Uniform continuous of $f(x,y) = \sin(x^2+y)$

Question

Let $f(x,y) = \sin(x^2+y)$. Show that $f$ is not uniformly continuous over $\mathbb{R^2}$.

I know that $g(x) = \sin(x^2)$ is not uniformly continuous over $\mathbb{R}$, but does it imply something about $\mathbb{R^2}$? How can we use the definition of uniformly continuity in $\mathbb{R}$ to uniformly continuity in $\mathbb{R^2}$?

Answer 1

Whenever a function is uniformly continuous, its restriction to a subset of its domain is uniformly continuous too. But, as you know, the restriction of $f$ to $\mathbb R\times\{0\}$ is not uniformly continuous and so neither is $f$.

Answer 2

If it is uniformly continuous then given $\epsilon >0$ there exists $\delta >0$ such that $\|(x,y)-(u,v)\| <\delta $ implies $|f(x,y)-f(u,v)| <\epsilon$. In particular we can take $y=v=0$ to conclude that $|f(x,0)-f(u,0)| <\epsilon$ whenever $|x-u|<\delta$. But you already know that this is not true.