4
$\begingroup$

I recently asked for some direction with how to connect a convergent function to proving it is uniformly continuous and I was directed to an awesome post that explained a lot for me. So now I tried to prove this problem which pretty much asks a specific case of that and I was wondering if some people could look it over for errors.

Suppose $f$ is continuous on $[0, \infty)$ and $f(x) \rightarrow 2$ as $x \rightarrow \infty$. Prove $f$ is uniformly continuous on $[0, \infty)$.

Since $f(x) \rightarrow 2$ as $x \rightarrow \infty$, then for $\epsilon>0, \exists k $ s.t. $ \forall x \geq k$ we have $|f(x) - 2| < \frac{\epsilon}{2}$.

Also, note that since $f$ is continuous on $[0, \infty)$ it is uniformly continuous on $[0, k+2]$ and so for $\epsilon >0$, $\exists \delta_1 > 0$ s.t. if $x,y \in [0, k+2]$ and $|x-y| < \delta_1$ then $|f(x) - f(y)| < \epsilon$.

Now given $\epsilon >0$, take $\delta = \min(\delta_1, \frac{1}{2}) > 0$. Suppose $x,y \in [0, \infty)$ and $|x-y| < \delta$.

Then if $x > k + 1$, since $|x-y| < \frac{1}{2}$ we have $k+1 - y < x - y < \frac{1}{2}$ and so $y > k$. Together, we have $x, y > k$ and $|x-y| < \delta$, so $|f(x) - f(y)| = |(f(x) - 2) + (-f(y) + 2)| \leq |f(x) -2| + |f(y) - 2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

If $x \leq k + 1$, since $|x - y| < \frac{1}{2}$, then we have $y < k + \frac{3}{2} < k+2$. Together then, we have $x, y \in [0,k+2]$ and $|x-y| < \delta \leq \delta_1$ and so $|f(x) - f(y)| < \epsilon$.

Thus we have shown for $\epsilon >0$, there is a $\delta > 0$ s.t. if $x,y \in [0, \infty)$ and $|x-y| < 0$, then $|f(x) - f(y)| < \epsilon$.

$\endgroup$
  • 1
    $\begingroup$ Your proof looks fine for me. Good job. $\endgroup$ – Crostul Apr 5 '17 at 13:02
  • $\begingroup$ Thats great to hear, thank you! $\endgroup$ – student_t Apr 5 '17 at 13:02
1
$\begingroup$

Since $\displaystyle \lim_{x \to \infty}f(x)=2$ so for a given $\epsilon >0$ ,$\exists$ a real number $k$ such that $|f(x)-2|<\epsilon/2$ whenever $x \ge k$. Now let , $x,y\in [k,\infty)$.

Then , $|f(x)-f(y)|\le |f(x)-2|+|f(y)-2|<\epsilon$ , for every $x,y \in [k,\infty)$ . So particularly ,$|f(x)-f(y)|<\epsilon$ , whenever $|x-y|<\delta$. So $f$ is uniformly continuous in $[k,\infty)$.

Again $[0,k]$ is compact domain, so $f$ is uniformly continuous in $[0,k]$. Hence $f$ is uniformly continuous in $[0,\infty)$.

$\endgroup$
  • $\begingroup$ This is much more elegant. Thank you for this. $\endgroup$ – student_t Apr 5 '17 at 15:59
  • $\begingroup$ How do we know such a $\delta$ exists? $\endgroup$ – student_t Apr 5 '17 at 17:19
  • $\begingroup$ @danny Good question. See the edition. I think you will get it. Mainly see the boldmark. $\endgroup$ – Empty Apr 5 '17 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.