Explain that for every $\epsilon > 0$ there exists $K > 0$ such that $ |f(x)-f(y)|<\varepsilon \quad \text {as long as } x, y>K $

Question

Assume

$$ \begin{aligned} &f(x) \rightarrow a \quad \text { for } x \rightarrow-\infty\\ &f(x) \rightarrow b \quad \text { for } x \rightarrow \infty \end{aligned} $$

for two real numbers $a$ og $b$.

a) Explain that for every $\epsilon > 0$ there exists $K > 0$ such that $$ |f(x)-f(y)|<\varepsilon \quad \text {as long as } x, y>K $$

My answer

To be honest I'm not sure what the problem is asking. In the following I'm actually guessing.

$$|f(x)-f(y)| = |f(x)-f(y)+a-a| = |f(x)-a+a-f(y)| \le |f(x)-a| + |a-f(y)|$$

Q1: Is there some definitions and/or theorems I can look up?

b) Show that $f$ is a uniformly continuous.

I do not have a good grasp of uniform continuity. Where would you guess suggest me to start.

Answer 1

For the first part, it came from the following definition of limit

$$\displaystyle \lim_{x\to +\infty}f(x) = b$$ if and only if, $$\text{for every $\epsilon>0$, there exist a $K>0$ such that if $x>K$, then $\lvert f(x)-b\rvert<\epsilon$}$$

With this definition, you are on the right track.

Let $\epsilon>0$. From the definition of limit, there exist $K_b>0$ such that if $x>K_b$, then $\lvert f(x)-b\rvert<\dfrac\epsilon2$.

Then if $x, y>K_b$ we have $$\left\lvert f(x)-f(y)\right\rvert=\left\lvert f(x)-b+b-f(y)\right\rvert\leq\left\lvert f(x)-b\right\rvert+\left\lvert b-f(y)\right\rvert < \frac\epsilon2 + \frac\epsilon2=\epsilon$$ $$\left\lvert f(x)-f(y)\right\rvert<\epsilon$$

A similar argument could be made with the limits to $-\infty$

From the definition of limit, there exist $K_a<0$ such that if $x<K_a$, then $\lvert f(x)-a\rvert<\dfrac\epsilon2$.

Then if $x, y<K_a$ we have $$\left\lvert f(x)-f(y)\right\rvert=\left\lvert f(x)-a+a-f(y)\right\rvert\leq\left\lvert f(x)-a\right\rvert+\left\lvert a-f(y)\right\rvert < \frac\epsilon2 + \frac\epsilon2=\epsilon$$ $$\left\lvert f(x)-f(y)\right\rvert<\epsilon$$

---

For part b), first the function need to be continuous. We could easily create a monstruous function, nowhere continuous that respect the limit conditions.

If the function is continuous, let'a take $\epsilon>0$ and we are looking for $\delta>0$ such that if $\lvert x - y \rvert<\delta$ then $\lvert f(x) - f(y)\rvert<\epsilon$.

From the first part, there exist $K_a<0$ and $K_b>0$ such that $$\lvert f(x) - f(y)\rvert<\epsilon \qquad \text{for all }x,y<K_a\tag{1}$$ $$\lvert f(x) - f(y)\rvert<\epsilon \qquad \text{for all }x,y>K_b\tag{2}$$ No mather the value of $\delta$ we have, the uniformly continuous definition will be satisfied for those parts.

It remains to show that the function is uniformly continuous on the closed interval $\left[K_a-1,K_b+1\right]$.

A continuous function on a compact set is uniformly continuous. This is known as Heine-Cantor theorem. If you are allow to use it, you are done. If not, look up the proof of the theorem and apply it to your case.

Since the function is uniformly continuous on $\left[K_a-1,K_b+1\right]$, there exist a $\delta<1$ such that for $x,y\in\left[K_a-1,K_b+1\right]$, $$\lvert x - y \rvert<\delta \implies\lvert f(x) - f(y)\rvert<\epsilon\tag{3}$$ This is the $\delta$ we were looking for. If $\lvert x - y \rvert<\delta$, three cases are possible.

• $x,y < K_a$, then it follows from equation $(1)$;
• $x,y\in\left[K_a-1,K_b+1\right]$, then it follows from equation $(3)$;
• $x,y > K_b$, then it follows from equation $(2)$.

The function is uniformly continuous.