# uniform convergence on union interval

Theorem. If  $a  and the sequence$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 UniformConvergenceOnUnionInterval 2013-03-22 17:27:09 2013-03-22 17:27:09 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 40A30 MinimalAndMaximalNumber