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

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

For every $$\varepsilon>0$$ there exist $$0<\delta_{\varepsilon}$$ and $$a so that for every $$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?

• You need to add the condition $\lim_{\infty}f$ exists. – hamam_Abdallah May 24 '20 at 15:42

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, 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, 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 $$[a,\infty)$$.

$$f: x\mapsto x^2$$ 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

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

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