# 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.

• What book are you using? It seems strange this would appear in a PDE book. Aside from that question, do you know the definition of uniformly continuous? Sep 24, 2013 at 2:13
• Uniformly continuous is a stronger condition than continuous. Sep 24, 2013 at 2:13
• The book is Partial Differential Equations, Asmar. And I am not sure the definition of uniformly continuous helped, but if you can tell me how that fits I am most grateful. Like I said, I looked up the proofs and none of them seem to be what my prof will be looking for. Sep 24, 2013 at 2:15
• Here's a rough idea of where things break down: When you try to show $f(x) = x^2$ is continuous, you end up choosing your $\delta$ to be the minimum of $1$ and some number that is inversely proportional to $\epsilon$. So you see that if you try to make $\delta$ large, you end up forcing $\epsilon$ to zero. Sep 24, 2013 at 2:16
• I'm not sure that helps a lot. But thanks. :-( Sep 24, 2013 at 2:17

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.

• That's the thing. I get that kind of proof of continuity, but I feel like there must be some other answer my text (and the professor) are seeking here. Sep 24, 2013 at 2:19
• @Jesse This is a full proof that it's not uniformly continuous, since it would force $\delta =$, a contradiction.
– user61527
Sep 24, 2013 at 2:20
• Well, then it is what it is. I can't understand why the text spends a whole section on proof of Fourier representation theorem, then, ahead of this, or the Dirichlet kernel, or Riemann Lebesque Lemma. It seems to have nothing to do with it. Is our prof or Asmar (the author of the text) messing with us? Sep 24, 2013 at 2:22

Here is a more general approach to solving this problem.

Let me formulate and prove a theorem:

Theorem:

Let $$E = [a,+\infty),$$ function $$f:E \rightarrow \mathbb{R}$$ is differentiable on $$E$$ and $$\displaystyle{\lim_{x \to \infty}} f'(x) = \infty.$$ Then $$f$$ is not a uniformly continuous function.

Proof:

Let the function $$f$$ be uniformly continuous. Let $$\epsilon = 1$$ and $$\delta>0$$ satisfies the deffinition of uniform continuous. From the $$\displaystyle{\lim_{x \to \infty}} f'(x) = \infty$$ follows that $$\exists m\geqslant a: \forall x \geqslant m \\ \left|f'(x)\right| > \frac{2}{\delta}$$ Let $$x_1=m, x_2=m+\frac{\delta}{2}.$$ Using Lagrange theorem for $$[x_1,x_2]$$ we have $$\left|f(x_1)-f(x_2)\right| = \frac{\delta}{2}\left|f'(\zeta)\right| \\ \text{for some } \zeta \in[x_1,x_2]$$ Since $$\zeta \geqslant m$$ then $$\left|f'(\zeta)\right|>\frac{2}{\delta}$$ from where $$\left| f(x_1) - f(x_2)\right| > 1 = \epsilon.$$ This contradicts the uniform continuity of the function $$f$$. Thus, $$f$$ is not uniformly continuous. ∎

So, your function is $$f(x)=x^2$$. Its derivative is $$\frac{d(x^2)}{dx} = 2x.$$ The limit of this derivative at infinity is $$\displaystyle{\lim_{x \to \infty}} (2x) = \infty.$$ So, by the theorem this function is not uniformly continuous on the $$[0,+\infty)$$. The same arguments can be used for $$(-\infty,0]$$. And the statement that if a function is uniformly continuous on the [a, c] as it is on [c,b], then it is uniformly continuous on the [a, b] finally can be used to prove that your function is not uniformly continuous on the real line $$\mathbb{R}.$$

On the contrary, suppose that $$f(x) = x^2$$ is uniformly continuous on the real line. Then by the definition: for any $$\epsilon > 0$$ and for any $$x,y \in \mathbb{R}$$ we have a $$\delta$$ such that $$|x-y| < \delta \implies |f(x) - f(y) | < \epsilon$$. Our claim here is now $$|x^2 -y^2 | < \epsilon$$. That is the distance between $$x^2$$ and $$y^2$$ is at most $$\epsilon$$ everywhere on the vertical axis as long as we keep $$x$$ and $$y$$ at most $$\delta$$ away from each other. The problem arises if we let $$x$$ and $$y$$ getting larger and larger.

So for the given $$\epsilon$$ we have fixed $$\delta$$ and we may play with $$x$$ and $$y$$ values. Let's make them large enough to satify $$\frac{\delta}{2} < |x-y | < \delta$$ and $$x > \frac{\epsilon}{\delta}$$, $$y > \frac{\epsilon}{\delta}$$. Then: $$|x-y| < \delta$$ but $$|x^2 - y^2| = |x-y||x+y| > \frac{\delta}{2}\frac{2\epsilon}{\delta} \implies |x^2 - y^2| > \epsilon$$

Which is a contraction with the definition which is assumed true at the beginning. Therefore $$f(x) = x^2$$ can not be uniformly continuous.