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I've been reading up on some Analysis for my comp exams, and I couldn't find in my texts a proof of $L^\infty$ being Banach. Someone pointed me to the following exercise in Royden & Fitzpatrick. Now, I've found other proofs of $L^\infty$ being Banach online, but this problem is now really bothering me:

Let ${f_n}$ be a sequence in $L^\infty(E)$ and $\sum_{k=1}^\infty a_k$ a convergent series of positive numbers such that $||f_{k+1}-f_k||_\infty \leq a_k$ for all $k$.

Then, $\exists$ a subset $E_0$ of $E$ of measure zero and

$|f_{n+k}(x)-f_k(x)|\leq||f_{n+k}-f_k||_\infty \leq \sum_{j=n}^\infty a_j$ for all $k,n$ and all $x\in E$~$E_0$

a) Conclude that there is a function $f\in L^\infty(E)$ such that ${f_n}$->$f$ uniformly on $E$~$E_0$. b) Now, show that $L^\infty(E)$ is a Banach space.

ATTEMPT:

Thanks to the comments. The first part of the inequality:

$|f_{n+k}(x)-f_k(x)|\leq||f_{n+k}-f_k||_\infty \leq \sum_{j=n}^\infty a_j$

is immediate from the definition of the sup norm (keeping in mind that Lp spaces are really equivalence classes where equality means values coincide a.e.) and the second part is obvious by summing up the left and right-hand sides of the given inequality.

Now, does this inequality give me uniform convergence with any indices $l,q>n$ and $\epsilon = \sum_{j=n}^\infty a_j $?

If so, why does all this give me b), that $L^\infty$ is complete?

Any help is much appreciated.

Mike
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    The subset $E_0$ comes from the fact that the $L^{\infty}$ norm is the max value that $|f(x)|$ takes on a set of positive measure. Thus while $||f||_{\infty}$ may exist, you can only be sure that $|f(x)|$ also exists if you're allowed to subtract your choice of any set of measure zero from $E$. – Set Dec 13 '15 at 22:02
  • @Thoth Thanks, I see. So a simple application of the definition on the difference $f_{n+k}-f_k$ will give me the first part of the inequality: $|f_{n+k}(x)-f_k(x)|\leq||f_{n+k}-f_k||\infty \leq \sum{j=1}^\infty a_j$

    Do you have any idea why both of these are bounded by the sum $\sum_{j=1}^\infty a_j$? Or why this forces the existence of a function in $L^\infty$ that this sequence ${f_n}$ converges to uniformly?

     – Mike Dec 13 '15 at 22:08
  • Sum both sides of the inequality you are given at the start from $k$ to $k+n$ and use the sub-additivity of norms (i.e. the triangle inequality). – Set Dec 13 '15 at 22:11
  • @Thoth I see, the sum was a silly question it's obvious. I'll edit the question to make it more specific now. – Mike Dec 13 '15 at 22:21
  • As for why this implies it's a Banach space I'm not quite sure, what page in Rodyen and Fitz is this problem on? – Set Dec 13 '15 at 22:22
  • @Thoth They are exercises 32 and 33 on section 7.3 On my edition it's page 150. – Mike Dec 13 '15 at 22:22
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    the $j$ indices of your sums should start at $n$, not $1$ (just looked at my copy of the book). – Set Dec 13 '15 at 22:32
  • @Thoth Bah you're right, that's what's been bothering me with uniform convergence. In that case, I think I can call that sum to be my $\epsilon$ and for any $k,l>n$ indices uniform convergence works.

    I'll correct it.

     – Mike Dec 13 '15 at 22:35

1 Answers1

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Yes, the inequality give uniform convergence. Since for all $x\in E-E_0$, $$ \varlimsup_{n\to\infty}|f_{n+k}(x)-f_k(x)|=|f(x)-f_k(x)|\leqslant\varlimsup_{n\to\infty}\sum_{j=n}^\infty a_j\leqslant\epsilon $$ $f_k(x)\to f(x)$ uniformly on $E-E_0$. Since any Cauchy sequence converges in $L^\infty(E)$, it is complete which means is Banach.

Eugene Zhang
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  • Thanks, great. So, what we chose was an arbitrary Cauchy sequence and we showed it converges in the space. Was the uniform convergence just an extra? We don't need it to show completeness.

    Also, I made a typo, instead of sup norms I should have max norms in the first part of the question. Going through it, I don't see any of the results being affected by the typo.

     – Mike Dec 13 '15 at 23:28
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    Yes, uniform convergence is not need here. It is used to show $f$ is continuous. Also sup norms and max norms are basically the same thing. – Eugene Zhang Dec 13 '15 at 23:33