Prove that a function is uniformly continuous in $[a,\infty)$

Question

Let $f:[a,\infty)\to\mathbb{R}$ be a continuous function.

For every $\varepsilon>0$ there exist $0<\delta_{\varepsilon}$ and $a<c_{\varepsilon}\in\mathbb{R}$ so that for every $x_{1},x_{2}>c_{\varepsilon}$ so that $|x_{1}-x_{2}|<\delta_{\varepsilon}$ it holds that $|f(x_{1})-f(x_{2})|<\varepsilon$.

Prove that $f$ is uniformly continuous in $[a,\infty]$. Hint: $[a,\infty]=[a,c_{\varepsilon/2}]\cup[c_{\varepsilon/2},\infty)$ and use the Heine-Cantor Theorem.

I get that the Heine-Cantor Theorem proves uniform continuity in $[a,[c_{\varepsilon/2}]$, but don't get how we get to that epsilon and the general idea of the proof. Could you give me a more obvious hint or where to start?

Answer 1

The definition that a function $f(x)$ is uniformly continuous on $[a,\infty)$ is that for any real number $\epsilon>0$, there exist a real number $\delta >0$ such that if the distance between two points in the domain is close, $|x_i-x_j|<\delta $, we have that the distance between the evaluation of the function at those two points is close ,$|f(x_i)-f(x_j)|<\epsilon$.

So given $\frac{\epsilon}{2} >0$, from the condition given to us in the question, we know that there exist $c_{\frac{\epsilon}{2}} \in (a, \infty)$ and $\delta_{\frac{\epsilon}{2}}>0$ such that if $|x_k-x_j|<\delta_{\frac{\epsilon}{2}}, x_k,x_j >c_{\frac{\epsilon}{2}}$, then $|f(x_k)-f(x_j)|<\frac{\epsilon}{2}$.

Also, on the interval $[a,c_{\frac{\epsilon}{2}}]$, we know that the function is uniformly continuous, so we have that there is an $\delta'_{\frac{\epsilon}{2}}>0$ such that if $|x_i-x_j|<\delta'_{\frac{\epsilon}{2}}, x_i,x_j \in [a,c_{\frac{\epsilon}{2}}]$, then $|f(x_i)-f(x_j)|<\frac{\epsilon}{2}$.

Now if we choose $\delta := \min(\delta'_{\frac{\epsilon}{2}},\delta_{\frac{\epsilon}{2}})$ and now we are ready to verify the condition for the function $f(X)$ to be uniformly continuous on the interval $[a, \infty)$

So we choose two arbitary numbers in the domain of the function such that their distance is less less than $\delta$, like: $x_s<x_d \in [a,\infty), |x_s-x_d|<\delta$, We have three cases for the locations of the two numbers:

if $x_s,x_d \in [a,c_{\frac{\epsilon}{2}}]$, then $|f(x_s)-f(x_d)|<\frac{\epsilon}{2}<\epsilon$

if $x_s,x_d \in [c_{\frac{\epsilon}{2}},\infty)$, then $|f(x_s)-f(x_d)|<\frac{\epsilon}{2}<\epsilon$

Now if could be that the two numbers where one belongs to $[a,c_{\frac{\epsilon}{2}}]$ and the other belongs to $[c_{\frac{\epsilon}{2}},\infty)$, but since we assume that $x_s<x_d$, we have that:

$x_s \in [a,c_{\frac{\epsilon}{2}}], x_d \in [c_{\frac{\epsilon}{2}},\infty)$, and then $|f(x_s)-f(x_d)| = |f(x_s)-f(c_{\frac{\epsilon}{2}})+f(c_{\frac{\epsilon}{2}})-f(x_d)| \leq |f(x_s)-f(c_{\frac{\epsilon}{2}})|+|f(c_{\frac{\epsilon}{2}})-f(x_d)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

Hence, in all cases for possible location that the two numbers $x_s,x_d$ that could be, we have shown the definition for uniformly continuous is satisfied.

Hence the function is uniformly continuous on the interval $[a,\infty)$.

Answer 2

$f: x\mapsto x^2$ is continuous at $[0,+\infty)$ but is not uniformly continuous at $ [0,+\infty)$, since

with $$u_n=n \;\ \text{ and } \; v_n=n+\frac 1n$$

we have

$$\lim_{n\to +\infty}(v_n-u_n)=0$$

but

$$\lim_{n\to+\infty}(f(v_n)-f(u_n))=$$

$$\lim_{n\to+\infty}(2+\frac{1}{n^2})=2\ne 0$$