uniform convergence on union interval

Theorem. If $a<b<c$ and the sequence $f_{1},\,f_{2},\,f_{3},\,\ldots$ of real functions converges uniformly both on the interval $[a,\,b]$ and on the interval $[b,\,c]$ , then the function sequence converges uniformly also on the union (http://planetmath.org/Union) interval $[a,\,c]$ .

Proof. We have the limit functions $\displaystyle f_{ab}:=\lim_{n\to\infty}f_{n}$ on $[a,\,b]$ and $\displaystyle f_{bc}:=\lim_{n\to\infty}f_{n}$ . It follows that

$f_{ab}(b)=\lim_{n\to\infty}f_{n}(b)=f_{bc}(b).$

Define the new function

$\displaystyle f(x):=\begin{cases}f_{ab}(x)\quad\forall x\,\in[a,\,b],\\ f_{bc}(x)\quad\forall x\,\in[b,\,c].\end{cases}$

Choose an arbitrary positive number $\varepsilon$ . The supposed uniform convergences on the intervals $[a,\,b]$ and $[b,\,c]$ imply the existence of the numbers $n_{1}(\varepsilon)$ and $n_{2}(\varepsilon)$ such that

$|f_{n}(x)-f(x)|<\varepsilon\;\;\forall x\,\in[a,\,b],\quad\mbox{when}\;n>n_{1}% (\varepsilon)$

and

$|f_{n}(x)-f(x)|<\varepsilon\;\;\forall x\,\in[b,\,c],\quad\mbox{when}\;n>n_{2}% (\varepsilon).$

If one takes $n>\max\{n_{1}(\varepsilon),\,n_{2}(\varepsilon)\}$ , then one has simultaneously on both intervals $[a,\,b]$ and $[b,\,c]$ , i.e. on the whole greater interval $[a,\,c]$ , the condition

$|f_{n}(x)-f(x)|<\varepsilon.$

Title	uniform convergence on union interval
Canonical name	UniformConvergenceOnUnionInterval
Date of creation	2013-03-22 17:27:09
Last modified on	2013-03-22 17:27:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 40A30
Related topic	MinimalAndMaximalNumber