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.


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?


(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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    $\begingroup$ 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

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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