If $\sup(f(x)-L(x))$ is finite for all $L \in V^*$ then for every $C\in\Bbb R$ there exists a compact set $K$ such that $f\le C$ on $V\setminus K$. Let $V$ be an $n$-dimensional real vector space. Let $f:V\rightarrow \mathbb R$ be a function (not necessarily linear). Suppose that  for all $L\in V^*$ it follows that $\sup_{x\in V}(f(x)-L(x))<\infty$. I want to show that

Then for any $C\in \mathbb R$, there is $K\subset V$ compact such that $f(x)<C$ on $V-K$

It's obvious that $f$ must be bounded by some $M$, so if $C>M$ we can take $K$ to be the empty set. But I am not sure what to do when $C<M$
 A: This is actually straightforward, if perhaps a bit intricate.
Just to give it a name, let's say
$\newcommand\small{\sup_x(f(x)-Lx)}$


Def. $f$ is small if $\small<\infty$ for  every $L\in V^*$.


And note that a simple application  of looking up definitions shows that 


Exercise. The following are equivalent: (i) $\lim_{x\to\infty}f(x)=-\infty$, (ii) for every $C$ there exists a compact $K$ such that $f\le  C$ on $V\setminus K$.


The other day I said silly things about how there are no small functions. In fact of course


Triviality If $f$ is bounded above and $\lim_{x\to\infty}f(x)/|x|=-\infty$ then $f$ is small.


And in fact this answers the question:


Thm. $f$ is small if and only if $f$ is bounded above and $\lim_{x\to\infty}f(x)/|x|=-\infty$.


For the less trivial direction, suppose that $f$ is small but the conclusion does not hold. Since $L=0$ shows that $f$ is bounded above, we must have $f(x)/|x|\not\to-\infty$, which says that there exist a sequence $x_n\to\infty$ and $c\in\Bbb R$ with $$f(x_n)\ge c|x_n|.$$
Now since the unit sphere of $V$ is compact, passing to a subsequence we can assume $$\frac{x_n}{|x_n|}\to\zeta.$$Let $\alpha=c-1$ and choose $L\in V^*$ with $$L\zeta=\alpha.$$Now $$f(x_n)-Lx_n\ge |x_n|\left(c-L\left(\frac{x_n}{|x_n|}\right)\right);$$since $L$ is continuous we have $c-L\left(\frac{x_n}{|x_n|}\right)\to c-\alpha=1$, which shows that $$f(x_n)-Lx_n\to\infty,$$so $f$ is not small.
