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.


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.

  • 1
    $\begingroup$ 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 $E$. $\endgroup$
    – Set
    Dec 13, 2015 at 22:02
  • $\begingroup$ @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? $\endgroup$
    – Mike
    Dec 13, 2015 at 22:08
  • $\begingroup$ 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). $\endgroup$
    – Set
    Dec 13, 2015 at 22:11
  • $\begingroup$ @Thoth I see, the sum was a silly question it's obvious. I'll edit the question to make it more specific now. $\endgroup$
    – Mike
    Dec 13, 2015 at 22:21
  • 1
    $\begingroup$ the $j$ indices of your sums should start at $n$, not $1$ (just looked at my copy of the book). $\endgroup$
    – Set
    Dec 13, 2015 at 22:32

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Mike
    Dec 13, 2015 at 23:28
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 13, 2015 at 23:33

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