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.


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

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).

  • $\begingroup$ 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

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 $(a,+\infty)$.

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

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