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

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 $$\{M_{f_n}\}_{n \in \mathbb{N}}$$.

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!

• Have you tried $f_n(x) = n$ when $x \geqslant n$, $f_n(x) = 2x - n$ when $\tfrac{1}{2}n \leqslant x < n$, and $f_n(x) = 0$ when $x < \tfrac{1}{2}n$? Apr 26, 2021 at 10:10

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".