A calculus problem by Hamza A

Let $m = \max|F|$ and $G(x) = \dfrac{dF}{dx}$ .

First let's prove a lemma, namely that $m > \dfrac{1}{\sqrt{\pi}}$ .

Observe that since $F^2(x) \le m^2$ , then $\int_{0}^{\pi} F^2 dx = 1 \le \int_{0}^{\pi} m^2 dx = m^2 \pi$ .

The only way equality can occur is if $F(x) = m = \dfrac{1}{\sqrt{\pi}}$ for all $0 \le x \le \pi$ . However, this is impossible, given that $F(0) = F(\pi) = 0$ and $F$ is a smooth function. Hence $m^2 \pi > 1$ , so $m$ is strictly greater than $\dfrac{1}{\sqrt{\pi}}$ . $\square$

So let's prove the theorem, that the minimum value of $\int_{0}^{\pi} G^2 dx$ is $1$ . $\int_{0}^{\pi} G^2 dx = \frac{1}{m^2\pi - 1}\int_{0}^{\pi} G^2 dx \int_{0}^{\pi} \left(m^2 - F^2\right)dx$

By the Cauchy-Schwarz Inequality for integrals, we have: $\int_{0}^{\pi} G^2 dx \int_{0}^{\pi} \left(m^2 - F^2\right)dx \ge \left(\int_{0}^{\pi} |G| \sqrt{m^2 - F^2} dx \right)^2$ Now let $a_0 = 0, a_n = \pi$ and let $a_1, a_2, a_3 ... a_{n-1}$ be the local minimum/maximum points of $F$ . Without loss of generality, let's assume that $G(x) \ge 0$ for $0 \le x \le a_1$ . So in general, for $a_{2k} \le x \le a_{2k+1}$ , $G(x) \ge 0$ and $F$ is increasing; while for $a_{2k-1} \le x \le a_{2k}$ , $G(x) \le 0$ and $F$ is decreasing. As a corollary, for all $k$ , we have $F(a_{2k}) < F(a_{2k+1}) > F(a_{2k+2})$ .

Therefore, the last integral can be rewritten as follows: $\int_{0}^{\pi} |G| \sqrt{m^2 - F^2} dx = \sum_{k = 0}^{\lfloor (n-1) / 2 \rfloor} \left(\int_{a_{2k}}^{a_{2k+1}} G \sqrt{m^2 - F^2} dx \right) + \sum_{k = 1}^{\lfloor n / 2\rfloor} \left(\int_{a_{2k-1}}^{a_{2k}} (-G) \sqrt{m^2 - F^2} dx \right)$ $= \sum_{k = 0}^{\lfloor (n-1) / 2 \rfloor} \left(\int_{F(a_{2k})}^{F(a_{2k+1})} \sqrt{m^2 - y^2} dy \right) + \sum_{k = 1}^{\lfloor n / 2\rfloor} \left(\int_{F(a_{2k})}^{F(a_{2k-1})} \sqrt{m^2 - y^2} dy \right)$ Let $H(c,x) = \displaystyle\int_{c}^{x} \sqrt{m^2 - y^2} dy$ . Note that $\dfrac{dH}{dx}$ is positive. Therefore, if $c \le x$ then $H(c, x) \ge 0$ and vice versa.

Let $s$ be the minimum value of $F$ and $t$ be the maximum value of $F$ on $0 \le x \le \pi$ . By our assumptions about the local minima/maxima, $t = F(a_{2p+1})$ and $s = F(a_{2q})$ for some $p,q$ . Now consider all the intervals $\left[F(a_{2k}), F(a_{2k+1})\right]$ . I will leave it as an exercise to prove that the union of these intervals is the entire interval $\left[s, t\right]$ . Therefore, $\sum_{k = 0}^{\lfloor (n-1) / 2 \rfloor} \left(\int_{F(a_{2k})}^{F(a_{2k+1})} \sqrt{m^2 - y^2} dy \right) \ge \int_{s}^{t} \sqrt{m^2 - y^2} dy = H(s, t)$ By similar reasoning, we can show that: $\sum_{k = 1}^{\lfloor n / 2 \rfloor} \left(\int_{F(a_{2k})}^{F(a_{2k-1})} \sqrt{m^2 - y^2} dy \right) \ge \int_{s}^{t} \sqrt{m^2 - y^2} dy = H(s, t)$ Putting the two sums together, we get: $\int_{0}^{\pi} |G| \sqrt{m^2 - F^2} dx \ge 2H(s,t)$ By definition of $m$ , either $s = -m$ or $t = m$ . If $s = -m$ , we know that $t \ge 0$ because $F(0) = 0$ . Then $H(s, t) = H(-m, 0) + H(0, t) \ge H(-m, 0) = \dfrac{m^2 \pi}{4}$ (the last equality is by definition of $H$ ). Similarly if $t = m$ , we also get $H(s, t) \ge \dfrac{m^2 \pi}{4}$ . Therefore, $\int_{0}^{\pi} |G| \sqrt{m^2 - F^2} dx \ge \dfrac{m^2 \pi}{2}$

Putting it all together gives us: $\int_{0}^{\pi} G^2 dx \ge \frac{1}{m^2\pi - 1} \left(\int_{0}^{\pi} |G| \sqrt{m^2 - F^2} dx \right)^2 \ge \frac{1}{m^2\pi - 1} \left(\dfrac{m^2 \pi}{2}\right)^2 = \frac{m^4 \pi^2}{4(m^2\pi - 1)}$

Now let's set a boundary on $\dfrac{m^4 \pi^2}{4(m^2\pi - 1)}$ :

$\dfrac{m^4 \pi^2}{4(m^2\pi - 1)} = 1 + \dfrac{(m^2\pi - 2)^2}{4(m^2\pi - 1)} \ge 1 + 0 = 1$

(Note that we need the fact that $m^2\pi > 1$ here).

Therefore, $\displaystyle\int_{0}^{\pi} G^2 dx \ge 1$

This value is achieved when $F(x) = \sqrt{\dfrac{2}{\pi}} \sin(x)$ .

QED

This is easily done using Fourier analysis. Since $F$ is a smooth function on $[0,\pi]$ , we can write $F$ and $F'$ as the convergent series $F(x) \; = \; \sum_{n=1}^\infty a_n \sin nx \qquad \qquad F'(x) \; = \; \sum_{n=1}^\infty na_n \cos nx$ Since $\int_0^\pi \sin mx \sin nx\,dx \; = \; \int_0^\pi \cos mx \cos nx\,dx \; = \; \tfrac12\pi \delta_{m,n}$ for all integers $m,n \ge 1$ , where $\delta_{m,n}$ is the Kronecker delta function $\delta_{m,n} \; = \; \left\{ \begin{array}{lll} 1 & \qquad & m = n \\ 0 & & m \neq n \end{array} \right.$ we deduce from Parseval's Identity that $1 \; = \; \int_0^\pi F(x)^2\,dx \; = \; \tfrac12\pi \sum_{n=1}^\infty |a_n|^2 \qquad \qquad \int_0^\pi F'(x)^2\,dx \; = \; \tfrac12\pi\sum_{n=1}^\infty n^2 |a_n|^2$ Since $\sum_{n=1}^\infty n^2 |a_n|^2 \; \ge \; \sum_{n=1}^\infty |a_n|^2$ it follows that the minimum value of $\int_0^\pi F'(x)^2\,dx$ is $\boxed{1}$ , and this minimum value is achieved with $a_n = \sqrt{\tfrac{2}{\pi}}\delta_{n,1}$ , so that $F(x) = \sqrt{\tfrac{2}{\pi}}\sin x$ .