# one-sided continuity by series

Theorem.  If the function series

 $\displaystyle\sum_{n=1}^{\infty}f_{n}(x)$ (1)

is uniformly convergent on the interval$[a,\,b]$,  on which the $f_{n}(x)$ are continuous from the right or from the left, then the sum function $S(x)$ of the series has the same property.

Proof.  Suppose that the terms $f_{n}(x)$ are continuous from the right.  Let $\varepsilon$ be any positive number and

 $S(x)\;:=\;S_{n}(x)+R_{n+1}(x),$

where $S_{n}(x)$ is the $n^{\mathrm{th}}$ partial sum of (1) ($n\,=\,1,\,2,\,\ldots$).  The uniform convergence implies the existence of a number $n_{\varepsilon}$ such that on the whole interval we have

 $|R_{n+1}(x)|<\frac{\varepsilon}{3}\quad\mathrm{when}\;\;n>n_{\varepsilon}.$

Let now  $n>n_{\varepsilon}$  and  $x_{0},\,x_{0}\!+\!h\in[a,\,b]$  with  $h>0$.  Since every $f_{n}(x)$ is continuous from the right in $x_{0}$, the same is true for the finite sum $S_{n}(x)$, and therefore there exists a number $\delta_{\varepsilon}$ such that

 $|S_{n}(x_{0}\!+\!h)-S_{n}(x_{0})|<\frac{\varepsilon}{3}\quad\mathrm{when}\;\;0%

Thus we obtain that

 $\displaystyle|S(x_{0}\!+\!h)-S(x_{0})|$ $\displaystyle\;=\;|[S_{n}(x_{0}\!+\!h)-S_{n}(x_{0})]+R_{n+1}(x_{0}\!+\!h)-R_{n% +1}(x_{0}|$ $\displaystyle\;\leqq\;|S_{n}(x_{0}\!+\!h)-S_{n}(x_{0})|+|R_{n+1}(x_{0}\!+\!h)|% +|R_{n+1}(x_{0})|$ $\displaystyle\;<\;\frac{\varepsilon}{3}\!+\!\frac{\varepsilon}{3}\!+\!\frac{% \varepsilon}{3}\;=\;\varepsilon$

as soon as

 $0

This means that $S$ is continuous from the right in an arbitrary point $x_{0}$ of  $[a,\,b]$.

Analogously, one can prove the assertion concerning the continuity from the left.

Title one-sided continuity by series OnesidedContinuityBySeries 2013-03-22 18:34:03 2013-03-22 18:34:03 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 40A30 msc 26A03 one-sided continuity of series with terms one-sidedly continuous OneSidedContinuity SumFunctionOfSeries