Let $f_n$ be a sequence of continuous functions defined in a compact interval C, so that $f_n$ converges to $f$ pointwise. Show that $f_n$ is uniformly convergent to $f$ if and only if:
1) $f$ is continuous on C.
2) $\forall\epsilon>0 ,\exists \delta>0,\exists N\in \mathbb{N}$ so that $n\geq N$ and $|f_k(x)-f(x)|<\delta$ implies $|f_{k+n}(x)-f(x)|<\epsilon, \forall x\in C$ and $\forall k\in \mathbb{N}$.
I think I'm able to prove it if $\delta\geq\epsilon$, in the same way that Dini's theorem is proved, showing that if $|f_k(x)-f(x)|<\epsilon$ then $|f_k(x)-f(x)|<\delta$, so $|f_{k+n}(x)-f(x)|<\epsilon$. Then we can proceed in the same way as the standard proof, since $E(_{jN})$ is ascending ($x\in E_k \rightarrow x\in E_{k+N}$). Is this correct?
I can't figure out the case $\delta<\epsilon$, any help would be aprecciated.
I suppose that your problem is about the implication: 1) and 2) implies uniformly convergence.
You can proceed also as follow:
Using you notation and fixing $\epsilon>0$:
for any $x \in C$ there exists $k_x$ such that $|f_{k_x}(x)-f(x)|<\delta$, an open neighborhood $U_{x}$ of $x$ such that $|f_{k_x}(y)-f(y)|<\delta$ is verified in $U_{x}$, and thanks to 2) there exists $N_x$ such that:
$$|f_k(x)-f(x)|<\delta$$
implies
$$|f_{n}(x)-f(x)|<\epsilon, \forall n \geq N_x$$
Thus for all $y \in U_{x}$ $$|f_{n}(y)-f(y)|<\epsilon, \forall n \geq N_x$$
Thanks to the fact that $C$ is compact, you can exctract a finite number of such $U_{x_i}$ and then take $N=\max_{i} N_{x_i}$. Then you have that for all $x \in C$, for all $n \geq N$:
$$|f_n(x)-f(x)|<\epsilon$$