Proving $ f(x) = x^2 $ is not uniformly continuous on the real line

Question

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.

Answer 1

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.

Answer 2

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. ∎

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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 $\mathbb{R}.$

Answer 3

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.