# Prove $f$ isn't uniformly continuous

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.

• $\lvert x-y\rvert < \delta$ by assumption, not $\lvert x-y\rvert > \delta$ (the last inequality is not true). – 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).
• $x-y = \delta/2$ by choice of $y$. Now, just plug in the value in the first inequality. – Clement C. Aug 25 '14 at 11:33
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 $(a,+\infty)$.