Proof that $x^2$ is not uniformly continuous on $\mathbb{R}$

Question

I am trying to prove that $x^2$ is not uniformly continuous on $\mathbb{R}$. Is the following correct? (I am able to write a proof for an arbitrary $\epsilon>0$, but I also want to write one for a particular $\epsilon>0$.

Set $\epsilon=0.7$. Consider an arbitrary $\delta>0$. Set $a=b+\frac{\delta}{2}$. Set $b=\frac{1}{\delta}-\frac{\delta}{4}$. Then $|a-b|=\frac{\delta}{2}< \delta $ and $|a^2-b^2|=|(a-b)(a+b)|=|(\frac{\delta}{2})(2b+\frac{\delta}{2})|=|(\frac{\delta}{2})(\frac{2}{\delta})|=1>0.7$.

Answer 1

To prove that $x^2$ is not uniformly continuous, you need to show that there exists an $\varepsilon>0$ such that for every $\delta>0$, there exists $x,y$ such that $|x-y|<\delta$ and $|x^2-y^2|\ge\varepsilon$. So it suffices to set $\varepsilon=0.7$, and prove that, for every $\delta>0$, there exists $x,y$ such that $|x-y|<\delta$ and $|x^2-y^2|\ge0.7$. You don't need to prove that for every $\varepsilon,\delta>0$ there exists $x,y$ such that $|x-y|<\delta$ and $|x^2-y^2|\ge\varepsilon$. That is a strictly stronger assertion than $x^2$ not being uniformly continuous (although that stronger assertion happens to be true in this case).

Your argument is correct, though perhaps the following argument, with $\varepsilon=1$, is cleaner:

Let $\delta>0$ and set $k=1/\delta$. Then, if $x=k+\delta/2$ and $y=k$, we have $|x-y|=\delta/2<\delta$, and $|x^2-y^2|=k\delta+\dfrac{\delta^2}{4}>k\delta=1$.

To prove the stronger assertion, we can modify the above argument by setting $k=\varepsilon/\delta$.

Answer 2

A neater solution is to take $a_n:=\sqrt n$ and $b_n:=\sqrt {n+1}$. Then $|a_n-b_n|\rightarrow 0$ while for every $n$ we have $|f(a_n)-f(b_n)|=1$.