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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

Our
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  • Perhaps you should try to be more specific. Bear in mind that nobody "understands" a proof when they read it through; that's just not how it works. – Thompson Oct 30 '16 at 15:58

1 Answers1

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I'll go line by line.

"If $\epsilon > 0$ is given, there... for all $x \in S$."

This uses the uniform convergence of the sequence $f_n$.

"Since $f_N$ is continuous at $p$, there is a neighbourhood..."

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.

"Therefore, for all $x \in N(p) \cap S$..."

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."

Sean Roberson
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  • but we are trying to prove the uniform convergence of a sequence $f_n$, so how can it be used within its proof itself ? – Our Oct 31 '16 at 06:27