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Suppose $M=\left \{ f\in L^1([0,1])\, |\, 0<f(x)<\frac 1{\sqrt x} \text{almost everywhere on} \, (0,1) \right\}$.

Is it true or not, that $M$ is totally bounded?

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I don't know if you've seen the comment to the deleted answer where I said I messed up. Sorry about that. I don't have time to write up an answer right now, but you should be able to adapt the proof of theorem 5 in Hanche-Olsen and Holden's notes on the Kolmogorov-Riesz compactness theorem. This may be overkill, but should lead to an (affirmative) answer. –  t.b. Apr 8 '12 at 15:51
    
Wow. Thx, I'll think about it. It's interesting result. My intuition tells me, that $M$ isn't totally bounded but I'm completely stucked. –  sas Apr 8 '12 at 15:59
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Isn't this just false? Take a sequence $(A_n)$ of subsets of $(0,1)$ of measure $1/2$ such that the symmetric difference of any two has measure $1/2$. Then the sequence of functions $f_n$ which take value $1-\epsilon$ on points in $A_n$ and $\epsilon$ on points not in $A_n$ gives an infinite collection whose pairwise distances are large, no? –  user83827 Apr 8 '12 at 16:04
    
@ccc Where did you get such subsets? –  Norbert Apr 8 '12 at 16:08
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Well I suppose one easy construction would be something like $A_n$ is the set of guys whose $n$th decimal digit is even. –  user83827 Apr 8 '12 at 16:10
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2 Answers

up vote 7 down vote accepted

It is not totally bounded.

For an abstract reason why not, note that $M$ contains a ball of the $L^\infty$ norm. If $M$ were totally bounded then the inclusion $L^\infty([0,1]) \hookrightarrow L^1([0,1])$ would be compact. But it isn't.

We can use the example given there to make a concrete proof. Let $e_n(x) = \sin(2 \pi n x)$. It's well known, and simple to check, that the $e_n$ are orthogonal in $L^2([0,1])$, i.e. $\int_0^1 e_n(x) e_m(x)\,dx = 0$ for $n \ne m$. We also have $\int_0^1 e_n(x)^2\,dx = \frac{1}{2}$. We can then observe that, for $n \ne m$, $\int_0^1 (e_n(x) - e_m(x))^2\,dx = 1$.

Now, noting that $\frac{1}{2}|e_n - e_m| \le 1$, we have $\frac{1}{2}|e_n - e_m| \ge \frac{1}{4}|e_n - e_m|^2$. Thus, for $n \ne m$, we have $$\frac{1}{2} \int_0^1 |e_n(x) - e_m(x)|\,dx \ge \frac{1}{4} \int_0^1 (e_n(x)-e_m(x))^2 \,dx = \frac{1}{4}.$$ So we have shown $\lVert e_n - e_m \rVert_1 \ge 1/2$.

Set $f_n = \frac{1}{2} + \frac{1}{4} e_n$. Then $\frac{1}{4} \le f_n \le \frac{3}{4}$ so $f_n \in M$, and we have $\lVert f_n - f_m \rVert_1 \ge \frac{1}{8}$ for all $n \ne m$. Thus if we have any cover of $M$ by balls of some radius $\epsilon < \frac{1}{16}$, each ball can contain at most one of the $f_n$, and hence the cover is not finite.

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So should I downvote this question and your answer because this homework question does not show any work...? –  1015 Jul 11 '13 at 15:31
    
It's been a very pleasant online interaction I learned a lot from, thanks. –  1015 Jul 11 '13 at 16:42
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Another useful set of functions in $M$ are the 'square waves'. Let $f_1$ be the indicator function of $\cup_{i=0}^{\infty} [i, i+\frac{1}{2}]$, and define $f_{n+1}(x) = f(2^n x)$. It is simple to show that $||f_n - f_m|| = \frac{1}{4}$ as long as $m\neq n$.

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A simpler characterization would be to let $f_n(x)$ be the $nth$ binary digit of $x \in [0,1)$ (resolve any ambiguity by taking the shorter representation). –  copper.hat Apr 8 '12 at 19:48
    
The functions that @copper.hat is referring to (at least one family of functions on [0,1] that picks off the nth binary digit) are called that Rademacher functions. These turn out I be pretty useful in many circumstances. –  Keaton Apr 8 '12 at 20:41
    
There is a wonderful book by Mark Kac entitled "Statistical independence in probability analysis and number theory" that explores non-obvious, but delightful connections between coin tosses (Rademacher functions), probability and numbers. –  copper.hat Apr 8 '12 at 23:07
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