Proving continuity of a linear function from $\mathbb R^2$ to $\mathbb R^2$ Consider a function
$$f(x,y) = (ax-y, x+y),a \in \mathbb{R}$$
How can you prove that $f(x,y)$ is continuous (everywhere) using the definition of:

$f$ is continuous at $x_0 \in \mathbb{R}^2$ if $\forall \epsilon > 0\ \exists \delta > 0$ such that for $x \in \mathbb{R}^2$ with $d(x,x_0) < \delta$ then $d(f(x), f(x_0)) < \epsilon.$

I am using the standard Euclidean metric i.e.
$d(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$
It is clear it is continuous as all components of $f$ are continuous, however, I am not sure how to obtain my delta.
I have:
$d(x,x_0)=\sqrt{(x-x_0)^2 + (y-y_0)^2} = \sqrt{x^2-2xx_0+x_0^2+y^2-2yy_0+y_0^2} < \delta$
$d(f(x),f(x_0))=\sqrt{(ax-y-ax_0+y_0)^2 + (y+y-y_0-y_0)^2}$
I was expecting to get $d(x,x_0) = kd(f(x),f(x_0))$, where $k=const. \in \mathbb{R}$ (some multiple of $a$). Then I could take $\epsilon=\delta$ and it would be clear. However, I am struggling to find this constant (if it even exists).
 A: I claim that $$\delta = \frac{\epsilon}{\sqrt{a^2 + 3 + |2-2a|}}$$
works. For all $a \in \mathbb R$, $a^2 + 3 + |2-2a| > 0$, so $\delta$ is well-defined.
Proof. Let us investigate continuity at $(x_0,y_0) \in \Bbb R^2$. Suppose $$d((x,y), (x_0,y_0)) = \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta$$ Then, $|x-x_0| < \delta$ and $|y-y_0| < \delta$.
Note that
$$\begin{align}
|(ax - ax_0 - y + y_0)^2 + (x-x_0 + y-y_0)^2| &= |(a^2 + 1)(x-x_0)^2 + 2(y-y_0)^2 + (2-2a)(x-x_0)(y-y_0)|\\ &\le |(a^2 + 1)(x-x_0)^2| + |2(y-y_0)^2| + |(2-2a)|(x-x_0)(y-y_0)|\\ &< (a^2 + 1)\delta^2 + 2\delta^2 + |2-2a|\delta^2\\ &= \delta^2 (a^2 + 3 + |2-2a|)\\ &= \epsilon^2
\end{align}$$
which means $$d(f(x,y), f(x_0,y_0))= \sqrt{(ax - ax_0 - y + y_0)^2 + (x-x_0 + y-y_0)^2} < \epsilon$$
from the above calculation. Thus, $f$ is continuous at $(x_0,y_0) \in \mathbb R^2$. Since $(x_0,y_0) \in \mathbb R^2$ is arbitrary, $f$ is continuous at every point of $\mathbb R^2$.
Remark. The choice of $\delta$ does not depend on $(x_0,y_0)$. Hence, $f(x,y)$ is uniformly continuous on $\Bbb R^2$.
