Exercise: Suppose that f:[0, ∞] → R is continuous, and that there is an real value L, such that f(x) → L as x → ∞. Prove that f is uniformly continuous on [0, ∞].
Attempt: I will try to use the definition and theorem.
Definition: a function f: E → R is uniformly continuous iff for every ε > 0, there is a δ > 0 such that |x-a| < δ and x,a are elements in E implies |f(x) – f(a)| < ε.
Theorem: Suppose a < b and that f: (a,b) → R. Then f is uniformly continuous on (a,b) iff f can be continuously extended to [a,b]; that is iff there is a continuous function g: [a,b] → R such that f(x) = g(x), and x is in (a,b).
Then, suppose that f:[0, ∞] → R, and that there is an real value L, such that f(x) → L as x → ∞.
Then let ε > 0. Then since [0,∞), we can try to make it of the form [0,N]. So it can look as the theorem. Suppose x >= N so that |f(x) – L | < ε. Then there is δ > 0 such that |x-a| < δ, and x,a are in [0,N], so that |f(x) – f(a)| < ε.
Can someone please help me? I don't know how to continue.
Thank you in advance.
Best Answer
The idea is to split your domain into two parts and then prove that your function is uniformly continuous on each of them. This implies that the function is uniformly continuous on the whole domain.
For first set we will use your convergence assumption:
$$\forall \varepsilon>0\exists X>0:\forall x\ge X \quad |f(x)-L|<\frac \varepsilon 2,$$ therefore $$\forall \varepsilon >0 \exists X >0 \quad \forall x,y\ge X \quad |f(x)-f(y)|=|f(x)-L+L-f(y)|\le \varepsilon,$$hence your function satisfies the requisites of a uniformly continuous function on $[X,\infty)$ for a given $\varepsilon $.
On the other hand, on the interval $[0,X]$ is also uniformly continuous (continuous on a compact is uniformly continuous), therefore, with the above remark, we obtain that your function is uniformly continuous on $[0,\infty)$.