Real Analysis: Uniform Continuity Proof Verification

Question

I currently attempting a proof of Uniform Continuity and am unsure if need an extra step. I have put the question and work I've done down below.

Q: Show that if $f$ is continuous on $[0,\infty)$ and $\lim \limits_{x \to \infty} f(x)=L$, then $f$ is uniformly continuous on $[0,\infty)$

Pf: Let $\epsilon>0$

Since it is stated that $\lim \limits_{x \to \infty} f=L$ exists we can state by def: For every $\epsilon$, there exist a $K>0$ s.th for any $x>K$ then $\lvert f(x)-L\rvert < \frac{\epsilon}{2}$

We will now split our domain into three cases and since our domain goes to infinity we will let $K$ distinguish our three cases.

(1) Suppose our interval is $[K,\infty)$ and $x\ge K$. Since we know $\lvert f(x) -L \rvert < \frac{\epsilon}{2}$ it follows that: For every $\epsilon > 0$ there exist $K>0$ s.th for every $x,u$ if $\lvert x-u \rvert <\delta$, then $\lvert f(x) -f(u)\rvert = \lvert f(x) - L + L -f(u)\rvert \le \lvert f(x)-L \rvert + \lvert L -f(u)\rvert \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
Therefore $f$ is uniformly continuous on $[K,\infty)$.

(2)Suppose our interval $I=[0,K]$. Since the interval is closed and bounded and $f$ is continuous on $I$, BOW uniform continuity theorem $f$ is uniformly continuous.

(3).......

I know I need to prove that $K \in (x,y)$ is uniformly continuous but I don't know if I need seprate step for this or if I can just draw from ideas from steps 1 and 2 to prove that $K \in (x,y)$ is uniformly continuous.

Any help for proving the last part and improving previous steps would be greatly appreciated.

Answer 1

Assume that $|f(x)-L|<\epsilon/3$ for all $x\geq K$. So the part $[K,\infty)$ has been controlled.

Now $f$ is continuous at $x=K$, so there is some $\delta_{1}>0$ such that $|f(x)-f(K)|<\epsilon/3$ for all $|x-K|<\delta_{1}$.

Now $f$ is continuous on $[0,K]$, so it is uniformly continuous on $[0,K]$, so there is some $\delta_{2}>0$ such that $|f(x)-f(y)|<\epsilon/3$ for all $x,y\in[0,K]$ and $|x-y|<\delta_{2}$.

By taking $\delta=\min\{\delta_{1},\delta_{2}\}$, for all $x,y\in[0,\infty)$ and $|x-y|<\delta$, we have several cases.

Case I. Both $x,y\in[0,K]$, then $|f(x)-f(y)|<\epsilon/3<\epsilon$.

Case II. One of $x\in[0,K]$ and $y\in(K,\infty)$, then $0\leq K-x=K-y+y-x<y-x<\delta$, then $|f(x)-f(y)|\leq|f(x)-f(K)|+|f(K)-L|+|f(y)-L|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$.

Case III. Both $x,y\in[K,\infty)$, then $|f(x)-f(y)|\leq|f(x)-L|+|f(y)-L|<\epsilon/3+\epsilon/3<\epsilon$.

Answer 2

For $e>0$ take $K_e\in \Bbb R^+$ such that $x\geq K_e\implies |f(x)-L|<e/2.$

Take $d_e>0$ such that $$x\in (-d_e+K_e,d_e+K_e)\cap [0,\infty)\implies |f(x)-f(K_e)|<e/2.$$ Since $f$ is uniformly continuous on $[0,K_e],$ take $d'_e>0$ such that $$\;(x,y\in [0,K_e] \land |x-y|<d'_e\;) \implies |f(x)-f(y)|<e.$$ Let $d''_e=\min (d_e,d'_e).$

If $x,y \in [0,K_e]$ then $|x-y|<d''_e\implies |x-y|<d'_e\implies |f(x)-f(y)|<e.$

If $x,y\in [K_e,\infty)$ then $$ |f(x)-f(y)|\leq |f(x)-L|+|L-f(y)|<e/2+e/2=e.$$

If $x\in [0,K_e]$ and $y\in [K_e,\infty)$ then $$|x-y|<d''_e\implies (\;x\in (-d''_e+K_e,K_e]\;\land \; y\in [K_e,d''_e+K_e)\;)\implies$$ $$\implies (\;x,y \in (-d_e+K_e,d_e+K_e)\;)\implies$$ $$\implies|f(x)-f(y)|\leq |f(x)-f(K_e)|+|f(K_e)-f(y)|<e/2+e/2=e.$$