Question

(a) Suppose f is continuously differentiable on the closed and bounded interval I = [0, 1]. Show that f is uniformly continuous on I. (b) Suppose g is continuously differentiable on the open interval J = (0,1). Give and example of such a function which is NOT uniformly continuous on J, and prove your answer.

Answer 1

@ Suppose fils continuously differentiable on the closed and bounded interval I= [0, 1]. Then we have show that f is uniformly continuous on I. Proof; since the interval 1 = [0, 1] is closed and f is continuons on I. f is bounded on 1 = [0, 1] . Let M, K2 El with 24442 since &' is bounded on I, there exists as positive real number k ouch that | $6915K V XEI Then & satisfills both the condition of Mean value Theorem on [ x2] and therefore there exists a point n E (2427 wuch that I'(W) = f(%) - for x-24 1 812) -f(1) LK (:if ( sk). 02-84 1 & 15 (1) 12K) & 1 f (x2) ~ for ISK11 for all x4,02 EI choose yo. Then } a sto such that If(R)-f(1) KE & XmEl sathfying 122-4 KS, where safe : f is uniformly continuous on 1= [0,1]. Mence Proved)

① suppose g is continuously differentiable on the open interval =60in, we have to give an example of such function which is not uniformly continuons on I. bet f = 1/2 xEJ = (011). step-1: f(x) = 1 X E(OI) is continuon ut nace (01) and f(0) 2/6 o 1 left limit = tim fx = lim fe-b) = lim ehe reight limit - lim f = lim fleth = lim cth = e. him. Son = lim +50=HC). f(x) = & is continuous on (o) atxe (OT) step-2: f(x) = 4, XE (Old is differentialle ut X=CE (0) 15 (2= tim sen- so _ A che hoch act . C aet. hoso -h c-cth -Ho he (C-5) to e(ch) Rf'. = lim f(cth)-fosch I t N o hayo h . hoh aft c-e-h ho hletha - - 4o eleth Lf (c) = Rf'(o) af is differentiable al XectCall).

estep. Bi f(0) = 1 h , XE(011) is not bounded on J. Let us assume that food is bounded then by definition, of (x) is breended if and only if g Myo Hit 100 CM V XE(9). Assume that x = Mt. Then f(x) = Mt! .. we have to 80 our assumption is wrong that f(a)= 1/2 is bounded on (01). estep-4: f(x) = 2 XE(011) is not uniformly continuon since fx is continuons and differentiable on (o in and not bouned on 7 = (011), therefore f(x) = 2 is not uniformly contínicom on (011) because I'm flag does not exists finitely.. Hence f(x) = 1 , XE(Old is not reniformly continues xot