condition for uniform convergence of sequence of functions

Proof of limits of functionsFernando Sanz Gamiz

Theorem 1.

Let  $f_{\mathrm{1}}\mathrm{,}{f}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }$  be a sequence of real or complexfunctions defined on the interval  $\mathrm{[}a\mathrm{,}b\mathrm{]}$.  The sequenceconverges uniformly to the limit function $f$ on theinterval  $\mathrm{[}a\mathrm{,}b\mathrm{]}$ if and only if

$$\underset{n\to \mathrm{\infty }}{lim}sup\{|f_n(x)-f(x)|,a\le x\le b\}=0.$$

Proof.

Suppose the sequence converges uniformly. By the very definition ofuniform convergence, we have that for any $ϵ$ there exist $N$such that

hence

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

$$\left|f(x_{n_i})-f_{n_i}(x_{n_i})\right|>ϵ\mathit{\hspace{1em}\hspace{0.25em}}\text{for all }i=1,2,\mathrm{\dots }$$

therefore

$$sup\{|f_n(x)-f(x)|,a\le x\le b\}>ϵ\mathit{\hspace{1em}\hspace{0.25em}}\text{ for infinitely many }n.$$

Consequently, it is not the case that

$$\underset{n\to \mathrm{\infty }}{lim}sup\{|f_n(x)-f(x)|,a\le x\le b\}=0.$$

∎

Theorem 2.

The uniform limit of a sequence of continuous complex or realfunctions $f_n$ in the interval $\mathrm{[}a\mathrm{,}b\mathrm{]}$ is continuous in $\mathrm{[}a\mathrm{,}b\mathrm{]}$

The proof ishere (http://planetmath.org/LimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous)