We set $$C(\mathbb{R}) = \left\{ f : \mathbb{R} \to \mathbb{R}: f \text{ is continuous and } \sup_{x \in \mathbb{R}}\left|f(x) \right| < +\infty \right\}, $$ and assume that $(f_n)_{n\geq 1}$ is a sequence in $C(\mathbb{R})$ that converges uniformly to $f \in C(\mathbb{R})$. Then it can be shown that $(f_n)_{n\geq 1}$ is pointwise equicontinuous on $\mathbb{R}$, i.e., for every $x \in \mathbb{R}$ and $\varepsilon > 0$, there exists $\delta >0$ (depending on $x$ and $\varepsilon$) such that, for all $n \geq 1$ and $\left| y-x \right| < \delta,$ we have $\left| f_{n}(y) - f_{n}(y)\right| < \varepsilon.$
My question is: In this setting, is it true that $(f_n)_{n\geq 1}$ is uniformly equicontinuous on $\mathbb{R}$? In other words, for every $\varepsilon > 0$, there exists $\delta >0$ (depending only on $\varepsilon$) such that, for all $n \geq 1$ and $\left| y-x \right| < \delta,$ we have $\left| f_{n}(y) - f_{n}(y)\right| < \varepsilon.$
This is not my homework. In fact, I am just curious about the result when reading books on Real Analysis, and I think the answer is negative but cannot find any counterexamples.
Any help would be appreciated.
Hint: Find a bounded continuous $f$ on $\mathbb R$ that is not uniformly continuous. Then definee $f_n(x) = f(x)$ for all $n.$