point preventing uniform convergence

Theorem. If the sequencef1,f2,f3,  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 x0 of the interval such that the function sequence converges uniformly on no closed sub-interval of  [a,b] containing x0.

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,a+b2]  and  [a+b2,b],  since otherwise it would do it on the union  [a,b]. Denote by  [a1,b1]  the first (fom left) of those half-intervals on which the convergence is not uniform. We have  [a,b][a1,b1]. Then the interval  [a1,b1]  is halved and chosen its half-interval  [a2,b2]  on which the convergence is not uniform. We can continue similarly arbitrarily far and obtain a unique endless sequence


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


The nested interval theorem thus gives a unique real number x0 belonging to each of the intervals  [a,b]  and  [an,bn]. Then  limnan=x0=limnbn.  Let us choose α and β such that  aαx0βb. There exist the integers n1 and n2 such that




This means that  fnf  not uniformly on  [an,bn][α,β], whence the function sequence does not converge uniformly on the arbitrarily chosen subinterval[α,β]  of  [a,b] containing x0.

Title point preventing uniform convergence
Canonical name PointPreventingUniformConvergence
Date of creation 2013-03-22 17:27:18
Last modified on 2013-03-22 17:27:18
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 6
Author pahio (2872)
Entry type Theorem
Classification msc 40A30
Related topic NotUniformlyContinuousFunction
Related topic LimitFunctionOfSequence