Proving that $f$ is uniformly continuous on $[0,\infty)$

Question

Prove: If $f$ is uniformly continuous on $[0,b]$ for all $b>0$ and $\lim_{x\to \infty}f(x)=L$, then $f$ is uniformly continuous on $[0,\infty)$.

I intend to answer that problem. Can you please look at my proof?

Answer 1

Let $\epsilon>0$. Then there exists $M>0$ such that for all $x>M$, $$|f(x)-L|<\frac{\epsilon}{2}.$$ Thus, $$x,y\in (M,\infty)\quad\implies\quad |f(x)-f(y)|\leq |f(x)-L|+|f(y)-L| <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ This means that $$x,y\in (M,\infty)\quad\implies\quad |f(x)-f(y)|<\epsilon.\quad \quad(1)$$ Using the assumption, it follows that $f$ is uniformly continuous on $[0,M+1]$. Then, there exists $\Delta>0$ such that for all $x,y\in [0,M+1]$, $$|x-y|<\Delta\quad\implies\quad|f(x)-f(y)|<\epsilon.\quad\qquad (2)$$

We take $$\delta=\min\left\{\frac{1}{2},\Delta\right\}.$$ Let $x,y\in [0,\infty)$ such that $|x-y|<\delta$. Then $|x-y|<\Delta$ and $|x-y|<\frac{1}{2}$. Without loss of generality, assume that $x\leq y$. We consider the following cases:

Case 1. Suppose $y\leq M+1$. Then $x,y\in[0,M+1]$ so that by using $(2)$, we get $|f(x)-f(y)|<\epsilon$.

Case 2. Suppose $y>M+1$. The subtracting by $\frac{1}{2}$ to both sides, we get $$y-\frac{1}{2}>M+\frac{1}{2}.\qquad (3)$$ Since $|x-y|<\frac{1}{2}$, we get $-\frac{1}{2}<x-y<\frac{1}{2}$ implying $$y-\frac{1}{2}<x.\qquad\quad (4)$$ Combining $(3)$ and $(4)$, we get $$x>y-\frac{1}{2}>M+\frac{1}{2}>M.$$ Thus, $x,y\in(M,\infty)$ and using $(1)$, $|f(x)-f(y)|<\epsilon$.

In either case, we have shown that for any $x,y\in[0,\infty)$, $$|x-y|<\delta\quad\implies\quad|f(x)-f(y)|<\epsilon.$$ This proves that $f$ is uniformly continuous on $[0,\infty)$.