Show that the function $x \rightarrow ||x||^2, \mathbb{R}^p\rightarrow \mathbb{R}, p\ge 1$ is not uniformly continuous.

Question

Show that the function $x \rightarrow ||x||^2, \mathbb{R}^p\rightarrow \mathbb{R}, p\ge 1$ is not uniformly continuous.

What i figured was that for $\forall \epsilon >0, \exists \delta>0, \forall x,y \in \mathbb{R}^p $that if $|x-y|<\delta$($\leftarrow$ not sure about this part) then $|f(x)-f(y)| <\epsilon$ thus $|\|x \|^2 - \|y\|^2 |= |x_1^2+x_2^2+\ldots +x_p^p - y_1^2-y_2^2+\ldots+y_p^2|$ however I do not know what I can do from here.

Answer 1

Consider the sequences $x_n=(n+\frac 1n,n+\frac 1n,...,n+\frac 1n)$ and $y_n=(n,n,...n)$ in $\Bbb R^p$.

Then $||x_n-y_n||=\underbrace{\sqrt{(n+\frac 1n -n)^2+(n+\frac 1n-n)^2+\cdots+(n+\frac 1n-n)^2}}_{p-\text{times}}=\frac{\sqrt p}n \Rightarrow \lim||x_n-y_n||=\lim \frac {\sqrt p}n =0.$

But $|f(x_n)-f(y_n)|=|||x_n||^2-||y_n||^2|=|(n+\frac 1n)^2+\cdots+(n+\frac 1n)^2-(n^2+\cdots+n^2)|=p|2+\frac 1{n^2}| \ge 2p \;\; \forall \;n \in \Bbb N.$