1
$\begingroup$

I already proved (followed by an hint) that $f(y)-f(x) > x(y-x)$ for all $y>x>0$. I need to prove $f$ isn't uniformly continuous on $(0, \infty)$.

What I did:

Lets assume by contradiction $f$ is uniformly continuous. Hence, for all $\varepsilon>0$ there is $\delta>0$ such that $\left|y-x\right| < \delta \implies \left|f(y)-f(x)\right| < \varepsilon$.

Let $\varepsilon > 0$.
Using our previous conclusion: $\left|f(y)-f(x)\right| > \left|x\right|\left|y-x\right| >\left|x\right|\delta$

Now, if we choose $x=\frac{\varepsilon}{\delta}$ then we have a contradiction and $f$ isn't uniformly continuous.

Is that right? I'd be glad to get a verification.

Thanks.

$\endgroup$
  • $\begingroup$ $\lvert x-y\rvert < \delta$ by assumption, not $\lvert x-y\rvert > \delta$ (the last inequality is not true). $\endgroup$ – Clement C. Aug 25 '14 at 11:22
1
$\begingroup$

You got to $$\left|f(y)-f(x)\right| > \left|x\right|\left|y-x\right|.$$ Now, keeping $x$ as a "free variable", take $y=x-\frac{\delta}{2}$ (so that $\lvert x - y\rvert< \delta$ indeed), to get $$\left|f(y)-f(x)\right| > \left|x\right|\frac{\delta}{2}$$ and let $x$ go to infinity to get that RHS greater than $\varepsilon$ (or any $x> \frac{2\varepsilon}{\delta}$ would work as well).

$\endgroup$
  • $\begingroup$ I just thought about it after reading your comment. Thanks! $\endgroup$ – Elimination Aug 25 '14 at 11:27
  • $\begingroup$ how can you go from the first inequality to the second? I dont understand how to go from xδ>x∣x−y∣ and |f(y)−f(x)|>|x||y−x| to the second inequality $\endgroup$ – k99731 Aug 25 '14 at 11:32
  • $\begingroup$ $x-y = \delta/2$ by choice of $y$. Now, just plug in the value in the first inequality. $\endgroup$ – Clement C. Aug 25 '14 at 11:33
  • $\begingroup$ @clementC. Oh thanks $\endgroup$ – k99731 Aug 25 '14 at 11:37
2
$\begingroup$

I believe that you are right, here's another approach.

Set $y_n=n^2+1/n$ and $x_n=n^2$. Clearly $x_n-y_n\to 0$ then if $f$ is uniformly continuous we'd have $f(y_n)-f(x_n)\to 0$ but $$f(y_n)-f(x_n) >n^2(n^2+\frac{1}{n}-n^2)=n\to +\infty$$ Contradiction ! then $f$ is not continuous on any interval $(a,+\infty)$.

$\endgroup$
  • $\begingroup$ Great! Forgot about the "two series contradiction trick". Thank you too $\endgroup$ – Elimination Aug 25 '14 at 11:29

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.