# On equibounded sequences in $L^\infty$

Let $$f_n: [0, 1] \to \mathbb R$$ be a sequence of positive functions in $$L^\infty$$ (hence a fortiori in $$L^1$$) that are equibounded in $$L^\infty$$ norm - that is $$\sup_{n \in \mathbb N} \|f_n\|_{L_\infty} \leq M$$ for some $$M > 0$$.

Is it true that there exists some absolute positive constant $$c < 1$$ such that

$$\inf_{n_k} \sup_{i, j > N} \|f_{n_i} - f_{n_j}\|_{L^1} \leq cM$$

for all such sequences $$f_n$$?

Where the first infimum is taken over all increasing sequences $$n_k$$ of naturals.

• Hm $f_n$ are positive, so in fact $c = 1$ works, but the question wants $c < 1$. Jul 17, 2021 at 3:51
• Oh, I missed that! Never mind. Jul 17, 2021 at 4:00
• Rademacher functions? Jul 17, 2021 at 4:19
• If that refers to what I think it is, then I believe that has $c = 1/2$. Jul 17, 2021 at 6:00
• I don't understand the purpose of the "$\lim_{N \to \infty}$" in the fomula. Isn't $\inf_{n_k} \lim_{N \to \infty} \sup_{i, j > N} \|f_{n_i} - f_{n_j}\|_{L^1} = \inf_{n_k} \sup_{i, j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1}$, since increasing sequences in $\mathbb{N}$ can start arbitrarily late? Jul 17, 2021 at 12:55

Edit: I improved the constant to $$c = \frac{2}{3}$$. (Later edit: But the optimal constant turns out to be $$c = \frac{1}{2}$$, see Yuval Peres' answer.)

Answer: Yes, we have $$\inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \label{1}\tag{\ast}$$ for each sequence $$(f_n)$$ in $$(L^1)_+$$ whose sup norm is bounded by $$M$$. So we can choose $$c = \frac{2}{3}$$.

To see this, let $$[0,\mathbf{1}] \subseteq L^\infty$$ denote the positive unit ball in $$L^\infty$$.

Lemma. Three functions $$f_1, f_2, f_3 \in [0,\mathbf{1}]$$ cannot have mutual $$L^1$$-distances that are all strictly larger than $$\frac{2}{3}$$.

Proof. Set $$g_1 = |f_1 - f_2|$$, $$g_2 = |f_1 - f_3|$$ and $$g_3 = |f_2 - f_3|$$. For any three numbers $$r_1,r_2,r_3 \in [0,1]$$, the sum of their three mutual distances in $$\mathbb{R}$$ is at most $$2$$.

Hence, $$\int g_1 + \int g_2 + \int g_3 \le 2$$, which shows that it can't happen that all three functions $$g_k$$ have norm strictly larger than $$\frac{2}{3}$$. $$\square$$

Proof of the claim. We may, and will, assume that $$M=1$$. Assume for a contradiction that we can find a sequence $$(f_n)$$ in $$[0,\mathbf{1}]$$ such that the infimum in the question is strictly larger than $$\frac{2}{3}$$.

Then there exists $$n_0$$ such that $$\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$$ for infinitely many $$n$$ (otherwise we could recursively construct a subsequence $$(f_{n_k})$$ such that the supremum in \eqref{1} is no more than $$\frac{2}{3}$$); let's denote the set of these $$n$$ by $$J$$.

For any two $$j,k \in J$$, it follows from the lemma that $$\|f_j - f_k\| \le \frac{2}{3}$$. Thus, you can take the elements of $$J$$ to be the indices of your wanted subsequence $$(f_{n_k})$$. Contradiction, since we assumed no such subsequence exists. $$\square$$

Remark. It's easy to see that the constant $$\frac{2}{3}$$ is optimal for the lemma (divide $$[0,1]$$ into three distjoint intervals $$I_k$$ of measure $$\frac{1}{3}$$ and define $$f_k = \mathbf{1} - \mathbf{1}_{I_k}$$), but I don't know whether it is optimal for the answer to the question.

• I think one can also argue that $\|f-g\|_1\simeq 1$ for such functions is only possible if essentially $f=\chi_A$, $g=\chi_{A^c}$ (off a set of small measure, and up to small (in $L^{\infty}$) perturbations), but then on a whole sequence the sets will start overlapping. But considering three functions at once is a neat trick that makes this less messy. Jul 17, 2021 at 14:42
• @ChristianRemling: Yepp, that was my first thought, too. I think it yields readily that the infimum is strictly smaller than $M$ for each fixed sequence $(f_n)$. But then I wasn't able to show that we get a uniform constant $c$ for all sequences (the usual trick to mix up sequences in order to show that a strict estimate for each sequence actually implies a uniform estimate, doesn't work here since we are allowed to take subsequences); so after a while I thought, it might be easier to just find an explicit estimate. Jul 17, 2021 at 15:01
• Very nice answer! Jul 18, 2021 at 1:21
• @NateRiver: Thanks for the extra points! :-) Jul 23, 2021 at 18:28

The sharp constant is $$c=1/2$$. As in Jochen's answer, we may and shall assume that $$M=1$$.

Proposition: Let $$f_n: [0, 1] \to \mathbb R$$ be a sequence of positive functions in $$L^\infty$$ such that $$\sup_{n \in \mathbb N} \|f_n\|_{L_\infty} \leq 1$$. Then

$$\inf_{\{n_k\}} \sup_{i, j \ge 1} \|f_{n_i} - f_{n_j}\|_{L^1} \leq 1/2 \,,$$ where the infimum is over all strictly increasing sequences $$\{n_k\}$$.

To see that this is sharp, consider the functions $$\{b_n\}$$ where $$b_n(x)$$ is the $$n$$th bit in the binary expansion of $$x$$. Note that $$\|b_n-b_m\|_1=1/2$$ for all $$n \ne m$$.

Lemma 1: For any $$k$$ numbers $$y_1,\ldots ,y_k$$ in $$[0,1]$$, we have $$\sum_{i=1}^{k-1}\sum_{j=i+1}^k |y_j-y_i| \le \lfloor k/2 \rfloor \cdot \lceil k/2 \rceil \le k^2/4 \,.$$ Proof: We may reorder the $$y_i$$ so that $$y_1 \le y_2\le\ldots \le y_k$$. Every interval $$[y_\ell,y_{\ell+1}]$$ is included in $$\ell(k-\ell)$$ intervals of the form $$[y_i,y_j]$$ with $$i \le \ell , and $$\max_\ell \ell(k-\ell)=\lfloor k/2 \rfloor \cdot \lceil k/2 \rceil$$.

Lemma 2. Given $$k$$ measurable functions $$f_1,\ldots, f_k$$ taking values in $$[0,1]$$, there exist $$i so that $$\|f_i-f_j\|_1 \le \frac{ \lfloor k/2 \rfloor \cdot \lceil k/2 \rceil}{{k \choose 2}} \le \frac{k}{2(k-1)} \,.$$.

Proof: By Lemma 1, for each $$x \in [0,1]$$ we have $$\sum_{i=1}^{k-1}\sum_{j=i+1}^k |f_j(x)-f_i(x)| \le \lfloor k/2 \rfloor \cdot \lceil k/2 \rceil \,,$$ so integrating gives $$\sum_{i=1}^{k-1}\sum_{j=i+1}^k \|f_j-f_i\|_1 \le \lfloor k/2 \rfloor \cdot \lceil k/2 \rceil \,.$$ Since the minimum of $${k \choose 2}$$ numbers is at most their average, the claim follows.

Proof of proposition: Let $$f_n: [0, 1] \to \mathbb R$$ be a sequence of positive functions in $$L^\infty$$ such that $$\sup_{n \in \mathbb N} \|f_n\|_{L_\infty} \leq 1$$. It suffices to show that for every $$c>1/2$$ there is a strictly increasing sequence $${n_k}$$ such that

$$\quad \sup_{i, j \ge 1} \|f_{n_i} - f_{n_j}\|_{L^1} \leq c \,.\label{2}\tag{\ast}$$ By changing the values on a set of measure zero, we may assume that each $$f_n$$ takes values in $$[0,1]$$. Fix $$c>1/2$$ and find $$k$$ such that $$\frac{k}{2(k-1)} . Define a graph on the positive integers where there is an edge $$\{i,j\}$$ iff $$\|f_i-f_j\|_1 > c$$. By Lemma 2, this graph does not contain a clique of $$k$$ nodes, so by Ramsey's Theorem [1] there is an infinite independent set (i.e., an anti-clique) in this graph, which proves \eqref{2}.

Remark: Related considerations in Hilbert space are in [2].

• Really great answer! Are you aware of any results of this type for $L^q(0,1)$ and $L^p(0,1)$ for general $p < q$? If $(f_n)$ is in the positive $L^p$-unit ball, then I guess I'd expect a similar estimate for the $L^q$-norm, maybe with the constant $\big(\frac{1}{2}\big)^{1/p - 1/q}$ (but after thinking about it a bit I couldn't find, for instance, a simple interpolation argument to prove it). Jul 19, 2021 at 10:33
• Thanks Jochen- of course my answer was inspired by yours. I think $p$ and $q$ are reversed in part of your comment. For functions in the positive unit ball in $L^\infty$ one can easily see that when switching from $L^1$ to $L^p$ the $1/2$ is replaced by $2^{-1/p}$. The positivity assumption is less natural once we leave $L^\infty$. In $L^\infty$ requiring positivity in the unit ball is equivalent to replacing the unit ball by the ball of radius $1/2$. Jul 20, 2021 at 2:10