$f: (\mathbb R^2,\|\cdot\|_\infty)\to \mathbb R$, $f(x , y) =2x+3y$ is uniformly continuous

Question

Show that

$$f: (\mathbb R^2,\|\cdot\|_\infty)\to \mathbb R, \ \ f(x , y) =2x+3y$$

is uniformly continuous. I actually know how to do this exercise or the concept atleast but im kinda stuck with the norm since its one of the very few times im working wth norms so I would like to get my exercise checked and corrected if done wrong. Thanks for your time in advance.

So my approach is let $\epsilon>0$ and $d=? $ I leave it blank will fill it later as i finish the exercise.

So let $x=(x_1,x_2)$ and $y=(y_1,y_2)$ and $k\in\{1,2\}$ such that

\begin{align} |x_k + y_k| &= \max \{|x_1 + y_1|,|x_2 + y_2|\} < d \\[.2cm] \implies(f(x_1,y_1)-f(x_2,y_2)) &= |2x_1+3y_1-2x_2-3y_2|\\&\leq 2|(x_1-x_2)|+3|(y_1-y_2)| \leq 5|x_k + y_k| \\&<5d=\epsilon \end{align} so i also add my $d=\frac{e}{5}$ here.

ps: added an extra line to make my proof even more clear.

Answer 1

You somehow screwed up the arguments of the function $f$. I would moreover, since you mentioned $\|\cdot\|_\infty$ explicitly in the definition of $f$, use that both $|x_1-y_1|\le\|x-y\|_\infty$ and $|x_2-y_2|\le\|x-y\|_\infty$ instead of this $k$-choice...