What is $\sup_\limits{E\subset[-\pi,\pi)}\left|\int_Ee^{it}\,dt\right|$?

Question

I couldn't think of a descriptive title other than the problem statement itself. What is $$\sup_\limits{E\subset[-\pi,\pi)}\left|\int_Ee^{it}\,dt\right|?$$

(The question seems possibly relevant to determining the best constant in the inequalities in my answer here.)

well it's certainly not bigger than $2\pi$, but of course you knew that — Rylee Lyman, May 18 '19 at 15:34
@RyleeLyman In fact it's clearly strictly smaller than $2\pi$, because of cancellation... — David C. Ullrich, May 18 '19 at 15:36

score 3 · Answer 1 · answered May 18 '19 at 15:47

3

Oh. This is simpler than I thought, by an argument not at all like what I had in mind:

First, replacing $E$ by a translate, wlog $$\int_Ee^{it}\,dt\ge0,$$so in particular $$\left|\int_Ee^{it}\,dt\right|=\int_E\cos(t)\,dt.$$And it's clear that the last integral is maximized by $E=[-\pi/2,\pi/2]$.

answered May 18 '19 at 15:47

David C. Ullrich

89,985

Can you expand on your WLOG? It seems like the integral is complex in general, so it's not obvious to me exactly what you mean. – preferred_anon May 18 '19 at 15:51
@preferred_anon If $F$ is $E$ translated by $\theta$ then $\int_F = e^{i\theta}\int_E$. Choose $\theta$ so $e^{i\theta}\int_E\ge0$. – David C. Ullrich May 18 '19 at 15:53
Ah! And multiplying by $e^{i\theta}$ doesn't change the modulus. Nice! – preferred_anon May 18 '19 at 15:55

What is $\sup_\limits{E\subset[-\pi,\pi)}\left|\int_Ee^{it}\,dt\right|$?

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