Show that the function $f(x)=\frac{1}{x}$is uniformly continuous on the set $A:=[a,\infty )$ where a is positive constant

Question

please check my proof

Let $\epsilon >0 \delta >0$ and a is arbitary positive constant in interval

$$|x-a|<\delta \leftrightarrow |f(x)-f(a)|<\epsilon $$

$$\leftrightarrow |\frac{1}{x}-\frac{1}{a}|<\epsilon $$

$$\leftrightarrow |\frac{a-x}{xa}|<\epsilon $$

$$\leftrightarrow \frac{1}{a}|\frac{a-x}{x}|<\epsilon $$

$$|\frac{a-x}{x}|<a\epsilon $$

choose $\delta =a\epsilon $

then

$|\frac{1}{x}-\frac{1}{a}|<\delta \leftrightarrow |\frac{1}{x}-\frac{1}{a}<a\frac{\epsilon }{a}=\epsilon $

then it is uniformly conitinouos

Answer 1

More generally:

(1). For $a,b\in \mathbb R,$ if $f:[a,b]\to \mathbb R$ is continuous then $f$ is uniformly continuous on $[a,b].$

(2). Corollary. If $f:[a,\infty)$ is continuous and $\lim_{x\to \infty}f(x)=0$ then $f$ is uniformly continuous on $[a,\infty).$

Proof of Corollary. Given $e>0,$ take $c\geq a$ such that $\forall x\in [c,\infty)\;(|f(x)|<e/2).$

Let $b=c+1.$

Take $d\in (0,1)$ such that $\forall x,y\in [a,b]\;(|x-y|<d\implies|f(x)-f(y)|<e/2).$

Now for all $x,y\in (b,\infty)$ we have $x,y\in [c,\infty)$ so$|f(x)-f(y)|\leq |f(x)|+|f(y)|<e/2+e/2=e.$

And for $a\leq x\leq b<y,$ if $|x-y|<d$ then $$y>x>y-d>b-d=c+1-d>c,$$ so $\{x,y\}\subset [c,\infty),$ so $|f(x)-f(y)|\leq |f(x)|+|f(y)|<e/2+e/2=e.$

So altogether we have $\forall x,y \in [a,\infty)\;(|x-y|<d\implies |f(x)-f(y)|<e).$

Answer 2

As pointed out in a comment, you're only looking at values near $x=a$. If you prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)-f(a)| \le \epsilon$ whenever $|x-a| \le \delta$ (which you did not, your proof is confused), then this would only imply the continuity of $f$ at $a$, not more.

I suggest another approach: note that $f$ is continuous in $[a,+\infty)$ (if $a>0$) and $\lim_{x \rightarrow +\infty} f(x) = 0 $. These two properties combined together give the uniform continuity of $f$ in $[a,+\infty)$. The main idea is to combine Heine-Cantor theorem (a continuous function on a compact set is uniformly continuous) with the vanishing condition. Let me know if you manage to conclude or you need a complete proof.