A sequence of function $(f_n)$ is said to converge unformly to $f$ if for every $\epsilon >0$ there exist $n_0 \in \mathbb{N}$ such that for every $n\geq n_0$ and for every $x \in X$ $|f_n(x)-f(x)|<\epsilon$.
From that definition, is it true that if $f_n \stackrel{u}{\rightarrow} f$ then we can find a sequence $(a_n)$ of positive real converging to 0 and $n_0\in \mathbb{N}$ such that for every $n\geq n_0$ and for every $x \in X$ $|f_n(x)-f(x)|<a_n$.
I have some idea on how to create $a_n$ that satisfied for every $n\geq n_0$ and for every $x \in X$ $|f_n(x)-f(x)|<a_n$, that is $a_n=||f_n-f||$ where $||f_n-f||=sup |f_n(x)-f(x)|$. It is obvious that $a_n$ converge to 0, but i cant guarantee that $a_n$ contains positive real for every n.
Is there any other way to create that kind of $(a_n)$?