any x 2 X. (Alternatively, one could argue as follows. Suppose that (xn) is a sequence in X such that xn ! x. Then xn ¡x+x0 ! x0 and so T(xn ¡x+x0) ! Tx0, since T is assumed to be continuous at x0. Thus Txn ¡ Tx + Tx0 ! Tx0, since T is linear. In other words Txn ¡ Tx ! 0, or Txn ! Tx.) Next we show that (iii) implies (ii). Let " > 0 be given. By (iii), there is k > 0 such that kTx0 ¡ Txk = kT(x0 ¡ x)k · kkx0 ¡ xk for any x; x0 2 X. Putting ± = "=k, we conclude that if kx0 ¡ xk < ±, then kTx0 ¡ Txk < ". Thus T is continuous at x 2 X. In fact, this estimate shows that T is uniformly continuous on X, and that therefore, continuity and uniform continuity are equivalent in this context. In other words, the notion of uniform continuity can play no special r^ole in the theory of linear operators, as it does, for example, in classical real analysis. Finally, we show that (ii) implies (iii). If T is assumed to be continuous at every point of X, then, in particular, it is continuous at 0. Hence, for given " > 0, there is ± > 0 such that kTxk < " whenever kx¡0k = kxk < ±. Now, for any x 6= 0, z = ±x=2kxk has norm equal to ±=2 < ±. Hence kTzk < ". But kTzk = ±kTxk=2kxk and so we get the inequality kTxk < 2"kxk ± which is valid for any x 2 X with x 6= 0. Therefore kTxk · 2"kxk ± holds for any x 2 X, and we conclude that T is bounded. (Alternatively, suppose that T is continuous at 0, but is not bounded. Then for each n 2 N there is xn 2 X such that kTxnk > nkxnk. Evidently xn 6= 0. Put zn = xn=nkxnk. Then kznk = 1=n ! 0, and so zn ! 0 in X. However, kTznk = k(Txn=nkxnk)k = kTxnk=nkxnk > 1 for all n 2 N, and so (Tzn) does not converge to 0. This contradicts the assumed continuity of T at 0.)
Posted on: Thu, 12 Sep 2013 05:45:34 +0000
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