Gap between the Uniform and Uniform convergence on compacts topologies

Question

Let $X$ and $Y$ be the set $C_0(\mathbb{R})$ (of functions vanishing at infinity) equipped with the topologies of compact-convergence and uniform convergence respectively. The second is clearly stronger than the first (as the closure of the compactly-supported functions differ in these cases) but what is an example of a function $f \in C_0(\mathbb{R})$ and a sequence $\{f_n\}$ there such that

• $\lim\limits_{n\to \infty} f_n(x) =f(x)$ in $X$
• $\lim\limits_{n\to \infty} f_n(x) \neq f(x)$ in $Y$?

Answer 1

I think you can start with any non-constant function $g$ in $C_0(\mathbb{R})$ and build such a sequence by defining $f_n(x) = g(x-n)$.

Now in the topology of the compact convergence this sequence converges to the function $f \equiv 0$, since for any compact set $K\subset \mathbb{R}$ you can find some integer $n$ such that $f_n(x)$ is pushed so far to the right that its value on $K$ gets arbitrarily small.

However in the topology of the uniform convergence this sequence never converges to $f\equiv 0$ (nor to any other function) since its supremum and infimum remain constant and by hypothesis at least one of them is nonzero.