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I'm reading page number 4 here. In particular the section where it deals with the case $p=\infty$, that is , showing that $L^{\infty}$ is complete.

http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf

Two questions:

1) Why is the convergence uniform? where it says "for $x \in N^{c}$ , $f_{n}$ is a Cauchy sequence of complex numbers. Thus $f_{n} \rightarrow f$ uniformly. Clearly we have pointwise convergence but why is it uniform?

2) I don't see why $||f_{n} - f||_{\infty} \rightarrow 0$. Can you please explain this step?

Thanks.

(3/2015) Edit: The original link appears to be broken. This document seems to provide a similar (maybe even identical) proof to the one the OP talks about with slight notational differences.

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closed as unclear what you're asking by Jack, user147263, user91500, Watson, John B May 24 '16 at 9:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You should read more closely the definition of the L^{\infty} norm and remember that f_n is assumed to be Cauchy with respect to it. $\endgroup$ – Qiaochu Yuan Nov 7 '10 at 20:59
  • $\begingroup$ Hint: Have you shown that $f_n(x)$ is Cauchy for $x\notin N$? A similar argument gives the uniform convergence on the set $N^c$. $\endgroup$ – user940 Nov 7 '10 at 21:05
  • $\begingroup$ Can't we simply say that since f_{n} is Cauchy with respect the norm $||.||_{\infty}$ then given $\varepsilon>0$ there exists a natural N such that for all n,m greater than N we have $||f_{n} - f_{m}|| < \varepsilon$. But by definition of $B_{n,m}$ we have $|f_{n}(x)-f_{m}(x)| < \varepsilon$. Passing to the limit we have $|f_{n}(x) - f(x)| <\varepsilon$. Now my question is: is the convergence uniform because the natural N only depends on $\varepsilon$? $\endgroup$ – student Nov 7 '10 at 21:09
  • $\begingroup$ @beginner: no. Reread the definition of the L^{\infty} norm. $\endgroup$ – Qiaochu Yuan Nov 7 '10 at 21:15
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The key is that we are working in $N^c$, where we have, morally speaking, defined $N$ to be the set of points where things go wrong. In detail, $N$ is the set of points $x$ where either 1) $f(x)$ is bigger than the limsup $\|f\|$, or 2) the series $(f_n)$ is not Cauchy at $x$.

Outside of $N$, the terms $f_n$ are bounded and the series $(f_n)$ is Cauchy.

For question 1), use the fact that $(f_n)$ is (uniformely!) Cauchy outside of $N$ and that $\mathbb C$ is complete to define a limit function $f$ on the set $N^c$. It should follow, pretty much by definition of $f$, that $f_n \to f$ uniformely on $N^c$.

For question 2), then we know from question 1 that $\|f_n - f\|_\infty \to 0$ on $N^c$. Next, extend the limit function $f$ to the whole set by setting it equal to 0 on $N$. What is the measure of $N$? How does that play into the definition of the $\| \cdot \|_\infty$ norm, and thus the question about limits?

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