9
$\begingroup$

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.

This norm and this inequality appear in the background of the classical Jordan's theorem saying that the Fourier series of a BV function converges: the point is that whereas the L¹ norm of the Dirichlet kernel goes to infinity, such isn't the case with this norm.

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 know what is the best constant $C$?
Improve this question
$\endgroup$
2
  • $\begingroup$ I asked the question on MSE, but go no real answer, despite a bounty → math.stackexchange.com/questions/4976792/… $\endgroup$
    – PseudoNeo
    Commented Oct 15 at 7:33
  • 3
    $\begingroup$ Your norm satisfies $\|F'\|_? = \sup F - \inf F$ and it is comparable to the supremum norm of the antiderivative: $\sup |F| \leqslant \|F'\|_? \leqslant 2 \sup |F|$ if $F(x) = 0$ for some $x \in [0, 1]$. In particular, if, given $f$, we choose $F = \int f$ in such a way that $\sup F = -\inf F$, then $\|f\|_? =2 \sup |F| = 2 \|F\|_\infty$. So your norm seems to be related (or maybe simply equal) to the norm of $W^{-1,\infty}([0,1])$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 15 at 8:15

1 Answer 1

Reset to default
8
$\begingroup$

I think $C=2$ is the best constant. Consider $\varepsilon>0$ and let $f,g$ continuous in $[0,1]$ defined as follows. \begin{equation*} f(x) = \begin{cases} -1, & 0\leq x \leq \frac12 - \varepsilon \\ 1, & \frac12 + \varepsilon \leq x \leq 1, \\ \text{linear}, & |x-\frac12| \leq \varepsilon \end{cases}, \end{equation*} and $g\in C^0[0,1]$ such that $\Vert g - (-\chi_{(1/4,1/2)}+\chi_{(1/2,1/4)})\Vert_{L^1[0,1]} < \varepsilon. $ Then $$ \Big| \int_0^1 f g \, dx \Big | = \frac12 + O(\varepsilon) $$ and $$ \Vert g \Vert_? \leq \Vert g- (-\chi_{(1/4,1/2)}+\chi_{(1/2,1/4)})\Vert_{L^1[0,1]} + \Vert -\chi_{(1/4,1/2)}+\chi_{(1/2,1/4)} \Vert_? = \frac14 + O(\varepsilon) $$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .