Is $ f \colon \mathbb{R^2} \to \mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $\mathbb{R^2}$

Question

Is $ f \colon \mathbb{R^2} \to \mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $\mathbb{R^2} ?$ I think it is not, I tried proving it by contradiction but can't find the right $\delta$

Answer 1

First note that $g:x \to x^2$ is not uniformly continuous on $\mathbb{R}$.

Then note that $e: \mathbb{R} \to \mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and $\pi_1: \mathbb{R}^2 \to \mathbb{R}$ given by $\pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).

So if $f$ were uniformly continuous, so would $\pi_1 \circ f \circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.