Fourier series and the isoperimetric inequality
Definition 1.1
A $\mathscr{C}^1$ mapping
\[\begin{align*} \gamma: [a,b] \to \mathbb{R}^2, \end{align*}\]such that $\gamma^\prime(s)\neq 0$, is called a parameterized curve. We call the image of $\gamma$ a curve (denoted by $\Gamma$). If $\gamma$ is 1-1, we call $\Gamma$ simple if $\gamma(a) = \gamma(b)$ we call $\Gamma$ closed.
Definition 1.2
We define the length of the curve $\Gamma$, parameterized by $\gamma(s)= (x(s), y(s))$ by
\[\begin{align*} \ell = \int^b_a \lVert \gamma^\prime(s) \rVert ds = \int^b_a (x^\prime(s)^2 + y^\prime(s)^2)^{1/2} ds \end{align*}\]Definition 1.3
Let $s: [c,d] \to [a,b]$ be a bijective and $\mathscr{C}^1$ mapping. Then we call
\(\begin{align*} \eta(t) := \gamma \circ s (t) \end{align*}\) a re-parameterization of $\Gamma$. If, further,
\[\begin{align*} \lVert \eta^\prime(t) \rVert = 1 \text{ for all } t \in [c,d] \end{align*}\]we call $\eta$ the arclength parameterization of $\Gamma$.
Remark
The length of the curve $\gamma: [a,b]\to \mathbb{R}^2$ is independent of parameterization.
First, since $s:[c,d]\to[a,b]$ is $\mathscr{C}^1$, it is continuous. Moreover, since $s$ is bijective, it is injective. A continuous injective function on an interval is strictly monotone.
To see this, suppose that $t_1<t_2<t_3$. We claim that $s(t_2)$ must lie strictly between $s(t_1)$ and $s(t_3)$. Suppose, for example, that
\[\begin{align*} s(t_2)>\max\{s(t_1),s(t_3)\}. \end{align*}\]Choose $y$ such that
\[\begin{align*} \max\{s(t_1),s(t_3)\}<y<s(t_2). \end{align*}\]By the Intermediate Value Theorem applied to $[t_1,t_2]$, there exists $u\in(t_1,t_2)$ such that $s(u)=y$. Applying the Intermediate Value Theorem again to $[t_2,t_3]$, there exists $v\in(t_2,t_3)$ such that $s(v)=y$. Thus,
\[\begin{align*} u\neq v \quad\text{but}\quad s(u)=s(v), \end{align*}\]which contradicts the injectivity of $s$. The case
\[\begin{align*} s(t_2)<\min\{s(t_1),s(t_3)\} \end{align*}\]is ruled out in the same way. Hence $s(t_2)$ always lies strictly between $s(t_1)$ and $s(t_3)$. Since $s(c)\neq s(d)$, either $s(c)<s(d)$ or $s(c)>s(d)$. In the first case, the above property implies that $s$ is strictly increasing, while in the second case it implies that $s$ is strictly decreasing. Therefore, $s$ is a monotone function.
Now, by the chain rule,
\[\begin{align*} \eta^\prime(t)=\gamma^\prime(s(t))s^\prime(t), \end{align*}\]and therefore
\[\begin{align*} \lVert \eta^\prime(t)\rVert = \lvert s^\prime(t)\rvert \lVert \gamma^\prime(s(t))\rVert. \end{align*}\]If $s$ is increasing, then $s(c)=a$, $s(d)=b$, and $s^\prime(t)\geq 0$. Thus, by the change of variable $u=s(t)$,
\[\begin{align*} \int_c^d \lVert \eta^\prime(t)\rVert dt &= \int_c^d s^\prime(t)\lVert \gamma^\prime(s(t))\rVert dt \\ &= \int_{s(c)}^{s(d)}\lVert \gamma^\prime(u)\rVert du \\ &= \int_a^b\lVert \gamma^\prime(u)\rVert du. \end{align*}\]If $s$ is decreasing, then $s(c)=b$, $s(d)=a$, and $s^\prime(t)\leq 0$. Hence $\lvert s^\prime(t)\rvert=-s^\prime(t)$, and
\[\begin{align*} \int_c^d \lVert \eta^\prime(t)\rVert dt &= -\int_c^d s^\prime(t)\lVert \gamma^\prime(s(t))\rVert dt \\ &= -\int_{s(c)}^{s(d)}\lVert \gamma^\prime(u)\rVert du \\ &= -\int_b^a\lVert \gamma^\prime(u)\rVert du \\ &= \int_a^b\lVert \gamma^\prime(u)\rVert du. \end{align*}\]Thus, in either case,
\[\begin{align*} \int_a^b \lVert \gamma^\prime(s) \rVert ds = \int_c^d \lVert \eta^\prime(t)\rVert dt, \end{align*}\]so the length of the curve is independent of parameterization.
Lemma 1
Let $\gamma: [a,b] \to \mathbb{R}^2$ be a $\mathscr{C}^1$ parametric curve. Now define a function $t: [a,b] \to [0,\ell]$ by
\[\begin{align*} t(s)= \int_a^s \lVert \gamma^\prime (x)\lVert dx, \end{align*}\]where $\ell = \int_a^b \lVert \gamma^\prime (x)\lVert dx$. Then $t(s)$ is a bijective function and $\eta(t) := (\gamma \circ s) (t)$ is an arclength parameterization, where $s(t)$ is the inverse of $t(s)$.
<Proof>
First, we want to show that $t(s)$ is bijective function. Suppose that $s_1, s_2 \in [a,b]$ are given , where $s_1 < s_2$. Then
\[\begin{align*} \int_a^{s_2} \lVert \gamma^{\prime}(x)\rVert dx - \int_a^{s_1} \lVert \gamma^{\prime}(x)\rVert dx = \int_{s_1}^{s_2} \lVert \gamma^{\prime}(x)\rVert dx > 0 \end{align*}\]since $\lVert\gamma^\prime(x)\rVert>0$. Thus, $t(s)$ is strictly monotonic increasing and hence 1-1.
Now suppose that $y\in [0,\ell]$ be given. Since $t(s)$ is continuous function, we can apply the Intermediate Value Theorem to the continuous $t$ on $[a,b]$. Thus, $t(s)$ is onto.
For the length of re-parmeterized curve $\eta(t)$,
\[\begin{align*} \lVert \eta^\prime(t) \rVert &= \lvert s^\prime(t) \rvert \cdot \lVert \gamma^\prime (s(t)) \rVert \\ &=\frac{1}{\lvert t^\prime(s(t))\rvert} \lVert \gamma^\prime (s(t))\rVert \\ &= \lVert \gamma^\prime (s(t))\rVert^{-1} \lVert \gamma^\prime(s(t))\rVert \\ &=1, \end{align*}\]where we used the Fundamental Theorem of Calculus,
\[\begin{align*} t^\prime(s)=\lVert\gamma^\prime(s)\rVert. \end{align*}\]$\therefore \eta(t)$ is an arclength parameterization.
\[\tag*{$\square$}\]Theorem 2.1
Let $\Gamma \subset \mathbb{R}^2$ be a simple closed curve. Let $\ell$ denote the length of $\Gamma, A$ the area of its enclosed region. Then
\[\begin{align*} A \leq \frac{\ell^2}{4\pi}, \end{align*}\]with equality if and only if $\Gamma$ is a circle.
<Proof> Observe that we can rescale the problem by a factor of $\delta>0$. Consider the mappling of the plane $\mathbb{R}^2$ to $\mathbb{R}^2$ where $(x,y)\mapsto (\delta x, \delta y).$ Then the length $\ell$ is scaled to $\delta \ell$ and its area $A$ is rescaled to $\delta^2 A$ since
\[\begin{align*} \int_a^b \sqrt{\delta^2 x^\prime(s)^2 + \delta^2 y^\prime(s)^2} ds &= \delta \ell \\ \frac{1}{2}\left\lvert\int_a^b \delta x(s)\delta y^\prime(s) - \delta y(s)\delta x^\prime(s)ds\right\rvert&= \delta^2 A. \end{align*}\]But the inequality does not change
\[\begin{align*} \delta^2 A \leq \frac{\delta ^2\ell^2}{4\pi} \iff A \leq \frac{\ell^2}{4\pi}. \end{align*}\]By taking $\delta = 2\pi / \ell$, it suffices to prove that if $\ell=2\pi$ then $A \leq \pi$, with equality only if $\Gamma$ is a circle. Note that any curve admits a parameterization by arclength by Lemma 1. Then, after a possible additional translation, we can consider a parameterization $\gamma(s)=(x(s), y(s))$ of $\Gamma$ by arc-length defined on $[0,\ell]=[0,2\pi]$. Then
\[\begin{align*} \frac{1}{2\pi} \int_0^{2\pi} x^\prime(s)^2 + y^\prime(s)^2 ds &=1 \\ \lVert x^\prime \rVert^2_{L^2} + \lVert y^\prime \rVert^2_{L^2} &=1 \end{align*}\]$x(s)$ and $y(s)$ are $2\pi$ periodic functions, and have Fourier coefficients $\{a_n\}$ and $\{b_n\}$, respectively. Their derivatives have coefficients $\{ina_n\}$ and $\{inb_n\}$, respectively. Since
\[\begin{align*} \hat{f}(n) &= \frac{1}{2\pi}\int_0^{2\pi} f(\theta) e^{-in\theta}d\theta \\ &=\frac{1}{2\pi}\left(\left[ f(\theta) \frac{-e^{-in\theta}}{in}\right]_0^{2\pi} + \frac{1}{in}\int_0^{2\pi} f^\prime(\theta) e^{-in\theta}d\theta \right)\\ &=\frac{1}{2\pi in}\int_0^{2\pi}f^\prime (\theta)e^{-in\theta}d\theta , \end{align*}\]Fourier coefficent of $f^\prime$ is $in\hat{f}(n)$.
Using Parseval’s identity on the above, we get
\[\begin{align*} 1 &=\lVert \{ina_n\}\rVert^2_{\ell^2} + \lVert \{inb_n\}\rVert^2_{\ell_2} \\ &=\sum_{n=-\infty}^\infty ina_n (-in\overline{a}_n) +\sum_{n=-\infty}^\infty inb_n(-in\overline{b}_n) \\ &= \sum_{n=-\infty}^\infty \lvert n\rvert^2(\lvert a_n\rvert^2 + \lvert b_n \rvert^2). \end{align*}\]Now, the area is by definition
\[\begin{align*} A = \frac{1}{2} 2\pi \left\lvert \frac{1}{2\pi}\int_0^{2\pi} x(s) y^\prime(s) - y(s)x^\prime(s) ds\right\rvert. \end{align*}\]To apply Parseval to the area term, we use the polarized Parseval identity. If $f$ and $g$ have Fourier coefficients $\hat{f}(n)$ and $\hat{g}(n)$, respectively, then
\[\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}f(s)\overline{g(s)}ds =\sum_{n\in\mathbb{Z}}\hat{f}(n)\overline{\hat{g}(n)}. \end{align*}\]This follows from Parseval’s identity by polarization. Since $x(s),y(s),x^\prime(s)$ and $y^\prime(s)$ are real-valued,
\[\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}x(s)y^\prime(s)ds &= \frac{1}{2\pi}\int_0^{2\pi}x(s)\overline{y^\prime(s)}ds \\ &= \sum_{n\in\mathbb{Z}}a_n\overline{inb_n} \\ &= \sum_{n\in\mathbb{Z}}a_n(-in\overline{b}_n), \end{align*}\]and similarly,
\[\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}y(s)x^\prime(s)ds &= \frac{1}{2\pi}\int_0^{2\pi}y(s)\overline{x^\prime(s)}ds \\ &= \sum_{n\in\mathbb{Z}}b_n\overline{ina_n} \\ &= \sum_{n\in\mathbb{Z}}b_n(-in\overline{a}_n). \end{align*}\]Using polarized Parseval on the area formula, we therefore get
\[\begin{align*} A &= \pi \left\lvert \sum_{n\in\mathbb{Z}} a_n(-in\overline{b}_n) - b_n(-in\overline{a}_n) \right\rvert \\ &\leq \pi \sum_{n\in\mathbb{Z}} \lvert n \rvert \lvert a_n\overline{b}_n-\overline{a}_nb_n \rvert. \end{align*}\]Since
\[\begin{align} \lvert a_n \overline{b}_n - \overline{a}_n b_n \rvert \leq \lvert a_n\overline{b}_n\rvert + \lvert \overline{a}_n b_n\rvert = 2\lvert a_n b_n\rvert \leq \lvert a_n \rvert^2 + \lvert b_n \rvert^2, \label{eq:5} \end{align}\]we bound the area
\[\begin{align} A &\leq \pi \sum_{n\in\mathbb{Z}} \lvert n \rvert \lvert a_n\overline{b}_n-\overline{a}_nb_n \rvert \label{eq:6}\\ &\leq \pi\sum_{n\in\mathbb{Z}} \lvert n \rvert^2 (\lvert a_n \rvert^2+ \lvert b_n \rvert^2) =\pi. \label{eq:7} \end{align}\]Equality in $\eqref{eq:6}$ and $\eqref{eq:7}$ could only occur if we have no terms $\lvert n \rvert\geq 2$. Moreover $x(s)$ and $y(s)$ are differentiable and real-valued functions and thus
\[\begin{align*} x(s) &= a_{-1}e^{-is} + a_0 + a_1 e^{is} \\ y(s) &= b_{-1}e^{-is} + b_0 + b_1e^{is}. \end{align*}\]by Theorem.
Since $x(s)$ and $y(s)$ are real-valued functions,
\[\begin{align*} \overline{a_n} &= \overline{\frac{1}{2\pi}\int_0^{2\pi} x(s) e^{-ins}ds} \\ &= \frac{1}{2\pi}\int_0^{2\pi} x(s) e^{ins}ds \\ &= a_{-n}, \\ \overline{b_n} &= \overline{\frac{1}{2\pi}\int_0^{2\pi} y(s) e^{-ins}ds} \\ &= \frac{1}{2\pi}\int_0^{2\pi} y(s) e^{ins}ds \\ &= b_{-n}. \end{align*}\]Equation $\ref{eq:7}$ implies that $2(\lvert a_1 \rvert^2 + \lvert b_1 \rvert^2)=1$.
Now by $\ref{eq:5}$, equality hold only if
\[\begin{align*} \lvert a_1 \overline{b}_1 - \overline{a}_1 b_1 \rvert = 2\lvert a_1 b_1\rvert =\lvert a_1 \rvert^2 + \lvert b_1 \rvert^2 = \frac{1}{2} \end{align*}\]and $(\lvert a_n \rvert - \lvert b_n \rvert)^2=0$ implies $\lvert a_n \rvert = \lvert b_n \rvert$, $\lvert a_1 \rvert = \lvert b_1 \rvert = 1/2$. So we get
\[\begin{align*} a_1= \frac{1}{2}e^{i\alpha} \quad \text{and } b_1 = \frac{1}{2}e^{i\beta} \end{align*}\]for some $\alpha, \beta \in \mathbb{R}$. Since $1 = 2\lvert a_1 \overline{b}_1 - \overline{a}_1 b_1\rvert$,
\[\begin{align*} \left\lvert \frac{1}{2} \left(e^{i(\alpha-\beta)} - e^{-i(\alpha-\beta)} \right)\right\rvert = 1, \end{align*}\]i.e., $\lvert \sin(\alpha-\beta)\rvert=1$. So,
\[\begin{align*} \alpha-\beta=\frac{\pi}{2}+k\pi \end{align*}\]for some $k\in\mathbb{Z}$. Putting all the pieces together,
\[\begin{align*} x(s) &= \frac{1}{2}e^{-i\alpha}e^{-is} + a_0 + \frac{1}{2}e^{i\alpha}e^{is} = a_0 + \cos(\alpha+s) \\ y(s)&= \frac{1}{2}e^{-i\beta}e^{-is} + b_0 + \frac{1}{2}e^{i\beta}e^{is} = b_0 + \cos(\beta+s) \\ &=b_0+ \cos\left(\alpha-\frac{\pi}{2}-k\pi+s\right) \\ &=b_0 \: \pm \sin(\alpha+s). \end{align*}\]$\therefore \Gamma$ is a circle.
\[\tag*{$\square$}\]Reference
- Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
- Math 139 Fourier Analysis Notes