# A periodic integral inequality

(This problem comes in connection with a geometric problem exposed here.)

Let $$\gamma(x,y)$$ be a (real) function on the unit disk such that

$$\frac{\partial^2\gamma}{\partial x \, \partial y} = 0\:\:\:\:\text{and}\:\:\:\:0<\gamma(x,y)<\pi,$$

then show that

\begin{align} \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\sqrt{1-\sin 2t \cos\gamma(t)}\leq 1 \end{align}

being $$\gamma(t)=\gamma\left(\cos t, \sin t\right)$$ the boundary value of $$\gamma(x,y)$$.

(For the case of $$\gamma(t)$$ sufficiently close to $$\pi/2$$, Robert Bryant's great answer to this question proves the result.)

An attempt of proof:

Following Bryant's idea a possible proof would come as follows. We define the function

$$$$f(\lambda)\equiv \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\sqrt{1-\sin 2t \sin\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right]},$$$$

being $$f(1)$$ the desired integral. It follows that

$$$$f'(0)=-\frac{1}{2} \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\left(\frac{\pi}{2}-\gamma(t)\right)\sin2t=0,$$$$

with the zero in the r.h.s coming only because $$\frac{\partial^2\gamma}{\partial x \partial y}=0$$. It also follows that $$f''(0)<0$$. So in order to prove that $$f(1)\leq f(0)=1$$ we need to show that $$f'(\lambda)$$ remains negative, i.e

$$$$f'(\lambda)=-\frac{1}{2} \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\frac{\sin2t \cos\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right] }{\sqrt{1-\sin 2t \sin\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right]}}\cdot\left(\frac{\pi}{2}-\gamma(t)\right)\leq 0 \tag{*}$$$$

for $$\lambda\in (0,1)$$. Now, the second derivative is in general not negative in the whole interval, and the integrand in $$f'(\lambda)$$ oscillates around $$0$$, so it is apparently highly non-trivial to show $$(*)$$

• how do Cauchy–Schwarz do the job? Nov 2, 2022 at 14:38
• I'm just curious if the function is just of the usual form $f(x)+g(y)$ (because the diameter of the disk is containing every other cross-section and the disk is convex) or I missing some fancy example that forces you to write the differential equation instead of this representation. Of course, the condition $0<\gamma<\pi$ is something non-trivial in both cases. Jan 24, 2023 at 2:38
• @fedja Indeed, I just wrote the partial derivatives $\gamma$ instead of $f(x)+g(x)$ to save notation, but both forms are equivalent and actually I use the second one in practical calculations. Jan 24, 2023 at 8:53

It is false in general. Let $$\gamma(x,y)=f(x)+g(y)$$. Splitting each term into the odd and even part, we get $$\gamma=A(x,y)+X(x)+Y(y)$$ where $$A$$ is even in each variable and $$X$$ and $$Y$$ are odd. The condition $$0<\gamma<\pi$$ implies that $$A(x,y)\pm(|X(x)|+|Y(y)|)\in(0,\pi)$$ when $$(x,y)\in \mathbb D$$. The natural attempt to prove the result would be to use the symmetries of $$\sin(2t)$$ and reduce the integration to $$[0,\pi/2]$$ with $$x=\cos(t), y=\sin(t)$$. Then, expanding $$\cos(A+X+Y)$$, it would suffice to show that for $$x,y>0, x^2+y^2=1$$, $$\mathcal E_{\varepsilon=(\varepsilon_x,\varepsilon_y)\in\{-1,1\}^2} \left[1+2xy(\cos A\sin X\sin Y+\varepsilon_x\sin A\cos X\sin Y \\+\varepsilon_y\sin A\sin X\cos Y-\varepsilon_x\varepsilon_y \cos A\cos X\cos Y)\right]^{1/2}\le 1$$ for all real numbers $$A,X,Y$$ satisfying the above assumptions. That would finish the story, but a computer check of random data quickly yields the counterexample $$A_0=2,\quad X_0=0.5, \\ \quad Y_0=-0.5,\quad 2x_0y_0=0.4$$ with the average $$E=1.01000235275203$$ and that is a killer because now we can make $$A_1(x)$$ and $$A_2(y)$$ with range in $$[\frac\pi 4,1]$$ each so that on the quarter-circle, $$A(x,y)=A_1(x)+A_2(y)$$ is $$A_0$$ near $$(x_0,y_0)$$ and quickly descends to $$\pi/2$$ when you leave that small neighborhood $$U$$ while $$X(x_0)=X_0$$, $$Y(y_0)=Y_0$$ and they quickly drop to $$0$$ when you leave the neighborhood $$U$$. In that case you essentially integrate $$1$$ outside $$U$$, $$E$$ on $$U$$, and the ranges are such that $$A(x,y)$$ stays in $$[\frac\pi 2,2]$$ and $$|X(x)|, |Y(y)|\le 0.5$$ for all $$x,y\in[-1,1]$$, so the sum $$f(x)+g(y)\in[\frac \pi 2-1, 3]\subset (0,\pi)$$ in the whole disk and not only on the boundary.
• Thank you. The argument is hard for me to follow, especially the second paragraph . What is $U$? As I see from the beginning, you think of a way that may prove the claim and then realize that it cannot, but that doesn't disprove it. Jan 24, 2023 at 9:03
• $U$ is a tiny arc around $(x_0,y_0)$. Yes, I initially thought of a way to prove it ,but once you have a counterexample to that way (found by a computer), you can immediately upgrade it to a counterexample to the original problem, which I did by checking that you can extend it from a small neighborhood $U$ of a problematic point to the full domain preserving the restrictions and keeping the integrand at 1 elsewhere except for a set of arbitrarily small measure (the descent region to keep the functions smooth; if you allow discontinuities, it is not even needed), so my post does disprove it. Jan 24, 2023 at 12:17