I'm studying the uniform convergence and continuity and I could not understand the proof that is given in the book, could you explain the proof explicitly ?
Particularly, Which method does it use in the proof ? What is it's strategy ? etc.
Theorem:
Assume $f_n \rightarrow f$ uniformly on an interval S.If each function $f_n$ is continuous at a point p in S, then the limit function f is also continuous at p
Proof:
We will show that for every $\epsilon > 0$ the is a neighbourhood N(p) such that $|f(x) - f(p)| < \epsilon$ whenever $x \in N(p) \cap S$.If $\epsilon > 0$ is given, there is am integer N such that $n \geq N$ implies $$|f_n(x)-f(x)| < \epsilon / 3$$ for all x in S.
Since $f_N$ is continuous at p, there is a neighbourhood N(p) such that $$|f_N (x) - f_N (p)| < \epsilon / 3$$ for all $$x \in N(p) \cap S$$
Therefore, for all $$x \in N(p) \cap S$$, we have $$|f(x) - f(p)|=|f(x) - f_N (x) + f_N (x) - f_N (p) + f_N (p) - f(p)| \\ \leq |f(x) f_N (x)| +|f_N (x) - f_N (p)| + |f_N (p) - f(p)| $$
Since each term on the right is $< \epsilon / 3$, we find $|f(x) - f(p)|< \epsilon$, which completes the proof.
I'm not sure whether it is ok, but you can also look the proof from here
I'll go line by line.
This uses the uniform convergence of the sequence $f_n$.
This line just uses the uniform continuity for each $f_n$. A threshold can be selected independent of the point in the interval. The neighborhood is basically selecting $\delta$ independent of the point.
This is my favorite trick. The triangle inequality is used over and over to force the estimate we want. We obtain that $|f(x) - f(p)| < \epsilon$ for any appropriate $x$. We chose $\epsilon/3$ because the continuity criterion holds for any $\epsilon$ - we might as well choose it to be another small number (as my prof says, "It's the same among friends."