Prove/disprove $(\int_{0}^{2 \pi} \!\!\cos f(x) d x)^{2}+(\int_{0}^{2 \pi}\!\!\! \sqrt{(f'(x))^{2}+\sin ^{2} f(x)}dx)^{2}\ge 4\pi^{2}$

Question

This problem has been posted on Math.SE but didn't receive any correct answer after a long time.

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Let $f(x)$ be a differentiable function on $[0,2\pi]$ s.t. $0\leq f(x)\leq 2\pi$ and $f(0)=f(2\pi)$. Prove or disprove that $$ \left(\int_{0}^{2 \pi} \cos f(x) d x\right)^{2}+\left(\int_{0}^{2 \pi} \sqrt{(f'(x))^{2}+\sin ^{2} f(x)} d x\right)^{2} \geq(2 \pi)^{2} $$

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It seems that when $f$ is an arbitrary constant, the left side equals $(2\pi)^2$ and seems to be the minimum. But how can I show that there's no other $f$ that makes the left side equal (or be less than) $(2\pi)^2$?

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A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$, we have its perimeter $\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$ and its area $\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$. From spherical isoperimetric inequality $L^2\ge S\left( 4\pi-S \right)$, we have $\left( 2\pi-S \right)^2+L^2\ge\left( 2\pi \right)^2$, and the equality holds iff $C$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry.

Now the question is, how to prove the inequality with only pure analysis methods?

Answer 1

An approach that should work is to derive the differential equation that any minimizer would have to satisfy and check that its solutions are the known ones for which equality holds. To fill in the details, one would need to show that a minimizer does exist and has the necessary regularity to make the derivation of the differential equation valid, but I think that such arguments are standard in the calculus of variations and should be applicable here.

So, assume that a minimizer $f$ exists and set $$ a = \frac1{2\pi}\int_0^{2\pi}\cos f(x)\,dx\quad\text{and}\quad b = \frac1{2\pi}\int_0^{2\pi}\sqrt{f'(x)^2+\sin^2 f(x)}\ dx. $$ Note that, if $b=0$, then $f(x)$ is an integer multiple of $\pi$ and we have equality. Set this case aside and assume that $b>0$. The assumption that $f$ be a minimizer implies that if $u$ is any $2\pi$-periodic function then $I'(0)=0$ where $$ \begin{align} I(t) &= \left(\int_0^{2\pi}\cos\bigl(f(x){+}tu(x)\bigr)\,dx\right)^2\\ &\qquad\qquad+\left(\int_0^{2\pi}\sqrt{(f'(x){+}tu'(x))^2+\sin^2 (f(x){+}tu(x)\bigr)}\ dx\right)^2. \end{align} $$ Calculation (using integration by parts and the fact that $u$ and $f$ can be regarded as $2\pi$-periodic) then yields that $I'(0)/(4\pi)$ equals $$ \int_0^{2\pi} \left(-b\left(\frac{f'(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}\right)'+\frac{b\sin f(x)\cos f(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}-a\sin f(x)\right)\,u(x)\ dx. $$ Consequently, setting $\lambda = a/b$, we have that $f$ must satisfy a second-order differential equation with parameter $\lambda$ $$ \left(\frac{f'(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}\right)'-\frac{\sin f(x)\cos f(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}+\lambda\,\sin f(x) = 0.\tag1 $$ Thus, there is a $3$-parameter family of solutions to this equation with parameter, and $f$ must belong to this $3$-parameter family.

Now, it just so happens that we already know a $3$-parameter family of solutions to this equation, namely $$ f(x) = \arccos\left(\frac{u{+}(p\cos x {+} q \sin x)\sqrt{(p\cos x {+} q \sin x)^2+1-u^2}}{1{+}(p\cos x {+} q \sin x)^2}\right), \tag2 $$ where $|u|<1$ and $p$ and $q$ are $3$ real numbers, and all of these solutions $f$ give $a^2+b^2 = 1$, i.e., equality in the desired inequality. One gets other $3$-parameter families by adding an integer multiple of $\pi$ to the above formula, but these can be considered equivalent. It turns out that this then gives all of the solutions except the ones that have $\sin f(x) \equiv 0$. (Allowing $|u|\ge1$ gives families of solutions that can be defined only over subintervals of $[0,2\pi]$, and these must be taken into account in the analysis as well.)

Again, in order to make this argument fully rigorous, one has to prove that a minimizer does exist in the first place and prove a regularity result for solutions of the above ODE at places where $\sin f(x)$ vanishes. I think that those are doable, but I haven't checked the details.

Answer 2

Don't trust me on this because I have never done calculus of variations (I'm still in high school), but I'm doing this just because I was bored. I also don't know how to insert special characters like pi or integral, but oh well.

Since we are trying to prove whether it is > or = 4pi² or not, we need to try our best to prove it wrong instead of right, and if we can't find a way to prove it wrong, then we are right. The best way to go about this is to find the least possible value of the left side of the equation. Let's look at the first integral. We have cos(f(x)), where f(x) is between 0 and 2pi. It can have any value between those numbers, but we want it to make the integral as small as possible. The first thing we should know is that cos(f(x)) will always have the range between -1 and 1. If we can make cos(f(x)) = 0 for [0,2pi], that will be perfect as the integral of it will be 0. In order for cos(f(x)) = 0 for [0,2pi], we need f(x) = pi/2 or 3pi/2 for [0,2pi] all the time. That means we will make f(x) always equal pi/2 or 3pi/2 (doesn't actually matter which one you use). cos(f(x)) now is cos(pi/2) or cos(3pi/2), which is just 0. Now if we take the integral from 0 to 2pi of 0, we get 0! Then we square the 0, which still gives us 0. We have made the first integral as small as possible. Now on to the next one. We have sqrt((f'(x))²+sin²(f(x))) this time. We need to remember that we have set f(x) to be a constant of pi/2 or 3pi/2, so we'll immediately plug it in. The equation is now sqrt((f'(x))²+sin²(pi/2)) which comes out to be sqrt((f'(x))²+1). Now we need to find out f'(x), which is easy because f(x) is a constant, which means f'(x) is 0. Now we have the integral from 0 to 2pi of sqrt(1) (or just 1). Now we can evaluate the integral, which comes out to be 2pi. Finally, since the integral is squared, we must square 2pi, which means it is 4pi². Now we add our two integrals, which are 4pi² and 0, and they give us 4pi² still. Hurray!! We have tried our best to get the least possible value, and our least possible value ended up with the lowest value we can have. 4pi² = 4pi².