# point preventing uniform convergence

Theorem. If the sequence$f_{1},\,f_{2},\,f_{3},\,\ldots$  of real functions converges at each point of the interval$[a,\,b]$ but does not converge uniformly on this interval, then there exists at least one point $x_{0}$ of the interval such that the function sequence converges uniformly on no closed sub-interval of  $[a,\,b]$ containing $x_{0}$.

Proof. Let the limit function of the sequence on the interval  $[a,\,b]$  be $f$. According the entry uniform convergence on union interval, the sequence can not converge uniformly to $f$ both half-intervals$[a,\,\frac{a+b}{2}]$  and  $[\frac{a+b}{2},\,b]$,  since otherwise it would do it on the union  $[a,\,b]$. Denote by  $[a_{1},\,b_{1}]$  the first (fom left) of those half-intervals on which the convergence is not uniform. We have  $[a,\,b]\supset[a_{1},\,b_{1}]$. Then the interval  $[a_{1},\,b_{1}]$  is halved and chosen its half-interval  $[a_{2},\,b_{2}]$  on which the convergence is not uniform. We can continue similarly arbitrarily far and obtain a unique endless sequence

 $[a,\,b]\supset[a_{1},\,b_{1}]\supset[a_{2},\,b_{2}]\supset\ldots$

of nested intervals on which the convergence of the function sequence is not uniform, and besides the length of the intervals tend to zero:

 $\lim_{n\to\infty}(b_{n}-a_{n})=\lim_{n\to\infty}\frac{b-a}{2^{n}}=0.$

The nested interval theorem thus gives a unique real number $x_{0}$ belonging to each of the intervals  $[a,\,b]$  and  $[a_{n},\,b_{n}]$. Then  $\lim_{n\to\infty}a_{n}=x_{0}=\lim_{n\to\infty}b_{n}$.  Let us choose $\alpha$ and $\beta$ such that  $a\leqq\alpha\leqq x_{0}\leqq\beta\leqq b$. There exist the integers $n_{1}$ and $n_{2}$ such that

 $|a_{n}-x_{0}|=x_{0}-a_{n}\leqq x_{0}-\alpha\quad\mbox{when}\;\;n>n_{1}$
 $|b_{n}-x_{0}|=b_{n}-x_{0}\leqq\beta-x_{0}\quad\mbox{when}\;\;n>n_{2}.$

Therefore

 $\alpha\leqq a_{n}\leqq x_{0}\leqq b_{n}\leqq\beta\quad\mbox{when}\;\;n>\max\{n% _{1},\,n_{2}\}.$

This means that  $f_{n}\to f$  not uniformly on  $[a_{n},\,b_{n}]\subset[\alpha,\,\beta]$, whence the function sequence does not converge uniformly on the arbitrarily chosen subinterval$[\alpha,\,\beta]$  of  $[a,\,b]$ containing $x_{0}$.

Title point preventing uniform convergence PointPreventingUniformConvergence 2013-03-22 17:27:18 2013-03-22 17:27:18 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 40A30 NotUniformlyContinuousFunction LimitFunctionOfSequence