Question

Definition: A function f : A → R is said to be uniformly continuous on A if for every e > O there is a δ > 0 such that *for all* z, y € A we have Iz-vl < δ nnplies If(r)-f(y)| < e. In other words a function is uniformly continuous if it is continuous at every point of its domain (e.g. every y A), with the delta response to any epsilon challenge not depending on which point is picked (sometimes it is nice to write 6, to emphasize when the choice might depend on y € A). 1. (a) Show the function f(x is uniformly continuous on R. (b) Show the function g(r) a2 is not uniformly continuous on R. This in particular shows that products of uniformly continuous functions are not necessarily uniformly continuous

Answer 1

We will show/prove uniform continuity of the given functions by the definition given in the problem (a) Choose epsilon > 0.Let delta = epsilon choose x_0 elementof R & x elementof R where R is set of real number Assume Vertical-bar x - x_0 Vertical-bar < delta Then Vertical-bar f(x) - f(x_0) Vertical-bar = Vertical-bar x - x_0 Vertical-bar < delta = epsilon Vertical-bar x - x_0 Vertical-bar < delta implies Vertical-bar f(x) - f(x_0) Vertical-bar < epsilon Hence, by definition f(x) = x is uniformly cont (b) To disprove anything, we have to give an example contradiction our result \r\n This method is called proof by contradiction Let e = 1 choose delta > 0 let x_0 = 1/delta & x = x_0 + delta/2 Then Vertical-bar x - x_0 Vertical-bar = delta/2 < delta but Vertical-bar x^2 - x_0^2 Vertical-bar = Vertical-bar (1/delta + delta/2)^2 - (1/delta)^2 Vertical-bar = 1 + delta^2/4 > 1 Now as we have chosen epsilon = 1, Vertical-bar x - x_0 Vertical-bar < delta implies Vertical-bar f(x) - f(x_0) Vertical-bar > epsilon therefore Not uniformly continuous & This proves that product of