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If $f \in C^0([0,1])$, one can define $$\Vert f \Vert_? = \sup_{J \subset [0,1]} \left\lvert \int_J f \right\rvert,$$ where $J$ runs among all subintervals of $[0,1]$.

This is a norm on $C^0([0,1])$ (and Lebesgue's density theorem shows that this generalizes on spaces like $L^1(I)$, where $I$ is an interval, but I don't really want to go to this generality).

The norm is strictly coarser than the $L^1$ norm, but it satisfies something reminiscent of it: if $f, g \in C^0([0,1])$ are such that $f$ is monotonic, then one gets $$ \left\vert \int_0^1 f g \right\rvert \leq C \Vert f \Vert_\infty \Vert g \Vert_?,$$ for some universal constant $C$. If I'm not mistaken, $C = 2$ works, thanks to the second mean value theorem for integrals.

If you had the patience to read all of this, my questions are the following:

  • is this norm (or a variant of it) well-known? has it got a name?
  • do you have a simple proof of the inequality with $f$ monotonic? one without that obscure second MVT for integrals? what is the best constant $C$?
PseudoNeo
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  • Is it easy to see or even true that $|f|? = \max(|f+|1,|f-|1)$? Where $f\pm=\max(0,\pm f)$ is the positive and negative parts of $f$. – Lutz Lehmann Sep 26 '24 at 15:49
  • It's false. If $f_n : x \mapsto \sin(2\pi n x)$, then $|f_n^+| \gg 1$ but $| f_n |_? = O(1/n)$, unless I'm mistaken. – PseudoNeo Sep 26 '24 at 16:05
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    @LutzLehmann I think if you use all measurable subsets, not just subintervals, then your formula would be true. – GEdgar Sep 26 '24 at 16:21
  • I expect that the second mean value theorem provides a natural and very simple way to prove the required inequality and it provides $C=2$. I expect to show that this bound is tight we can consider small positive $\varepsilon$ tending to $0$ and for each fixed $\varepsilon$ and each $x\in [0,1]$ put $f(x)=-1$, if $0\le x\le \tfrac 12-\varepsilon$, $f(x)=(x-\tfrac 12)\tfrac 1{\varepsilon}$, if $\tfrac 12-\varepsilon\le x\le \tfrac 12-\varepsilon$, $f(x)=1$, if $\tfrac 12+\varepsilon \le x\le 1$, and $g(x)=f(x)$. – Alex Ravsky Oct 02 '24 at 05:31
  • I crossposted on MO → https://mathoverflow.net/questions/480678/a-curious-norm-related-to-the-l%c2%b9-norm – PseudoNeo Oct 15 '24 at 07:34

1 Answers1

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I had some fun doing a bit of "abstract nonsense", which you may or may not consider as an answer to your first question. I hope it helps you nonetheless.

Let $X$ be a Banach space (generalizing $L^1([0, 1])$, which contains $C^0([0, 1])$ as a non-closed subspace) and suppose $\mathcal F^\ast\subset X^\ast$. Then the following defines a semi-norm on $X$: $$ \lvert f\rvert_{\mathcal F^\ast}:=\sup_{f^\ast\in \mathcal F^\ast}\left\lvert \langle f^\ast , f\rangle\right\rvert.$$ Your example is clearly of this form. For $\lvert \cdot\rvert_{\mathcal F^\ast}$ to be a norm, $\mathcal F^\ast$ should be "total" in the sense that $$\tag{1} \overline{\operatorname{span} \mathcal F^\ast}=X^\ast.$$ Indeed, $\lvert f\rvert_{\mathcal F^\ast}=0$ is equivalent to $f\bot \mathcal F^\ast$, and this implies $f=0$ if and only if $\mathcal F^\ast$ is total (NOTE: I am not 100% sure that's how sets satisfying (1) are called, but it gives the idea).

In your case, $\mathcal F^\ast=\{\mathbf 1_J\ :\ J\text{ is a subinterval of }[0, 1]\}$, which indeed is a total set in $L^\infty([0, 1])$, so that defines a norm on $L^1([0, 1])$. Since $C^0([0, 1])\subset L^1([0, 1])$, that is a norm on $C^0$, too, but I am rather sure that it is not Banach. (Take for example a sequence of continuous functions that approximates $\mathbf{1}_{[1/2, 1]}$, which is not in $C^0$. That sequence is going to be Cauchy. Hope it is clear what I mean).

Remark. All of this says absolutely nothing about your second question. That one is specific of continuous functions and as such is impermeable to all this abstract nonsense.

Giuseppe Negro
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  • I think the term is indeed total. Just to clarify, whe nyou say closure you mean weak-star closure and by orthogonal you mean $f$ belongs to the (pre)annihilator of $\mathcal F^*$, right? – Evangelopoulos Foivos Sep 26 '24 at 17:30
  • Yes, well, I don't really know if these are the right names, pre-annihilator, annihilator, and all of that, but I guess that's correct. I want an assumption that ensures that $\langle f^\star, f\rangle=0$ for all $f^\star$ implies that $f=0$. – Giuseppe Negro Sep 26 '24 at 17:50
  • Is F really total in that example? How do you approximate the indicator function of a positive-measure Cantor set, for instance? If it were, I feel "my" norm would be more or less equivalent to the L¹ norm. – PseudoNeo Sep 27 '24 at 11:06