Functions with at most linear growth at infinity: is the constant itself continuous?

Question

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!

Answer 1

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".