Theorem 1   Let $f_1,\,f_2,\,\ldots$ be a sequence of real or complex functions defined on the interval $[a,\,b]$ . The sequence converges uniformly to the limit function $f$ on the interval $[a,\,b]$ if and only if $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|, \,\, a \leq x \leq b\} = 0.$$

Proof. Suppose the sequence converges uniformly. By the very definition of uniform convergence, we have that for any $\epsilon$ there exist $N$ such that

hence

Conversely, suppose the sequence does not converge uniformly. This means that there is an $\epsilon$ for which there is a sequence of increasing integers $n_i, i=1,2,...$ and points $x_{n_i}$ with the corresponding subsequence of functions $f_{n_i}$ such that

therefore

Consequently, it is not the case that $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|, \,\, a \leq x \leq b\} = 0.$$

Theorem 2   The uniform limit of a sequence of continuous complex or real functions $f_n$ in the interval $[a,b]$ is continuous in $[a,b]$