When $f : \mathbb{R}^2\rightarrow \mathbb{R}$ is a nonnegative continuous function, then define $F (v) = \lim_{t\rightarrow \infty}\ \frac{f(tv)}{t}$. When $F$ is a norm on the vector space $\mathbb{R}^2$, then
given $\varepsilon$, there is $R>0$ s.t. $F(v)>R$ implies $|\frac{f(v)}{F(v)} -1 |<\varepsilon$.
How can we prove this ?
Proof : If not, then there is a sequence $v_n$ s.t. $\lim_n\ F(v_n)=\infty$ and $|\frac{f(v_n)}{F(v_n)} -1| \geq \varepsilon$.
Assume that $\frac{v_n}{F(v_n)}\rightarrow x$. But I can not finish the proof.
Sorry, I have to backtrack on my previous answer and change it quite radically, and I claim that not only the function $v\mapsto C_v$ I was describing above is not continuous, but actually the whole statement above is false: I'll produce a function $f$ such that there is a sequence of points $v_n$ with $F(v_n)$ going to infinity and yet $|f(v)/F(v)-1|\geq 1$ (I'll delete my previous answer as soon as I have some feedback on this one). Apologies for the mess.
Let $f$ be defined as follows: for $v=(x,y)$ with $x\not\in(0,2)$, $f(v)=|v|$. For $v$s of the form $(1,y)$, let $f(v)=3y$. For the two remaining stripes where the function is not defined, let $f$ be anything, as long as it is continuous: it is clear that such a function ca be found. Now, $F(v)=|v|$ for every point. But now, let us consider the points $v_n:=(1,n)$: $F(v_n)=\sqrt{1+n^2}$, but $f(v_n)=3n)$. In particular, for large enough $n$, we have $f(v_n)/F(v_n)=3n/\sqrt{1+n^2}>2$, so in particular $|f(v_n)/F(v_n)-1|>1$, as promised.