Proving $ f(x) = x^2 $ is not uniformly continuous on the real line This is  homework problem and the very premise has me stumped. It's in a text on PDE. 
The exercise says to show that $ f(x) = x^2 $ is not uniformly continuous on the real line. But every definition I know says that it is a continuous function, and unless you attach some special condition, like restricting the interval or making it a periodic function (perhaps saying $f(x-2) = f(x)$ or some such) it's by definition continuous. There's always a derivative since $f'(x) = 2x$. 
The preceding chapter is about Drichelet and the like, as an extension of Fourier series, so I am guessing that a Fourier expansion does something here but every proof of the proposition seems to have nothing to do with Fourier series in the slightest. 
So I am pretty lost here. This whole question seems utterly nonsensical. 

It is continuous. However, it is not uniformly continuous. 
Suppose it were; then for every $\epsilon$, there exists a $\delta$ for which $$|x - y| < \delta \implies |x^2 - y^2| < \epsilon$$
However, consider $\epsilon = 1$; if such $\delta$ existed and $y = x + \frac{\delta}{2}$, we would find that
$$|x^2 - (x + \frac{\delta}{2})^2| < 1$$
for every real $x$; however, this would imply that $$|x \delta + \frac{\delta^2}{4}| < 1$$ which is a clear contradiction, since we can  choose $x$ large.
