I am considering the family $\mathcal{F}$ of functions $f \colon \mathbb{R} \to \mathbb{R}$ which have at most linear growth at infinity, that is there exists a constant $M\_f$ such that
\begin{equation} |f(x)| \leq M\_f (1+|x|) . \end{equation}


In particular, for each $f \in \mathcal{F}$ we define $M\_f$ as the strictest constant to have this property:
\begin{equation} M\_f := \sup\_{x \in \mathbb{R}} \frac{|f(x)|}{1+|x|} . \end{equation}


We consider the family $\mathcal{F}$ as a subspace of the space of continuous functions $C(\mathbb{R},\mathbb{R})$, equipped with the topology of uniform convergence in compact sets.


**CLAIM:** I would like to prove that whenever a sequence $\{f\_n\}\_{n \in \mathbb{N}} \subset \mathcal{F}$ converges (uniformly in compact sets) to a function $f \in \mathcal{F}$, then $M\_{f\_n} \to M\_f$, or at least we can bound uniformly the sequence $\{M\_{f\_n}\}\_{n \in \mathbb{N}}$.


**Comments:**


1. If I restrict the domain to a compact set $K \subset \subset \mathbb{R}$, then I can prove the claim.
2. Further, in a compact set $K \subset \subset \mathbb{R}$, I can prove the following:
$$ M\_g^K \leq M\_f^K + d\_K(f,g) ,$$
where $M\_g^K$ is the constant restricted to $K$ and $d\_K$ is the uniform distance in $K$. A similar formula doesn't seem to hold in whole $\mathbb{R}$.


Can you prove the claim or find a counter-example? Thank you!



Functions with at most linear growth at infinity: is the constant itself continuous?
The answer is no to both your hopes: it can happen that neither $M\_{f\_n}\to M\_f$ nor $\sup\_n M\_{f\_n}<+\infty$ hold, although $M\_f<\infty$.


As a counter-example take
$$
f\_n(x)=\max(0,n(x-n)).
$$
(I'm too lazy to include a picture: $f\_n$ is zero for $x\leq n$, and then starts growing with slope $n$ for $x\geq n$).
Clearly $f\_n\to f\equiv 0$ locally uniformly and in particular $M\_f=0$, but $M\_{f\_n}=n$ (the slope of $f\_n$ at infinity).


Roughly speaking, the incompatibility comes from the fact that your topology (local uniform convergence) is local in space and thus "does not see infinity", while the functional $f\mapsto M\_f$ is allowed to "see up to infinity".


