Proving that $x^2$ is not unifromly continuous on $\mathbb{R}$

Question

Assuming $\delta > 0$, let $x_1 = \frac{1}{\delta}, x_2 = x_1 + \frac{\delta}{2}$. Thus: $$|x_1 - x_2| = \frac{\delta}{2} \lt \delta$$ Since I am struggling to prove that $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$, having any $\epsilon \gt 0$, following requirement has to be met: $$|x_1^2 - x_2^2| \lt \epsilon$$ As such: $|\frac{1}{\delta^2} - \frac{1}{\delta^2} + \frac{2\delta}{2\delta} + \frac{\delta^2}{4}| = |1 + \frac{\delta^2}{4}| \not \lt \epsilon$ when $\epsilon \lt 1$.

Question: is this proof valid? Since I discovered 1 as a constant, obviously independent from the $\delta$ change, there is infinite sequence of possible $\epsilon: 0 \lt \epsilon \le 1$ for which the uniform continuity requirement on the top is not completed? Seems to be enough for me.

Answer 1

A function $f: D \to R$ is uniformly continuous if for all $\epsilon > 0$ there exists some $\delta > 0$ such that for all $x,y \in D$, $|x-y|< \delta \implies |f(x) - f(y)| < \epsilon.$

Let's put this into quantifiers, $$\forall(\epsilon >0) \exists(\delta >0) \forall(x \in D) \forall(y \in D)\Big[|x-y| < \delta \implies |f(x) - f(y)| < \epsilon\Big].$$

So to be ``not uniformly continuous'' means

$$\exists(\epsilon >0) \forall(\delta >0) \exists(x \in D) \exists(y \in D)\Big[|x-y| < \delta \text{ and } |f(x) - f(y)| \geq \epsilon\Big].$$

So you showed that $$|f(x)-f(y)| = |1+\frac{\delta^2}{4}| \geq \epsilon$$ for any $\epsilon \leq 1.$ So just choose $\epsilon =1.$

Answer 2

Here is an other aproach using sequences.

Take $$x_n=n+\frac {1}{n} $$ and $$y_n=n .$$

thus $$\lim_{n\to+\infty}|x_n-y_n|=0$$

but $$x_n^2-y_n^2=2+\frac{1}{n^2}$$ and $$\lim_{n\to+\infty}|x_n^2-y_n^2|=2\neq 0.$$