# condition for uniform convergence of sequence of functions

Proof of limits of functionsFernando Sanz Gamiz

###### 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

 $\left|f_{n}(x)-f(x)\right|<\frac{\epsilon}{2},\,\,\,a\leq x\leq b\hskip 13.0pt% \mbox{ for }n>N$

hence

 $\sup\{\left|f_{n}(x)-f(x)\right|,\,\,\,a\leq x\leq b\}<\epsilon\hskip 13.0pt% \mbox{ for }n>N$

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

 $\left|f(x_{n_{i}})-f_{n_{i}}(x_{n_{i}})\right|>\epsilon\hskip 13.0pt\mbox{for % all }i=1,2,...$

therefore

 $\sup\{\left|f_{n}(x)-f(x)\right|,\,\,\,a\leq x\leq b\}>\epsilon\hskip 13.0pt% \mbox{ for infinitely many }n.$

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 $[a,b]$

The proof is here (http://planetmath.org/LimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous)

Title condition for uniform convergence of sequence of functions ConditionForUniformConvergenceOfSequenceOfFunctions 2013-03-22 17:07:49 2013-03-22 17:07:49 fernsanz (8869) fernsanz (8869) 6 fernsanz (8869) Proof msc 40A30 msc 26A15