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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 $$ \left(\int_0^{2 \pi} \cos f(x) \,d x\right)^2+\left(\int_0^{2 \pi} \sqrt{(f'(x))^2+\sin^2 f(x)} \, d x\right)^2 \geq(2 \pi)^2 $$


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) $(2\pi)^2$?


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 $L^2\ge S(4\pi-S)$, we have $(2\pi-S)^2+L^2\ge(2\pi)^2$, 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?

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    $\begingroup$ Its a nice inequality where did you get this? $\endgroup$ Jan 12, 2021 at 6:13
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    $\begingroup$ I suspect that $\int_0^{2\pi} f'(x)=0$ is to be used somehow $\endgroup$ Jan 12, 2021 at 12:23
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    $\begingroup$ I wonder if this can somehow be used here: math.stackexchange.com/q/117296/42969. $\endgroup$
    – Martin R
    Jan 12, 2021 at 13:48
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    $\begingroup$ It is found by accident when considering an equal-perimeter inequality on a sphere. Not sure if this helps. $\endgroup$
    – FFjet
    Jan 13, 2021 at 12:47
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    $\begingroup$ @FFjet: If there is a geometric interpretation of that inequality then I suggest that you add it to your question. That may help to understand the problem better and possibly to find an answer. $\endgroup$
    – Martin R
    Jan 13, 2021 at 18:26

4 Answers 4

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<This is not an answer!! This is just some of my thoughts, and also I am not good at English. If there are any things to fix, please be my guest.>

Prediction: For two functions $A(x)$ and $B(x)$, $$(\int_{\alpha}^{\beta} A(x) dx)^2+(\int_{\alpha}^{\beta} B(x) dx)^2 \geq (\int_{\alpha}^{\beta} \sqrt{A(x)^2+B(x)^2} dx)^2 $$ holds.

If this is true(which I am not able to prove...), the problem can be solved easily.

Let $A(x)=\cos f(x)$ and $B(x)=\sqrt{{f'(x)}^2+\sin^2 f(x)}$, and $\alpha=0, \beta=2\pi$.
Then, by the Prediction, $$(\int_{0}^{2\pi} \cos f(x) dx)^2+(\int_{0}^{2\pi} \sqrt{(f'(x))^2+\sin^2 f(x)} dx)^2\geq (\int_{0}^{2\pi} \sqrt{(f'(x))^2+1} dx)^2 $$ We know that the right hand side is the form of the length of a curve $f(x)$.
As $y=f(x)$ satisfies $f(0)=f(2\pi)$, the shortest length of the curve $y=f(x)$ in the interval $[0, 2\pi]$ would be just a simple line connecting $(0, f(0))$ and $(2\pi, f(2\pi)=f(0))$, so the length will be just the x-coordinate difference, which is $2\pi$. Therefore, the RHS will have the minimum of $(2\pi)^2=4\pi^2$, and the problem is solved.
However, I couldn't think of a sharp way to prove this Prediction, or even it is true at all. I managed to think of the integrals as sequences, and tried solving this. $$(\sum_{k=0}^{n} a_k)^2+(\sum_{k=0}^{n} b_k)^2 \geq (\sum_{k=0}^{n} \sqrt{(a_k)^2+(b_k)^2})^2$$ I think this equation can be solved by mathematical induction, but I am not sure.

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    $\begingroup$ prediction is false, just take $A=B=\sin(x -\frac{\alpha+\beta}2)$ $\endgroup$ Jan 16, 2021 at 9:54
  • $\begingroup$ @CalvinKhor Oh thanks! I wonder if the opposite direction is right...is it? $\endgroup$
    – Joshua Woo
    Jan 16, 2021 at 9:59
  • $\begingroup$ yes, it’s just triangle inequality for integrals of $\mathbb R^2$ valued functions ie $|\int \binom AB dx| \le \int|\binom AB | dx$ $\endgroup$ Jan 16, 2021 at 10:35
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Not an answer (for now), just observations:

We can at least prove that the left-side must be strictly positive. Moreover, this can be done without assuming that $0\leq f(x)\leq 2\pi$ and $f(2\pi)=f(0)$.

Proof: since any sum of squares is at least $0$, the following inequality holds for all differentiable functions $f:[0,2\pi]\to\mathbb{R}$ for which the integrals exist:

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2\geq 0$$

Suppose, for the sake of finding a contradiction, that there is a differentiable $f:[0,2\pi]\to\mathbb{R}$ for which

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2= 0$$

Since $x^2+y^2=0$ if and only if $x=y=0$, it follows that

$$\int_0^{2\pi}\cos[f(x)]dx=0\text{ and }\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx=0$$

From this result together with the observations that $[f'(x)]^2\geq 0$, $\sin^2[f(x)]\geq 0$, and $(\sin\circ f)^2$ is continuous, we can infer that

\begin{align} \int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx=0 &\implies \sin^2[f(x)]=0\text{ for every }x\in[0,2\pi]\\ &\implies \sin[f(x)]=0\text{ for every }x\in[0,2\pi]\\ &\implies \text{For every }x\in[0,2\pi]\text{, }f(x)=\pi n\text{ for some }n\in\mathbb{Z} \end{align}

From the continuity of $f$, we deduce that $f=(\pi n)_{[0,2\pi]}$ for some integer $n$ (the notation $c_{[a,b]}$ denotes the constant function mapping $[a,b]$ to $c$). This implies that $\cos\circ f=\pm 1_{[0,2\pi]}$, and consequently

$$\color{red}{0}=\int_0^{2\pi}\cos[f(x)]dx=\int_0^{2\pi}\pm 1dx\color{red}{=\pm 2\pi}$$

a contradiction. Thus, for every differentiable $f:[0,2\pi]\to\mathbb R$,

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2>0$$

I'm still thinking about how to prove the lower bound of $(2\pi)^2$. Stay tuned :)

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Also not an answer, but my train of thoughts. We can take advantage of the fact that for a complex number $z$ and its conjugate $\bar{z}$, it holds that $z\bar{z}=|z|^2$.

Thus, if we associate $$z(x)=\int_0^{2\pi}\left(\cos f(x)+i\sqrt{(f'(x))^{2}+\sin ^{2} f(x)}\right)dx=\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx,$$

the LHS becomes $z(x)\overline{z(x)}$ and thus we want to show that $$\left|\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx\right|^2 \geq (2\pi)^2$$

Now, somewhat fishy, but I believe since the function $f(x)$ is positive on the interval $(0, 2\pi)$, and also $\sqrt{1+(f'(x))^2}$ is positive, the following equality should hold: $$\color{gray}{\left|\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx\right|=\int_0^{2\pi}\left|\sqrt{1+(f'(x))^2}e^{i\phi(x)}\right|dx=\int_0^{2\pi}\sqrt{1+(f'(x))^2}dx=L}$$

Then, by the same argument as in the answer by @Joshua Woo, this is a length of the curve $f(x)$, which must be greater than the length of interval ($0$, $2\pi$), i.e., $L=2\pi$ (due to the condition $f(0)=f(2\pi)$), therefore we get the required inequality.

But I will be happy to learn where my argument is incorrect.

EDIT: Thanks for pointing out the error. I will keep the post struck-through just for the transparency.

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    $\begingroup$ As your argument goes, we are free to choose f(x) and phi(x) however we like, but what if f(x) = 1 and phi(x) = x, so that your integral is the integral of e^ix from 0 to 2pi, which is 0, so the ideal inequality isn't so. $\endgroup$ Jan 22, 2021 at 22:30
  • $\begingroup$ @Countingstuff $\phi(x)$ is not any function, it is the phase of the complex number, i.e., $\phi(x)=\arctan\frac{\sqrt{(f'(x))^2+\sin^2f(x)}}{\cos f(x)}$, so for $f(x)=1$, $\phi(x)=1$. Also, note that absolute value of $e^{i\phi(x)}$ is 1 regardless of $\phi(x)$. $\endgroup$
    – pisoir
    Jan 23, 2021 at 7:48
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    $\begingroup$ $\left|\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx\right|=\int_0^{2\pi}\left|\sqrt{1+(f'(x))^2}e^{i\phi(x)}\right|dx$ is wrong unless the argument $\phi(x)$ is constant. In general you only have $\le$ which does not help. $\endgroup$
    – Martin R
    Jan 23, 2021 at 10:05
  • $\begingroup$ @MartinR, I believe that linear dependence with the magnitude would sufficient. But I guess I can't guarantee that neither. $\endgroup$
    – pisoir
    Jan 24, 2021 at 16:15
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Maybe I am simplifying this too much, but if f(x) is not constant, then $$\sqrt{(f'(x))^2 + (\sin f(x))^2} > |\sin f(x)|$$ because $$f'(x) \neq 0$$ for some value(s) in the interval of integration. This leads to the following: $$\left(\int_0^{2\pi} \cos f(x) \: dx \right)^2 + \left( \int_0^{2\pi} \sqrt{(f'(x))^2 + \sin^2 f(x)} \: dx \right)^2 $$ $$ \geq \left(\int_0^{2\pi} \cos f(x) \: dx \right)^2 + \left( \int_0^{2\pi} \sqrt{\sin^2 f(x))^2} \: dx \right)^2 = (2 \pi)^2$$

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  • $\begingroup$ If the $\cos$ and the $\sin$ were the objects being squared, then the last equality would certainly be true. However, it’s not the functions that are being squared, but their integrals, so I don’t understand how you deduced that their sum is $(2\pi)^2$. Could you be kind enough to explain your reasoning? $\endgroup$ May 16, 2021 at 18:44

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