We have $f(\theta )= |x \cos \theta +y \sin \theta | $ and $e=|x|+|y|$ and define for $n\in\mathbb{Z}_{\ge 0}$ $$f_n= \sup \{f(2\pi k 2^{-n}): k=0,1,\dots,2^n \}.$$
We will show that $f_n $ is an increasing $e$-uniform Cauchy sequence.
I have to show that $|f_n-f_m|\leq \epsilon(|x|+|y|)$. Can anyone please help me? Here is my attempt:
Let $m<n$. We denote $|f_n-f_m|=| \bigvee_k \{f(2\pi k 2^{-n})- \bigvee_k\{f(2 \pi k^{-m}) \}|$. By the Birkhoff's inequality, we have
$$\leq \bigvee_k \Bigl\{|f(2\pi k 2^{-n})\}- \{f(2 \pi k^{-m}) |\Bigr\}$$
By the triangle inequality $||a|-|b|| \leq |a-b|$ we have
$$= \bigvee_k \Bigl\{ \big| |x \cos{(2\pi k 2^{-n})} + y \sin{(2\pi k 2^{-n})}| -|x \cos{(2\pi k 2^{-m})}+y \sin{(2\pi k 2^{-m})}| \big| \Bigr\}$$
$$\leq \bigvee_k \Bigl\{|x \cos{(2\pi k 2^{-n})} + y \sin{(2\pi k 2^{-n})}- x \cos{(2\pi k 2^{-m})}-y \sin{(2\pi k 2^{-m})}| \Bigr\}$$
$$= \bigvee_k \Bigl\{ |x (\cos{(2\pi k 2^{-n})} -\cos{(2\pi k 2^{-m})}) + y( \sin{(2\pi k 2^{-n})}- \sin{(2\pi k 2^{-m})})| \Bigr\}$$
$$= \bigvee_k \Bigl\{|-2x \sin{ (\pi k(2^{-n}+2^{-m}))} \sin{ (\pi k(2^{-n}-2^{-m}))}+2y\cos {(\pi k(2^{-n}+2^{-m})) \sin{(\pi k(2^{-n}-2^{-m}))}}| \Bigr\}$$
$$\leq \bigvee_k \Bigl\{|-2x \sin{ (\pi k(2^{-n}+2^{-m}))} \sin{ (\pi k(2^{-n}-2^{-m}))}|\Bigr\}+\bigvee_k\Bigl\{2y\cos {(\pi k(2^{-n}+2^{-m}))} \sin{(\pi k(2^{-n}-2^{-m}))}|\Bigr\}$$
$$\leq \bigvee_k\Bigl\{|-2x||\pi k(2^{-n}+2^{-m})||\pi k(2^{-n}-2^{-m})|\Bigr\} +\bigvee_k \Bigl\{ |2y||\pi k(2^{-n}-2^{-m})|\Bigr\}$$
$$\leq \bigvee_k\Bigl\{|-2x||\pi k(2^{-n}+2^{-m})||\pi k(2^{-n}-2^{-m})|\Bigr\} +\bigvee_k \Bigl\{ |2y||\pi k(2^{-n}-2^{-m})|\Bigr\}$$
$$\leq \left[|-2x||\pi (2^{-n}+2^{-m})||\pi (2^{-n}-2^{-m})|+|2y||\pi (2^{-n}-2^{-m})|\right]\sup|k|$$