Let $f$ and $f_1, f_2, \dots, f_n, \dots$ be continuous functions on $[a, b]$.
Suppose that $\lim_{n \to \infty} f_n(x) = f(x)$ for any $x \in [a, b]$.
I think the following proposition is true:
If $\bigcap_{n = 1}^\infty \{x \in [a, b] ; f(x) = \max \{|f(x) - f_n(x)| ; x \in [a, b]\} \} \ne \emptyset$, then $\{f_n\}$ converges uniformly to $f$.
If $\bigcap_{n = 1}^\infty \{x \in [a, b] ; f(x) = \max \{|f(x) - f_n(x)| ; x \in [a, b]\} \} = \emptyset$, then is it possible that there exists $\{f_n\}$ which doesn't converge uniformly to $f$?
As infinite intersection is not empty, this means exists $\exists x_0 \in [a,b]$ (1), in which we have maximum for all functions i.e.$\forall n \in \mathbb{N}, \max \{|f(x) - f_n(x)| = |f(x_0) - f_n(x_0)|$ (2). As sequence $f_n$ is convergence in all points from $[a,b]$, then it also converges in $x_0$ i.e. $|f(x_0) - f_n(x_0)| \to 0$ (3). By limit definition it means,that $\forall \epsilon > 0$ existence of $N(x_0)$ , that for $\forall n > N(x_0)$ we have $|f(x_0) - f_n(x_0)| < \epsilon$. From our property this estimation holds for all $[a,b]$, which gives us uniform convergence.