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Is it true that $$\left(\int_E\sin x \,\mathrm{d}x\right)^2+\left(\int_E\cos x \,\mathrm{d}x\right)^2\leqslant4$$ for any measurable set $E \subseteq [0,2\pi]$? I'm trying to work out a functional analysis problem and finally it's equivalent to prove this.

Angelo
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user725757
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  • Can you show this when one term is zero? Can you show that the left side is invariant when replacing the argument $x$ with $(x+c)$ in both terms simultaneously? – Lutz Lehmann Sep 24 '22 at 14:40
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    Or translate the claim to $\int_E e^{ix},dx\in B(0,2)$ and try geometric ideas in the complex plane. – Lutz Lehmann Sep 24 '22 at 14:43
  • See for example https://math.stackexchange.com/q/1190530/42969 or https://math.stackexchange.com/q/3082057/42969 – Martin R Sep 24 '22 at 15:07

2 Answers2

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Define $\phi(x)=\mathbb{1}_E(x)$ for $0\leq x<2\pi$; extend $\phi$ to $\mathbb{R}$ as a $2\pi$-periodic function.

Let $z=\int_E e^{ix}\,dx$. The quantity of interest is $|z|$. Notice that $|z|=e^{i\theta}z$ for some $\theta$ and so, \begin{align} 0\leq |z|&=\int_Ee^{i(x+\theta)}\,dx=\int^{2\pi}_0\phi(x)e^{i(x+\theta)}\,dx\\ &=\int^{2\pi+\theta}_\theta \phi(x-\theta)e^{ix}\,dx=\int^{2\pi}_0\phi(x-\theta)e^{ix}\,dx \end{align} where the last identity follows from the $2\pi$-periodicity of $x\mapsto \phi(x-\theta)e^{ix}$ ( If $f$ is $T$ periodic and integrable, then $\int^T_0f(t)\,dt = \int^{T+a}_af(t)\,dt$ for all $a$).

Since $\phi$ takes only values in $\{0,1\}$, there is a measurable set $E'\subset[0,2\pi)$ such that $\phi(x-\theta)=\mathbb{1}_{E'}(x)$ for $x\in [0,2\pi)$. $E'$ is the translation (mod $2\pi$) of $E$ (wrap the interval $[0,2\pi)$ around the unit circle $|z|=1$ and $E'$ is obtained from $E$ by rotating the circle by and angle $\theta$).

All this shows that it suffices to assume that $E\subset[0,2\pi)$ is such that $$\int_E e^{ix}\,dx=\int_E\cos x\,dx>0$$

Now find the measurable $E\subset[0,2\pi)$ that maximizes $E\mapsto\int_E\cos x\,dx$ to conclude that $|z|\leq 2$.

Mittens
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    Oliver Díaz, the measurable set that maximises the map $E\mapsto\int_E\cos xdx;$ is $;E!=!\left[0,\frac\pi 2\right]!\cup!\left[\frac{3\pi}2,2\pi\right]!.,$ Is it right? – Angelo Sep 24 '22 at 16:54
  • Oliver Díaz, why did you write $$\int_E e^{ix},dt=\int_E\cos x dx$$ instead of $$\int_E e^{ix},dt=\int_E\cos x dx+i\int_E\sin x dx;;?$$ – Angelo Sep 25 '22 at 00:18
  • @Angelo: In my starting argument, I show that it is possible to translate $E$ (modulo $2\pi$) to some set $E'\subset[0,2\pi)$ so that the integral $\int_{E'}e^{ix},dx\geq0$ and so, the imaginary part of the complex integral must be $0$. – Mittens Sep 25 '22 at 01:05
  • Oliver Díaz, in your proof you wrote $,d,$ instead of $,dx,.;$ Moreover, please could you write $$|z|=\int_E e^{ix},dt=\int_E\cos x,dx\geqslant0$$ instead of $$\int_E e^{ix},dt=\int_E\cos x,dx>0;;?$$ – Angelo Sep 25 '22 at 06:11
  • Oliver Díaz, I would like to get a proof without using complex numbers. Do you have any idea or hint ? – Angelo Sep 25 '22 at 06:15
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First note that $$ \int_0^{2\pi}1_E(x)\sin(x)\,dx=\int_0^\pi(1_E(x)-1_E(x+\pi))\sin(x)\,dx. $$ On the interval $[0,\pi]$ the sine is non-negative and the difference has values in $\{-1,0,1\}$ so that $$ \left|\int_E\sin x\,dx\right|\le\int_0^\pi\sin(x)\,dx=2. $$ Note that this upper bound does not depend on $E$, just that $1_E$ has values in $\{0,1\}$.

Next observe that $$ \left|\int_E\sin(x+c)\,dx\right|^2+\left|\int_E\cos(x+c)\,dx\right|^2 =\left|\int_E\sin(x)\,dx\right|^2+\left|\int_E\cos(x)\,dx\right|^2 $$ by trigonometric and binomial identities. As $$ \int_E\cos(x+\pi)\,dx=-\int_E\cos(x)\,dx $$ there is a value $c\in[0,\pi]$ with $$\int_E\cos(x+c)\,dx=0.$$ With the first observation the claim follows $$ \left|\int_E\sin(x)\,dx\right|^2+\left|\int_E\cos(x)\,dx\right|^2 =\left|\int_E\sin(x+c)\,dx\right|^2\le 4. $$

Lutz Lehmann
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  • Lutz, why is the existence of $c$ such that $\int_E\cos(x+c),dx=0$ a consequence of $ \int_E\cos(x+\pi),dx=-\int_E\cos(x),dx;;?$ – Angelo Sep 25 '22 at 14:11
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    You have a continuous function with values of opposite sign on the interval $[0,\pi]$. By the intermediate value theorem, there is a root inside this interval. – Lutz Lehmann Sep 25 '22 at 14:13
  • Lutz, have you extended $1_E(x)$ to $\mathbb R$ as a $2\pi-$periodic function in order to write $$\int_0^{2\pi}1_E(x)\sin(x),dx=\int_0^\pi(1_E(x)-1_E(-x))\sin(x),dx;;?$$ If you let $t=2\pi-x$, you should get $$\int_0^{2\pi}1_E(x)\sin(x),dx=\int_0^\pi(1_E(x)-1_E(2\pi-x))\sin(x),dx.$$Is it right? – Angelo Sep 25 '22 at 14:25
  • Yes, asked so directly it is clear that $-x$ is not in the original domain and is not according to the original idea. I'll correct this. The idea was to connect points that are opposite on the circle, splitting the interval this is $x+\pi$ to $x$. // Applying the bound to the last step will require some periodic continuation argument or some more complicated interval and integral rearrangement.. – Lutz Lehmann Sep 25 '22 at 15:23