I can not find a counterexample although I have the feeling it is not true.
Let $\ g: \mathbb{ R} \rightarrow \mathbb{R}$ continuous function
$ \forall x \in \mathbb{R} \ g(x+1) = g(x)$
$g(0) = 0$
Let $(f_n)_{n=1}^{\infty} \ f_n : \mathbb{R} \rightarrow \mathbb{R},\forall n \in \mathbb{N} \ f_n(x) = g(\frac{x}{n})$
Then $ \ (f_n)_{n=1}^{\infty} \ $ is uniform convergence in $ \ \mathbb{R} \ $ to $0$
So clearly we can see that there is a pointwise convergence $\ \forall x \in \mathbb{R} \ $ to $0$
And because there is a cycle I need to check only what is happening in $[0,1]$
foreach $ \epsilon \ \exists x\in [0,1] \ g(x) = \epsilon $ so there must be a such a $N \in \mathbb{N}$ so that even for the "worst" $x$ we can say that $|g(\frac{x}{n})| < \epsilon$
This is my unformal proof.
So, is the proof good ?
Am I right ?
If it is not true, please explain to me what am I doing wrong and give a counterexample.
Thanks in advanced !!
With the definition
$$f_n(x) = g\left(\frac{x}{n}\right),$$
is is clear that
$$f_n(\mathbb{R}) = \{ f(x) : x \in \mathbb{R}\} = \{ g(y) : y \in \mathbb{R}\} = g(\mathbb{R}) = g([0,\,1]).$$
So, for all $n$, we have $\sup\limits_{x\in\mathbb{R}} \lvert f(x) - g(0)\rvert = \max\limits_{y\in [0,\,1]} \lvert g(y) - g(0)\rvert$.
Hence the sequence converges uniformly on all of $\mathbb{R}$ if and only if $g$ is constant.
$f_n$ converges uniformly to $g(0)$ on all bounded subsets of $\mathbb{R}$, however.