one-sided continuity by series

Theorem.  If the function series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{f}_n(x)$(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 $\epsilon$ 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) ($n=\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }$).  The uniform convergence implies the existence of a number $n_{\epsilon }$ such that on the whole interval we have

Let now  $n>n_{\epsilon }$  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 ${\delta }_{\epsilon }$ such that

Thus we obtain that

$|S(x_0+h)-S(x_0)|$$\mathrm{\hspace{0.33em}}=|[S_n(x_0+h)-S_n(x_0)]+R_{n+1}(x_0+h)-R_{n+1}(x_0|$

$\mathrm{\hspace{0.33em}}\leqq |S_n(x_0+h)-S_n(x_0)|+|R_{n+1}(x_0+h)|+|R_{n+1}(x_0)|$

as soon as

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.