Let $f : \Omega \subseteq \mathbb{R}^n \to \mathbb{R}$ be a smooth and convex function. Let us assume that $\Gamma\_f = \mathrm{graph}(f) $ is a complete hypersurface of $\mathbb{R}^{n+1}$. Then I know that $\Gamma\_f$ must be **properly** embedded. 


I can prove this simple fact with an argument by contradiction, but I was wondering if there is a more direct and elegant proof. 


Any help will be very appreciated! 



On the properness of the graph of a convex function
A fairly elementary proof is as follows (I assume you are using the convention that $f$ convex means $\Omega$ is convex and $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$). For $x,y\in \Omega$ let $[x,y]\subset \Omega$ be the segment connecting them.


Idea, if $p=(x,f(x)),q=(y,f(y))\in \Gamma\_f$, then one has
$$
d(p,q)\leq \int\_0^1 \sqrt{|x-y|^2+|x-y|^2 |f'(tx+(1-t)y)|^2} dt
$$
where here $d(p,q)$ is the intrinsic distance in $\Gamma\_f$. 
That is,
$$
d(p,q)\leq |x-y| \int\_0^1 \sqrt{1+ |f'(tx+(1-t)y)|^2} dt
$$
Using $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ gives
$$
d(p,q) \leq |x-y| \int\_0^1 1+ |f'(tx+(1-t)y)| dt
$$
and so
$$
d(p,q) \leq |x-y|(1+f(x)+f(y)-2\min\_{z\in [x,y]} f(z)).
$$
Where this last estimate using the fact that $f|\_{[x,y]}$ is convex.


It follows, that if $d(p,p\_i)\to \infty$ for a sequence $p\_i=(y\_i, f(y\_i))$, then either $y\_i\to \infty$ or $f(y\_i)\to \infty$ and so one concludes $\Gamma\_f$ is proper as desired.


