Question

(b) 4 Let F: X Y be a linear map between two normed spaces. Prove that F is continuous at Ojf and only if F is uniformly continuous on X.

Answer 1

(b). Let Fixy be a linean map between normed turn spaces. we have to show that continuous "at o if f is and only if f is uniforening continuous on. X . Proof stehet of is continuous at o. we first show that F is continuous at each point of the if nex , and be a neighbourhood of FG), thane v-fx is a neighbourhood of o= F(o). Since. Ç is continuous at o. there is an nbid of oint HD (say u) such that FCU) civ-f(x) . . - F(x+u) c v.. since, xtu is open in a and *txotot x= xto ex+u, we get the fact that xt w is openi nad of x. Hence, F is continuous at ned, ie F is continuous on x:( since XeX was arbitrary) . Now, as step. 2 we show that there is a rad u of o ex sit. F(U) is bounded let us choose u an hod of oex sit. f(u) By (0,1) {(ball of radius 1 around o E y.)}. If W, is a rbd of oiny. then By (0, 1) citw for t sufficiently large . . Therefore, t(u) c tw for all such t, and hence T(U) is bounded . Step 3 we show that f maps bounded subsets of x onto bounded subsets of y . by step 2 .. there is a rbd u of o in x st. - Tullse for all uEU and a const c.. choose ryo st Bx (0,r) cu. let A bounded in , then 3 some positive t sit. Act. By(0,). o E X

Thus, FCA) C t. f(Bx (0,0)) c t.FCU). - Il F(x) || < tic # WE A . F(A) is bounded in y. . Step & F is bounded. By step 3, {xex I llallx=1} maps under of to a bounded subset of Y ie. Il Toally sm. for a constant M and Ilselly1 . Hence, zEx with a to IF (1122) SM - Il Fally s m llzlly - F is bounded. (Finally!) Step 5. f: xay is uniformly continuous. By Step 4 , there is some Mkoo sit - Ilfaelly <M llally .txex - Ilf(x-yilly = llfa-fyll yo 's Mllx-ylly - f is uniformly continuous. Converse part is obvious.