Question

Assume that. F_n converges to f uniformly on S if each f_n is continuous at a point c of S .show that the limit function f us also continuous at c

Answer 1

Solution:

Since $f_n → f$ uniformly on S and each $f_n$ is continuous on S, for  given $ε > 0$

there is a $\delta>0$ such that as $|y-x|<\delta$ , where $y \in S$ , we have

$|f(y)-f(x)|<\frac{\varepsilon}{2}$

For this $\delta>0,\exists$ a positive integer $N_{1}$ such that as $n \geq N_{1}$ , we have

$\left|x_{n}-x\right|<\delta$

Hence, as $n \geq N_{1}$ , we have

$\left|f\left(x_{n}\right)-f(x)\right|<\frac{\varepsilon}{2}\ ...(1)$

Also, since $f_{n} \rightarrow f$ uniformly on S, given $\varepsilon>0,\exists$ a positive integer $N \geq N_{1}$ such that as $n \geq N$ , we have

$\left|f_{n}(x)-f(x)\right|<\frac{\varepsilon}{2} \text { for all } x \in S$

$\Rightarrow \left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|<\frac{\varepsilon}{2} \ ... (2)$

By (1) and (2), we obtain that given $\varepsilon>0,\exists$ a positive integer N such that as $n \geq N$ , we have

$\begin{aligned} \left|f_{n}\left(x_{n}\right)-f(x)\right| &=\left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|+\left|f\left(x_{n}\right)-f(x)\right| \\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ &=\varepsilon \end{aligned}$

$\Rightarrow f_{n}\left(x_{n}\right) \rightarrow f(x)$

Hence, limit function f is also continuous at any point c.

Hence, proved.