Isoperimetric Inequality and Fourier Series
1. Formulation and Reduction
Theorem (Isoperimetric Inequality)
Suppose that \(\gamma\) is a piecewise \(C^1\) simple closed curve in the plane. If \(A\) is the area enclosed by \(\gamma\) and \(L\) is the length of \(\gamma\), then
\[ A \leq \frac{L^2}{4 \pi}. \]
Equality is attained if and only if \(\gamma\) is a circle.
Strategic Reductions The proof of the Isoperimetric Inequality can be accomplished via Fourier series. Key steps:
- Represent points in the plane as complex numbers. Use an arc length parametrization of \(\gamma\) so that\[ L = \int_{0}^L \left| \left| \frac{d \gamma}{dt}(t) \right| \right| dt = \int_0^{L} \left| \frac{d \gamma}{dt}(t) \right| dt \]where \(||\cdot||\) denotes length as a vector and \(|\cdot|\) denotes magnitude as a complex number.
- Green's Theorem and some computation gives us that\[ A = \int_{0}^L \frac{1}{2} \operatorname{Im} \left(\frac{d \gamma}{dt}(t) \overline{\gamma(t)} \right) dt. \]The Isoperimetric Inequality becomes equivalent to showing that\[{}\left| \int_{0}^L \operatorname{Im} \left(\frac{d \gamma}{dt}(t) \overline{\gamma(t)} \right) dt \right|{}\]\[{}\leq \frac{1}{2\pi} \left( \int_0^L \left| \frac{d \gamma}{dt}(t) \right| dt \right)^2.{}\]
- Because \(|d \gamma / dt| = 1\) anyway when using an arc length parametrization, the desired inequality is equivalent to\[{}\left| \int_{0}^L \operatorname{Im} \left(\frac{d \gamma}{dt}(t) \overline{\gamma(t)} \right) dt \right|{}\]\[{}\leq \frac{L}{2\pi} \int_0^L \left| \frac{d \gamma}{dt}(t) \right|^2 dt.{}\]
2. Proof via Fourier Series
[Step 1.] Write both sides of the inequality as inner products to use Parseval's identity.
Proposition (Parseval's Identity, Bilinear Form)
Suppose \(f\) and \(g\) are complex-valued Riemann integrable functions on \([0,L]\). Defining
\[ \hat{f}(n) := \frac{1}{L} \int_0^L f(t) e^{-\frac{2 \pi i n t}{L}} dt \]
and likewise for \(\widehat{g}(n)\), then
\[ \frac{1}{L} \int_0^L f(t) \overline{g(t)} dt = \sum_{n=-\infty}^{\infty} \hat{f}(n) \overline{\widehat{g}(n)}\]
Proof
\[ \hat{f}(n) \overline{\hat{g}(n)} = \frac{1}{4} \sum_{k=0}^3 i^k | \hat{f}(n) + i^k \hat{g}(n)|^2 \]
In general, for complex numbers \(a\) and \(b\),
\[ |a + i^k b| = a \overline{a} + a \overline{b} i^{-k} + i^k b \overline{a} + b \overline{b}\]
so
\[ i^k |a + i^k b| = a \overline{b} + i^k (|a|^2 + |b|^2) + (-1)^k b \overline{a}. \]
Summing over \(k=0,1,2,3\) makes all terms on the right-hand side cancel except \(a \overline{b}\). But Parseval's identity gives that
\[{}\frac{1}{L} \int_0^L \frac{1}{4} \sum_{k=0}^3 i^k | f(t) + i^k g(t)|^2 dt{}\]
\[{}= \sum_{n=-\infty}^\infty \frac{1}{4} \sum_{k=0}^3 i^k | \hat{f}(n) + i^k \hat{g}(n)|^2.{}\]
[Step 2.] Using Parseval's Identity in Bilinear Form gives that
\[{}\frac{1}{L} \int_{0}^L \operatorname{Im} \left(\frac{d \gamma}{dt}(t) \overline{\gamma(t)} \right) dt{}\]
\[{}= \sum_{n=-\infty}^\infty |\hat{\gamma}(n)|^2 \frac{2 \pi n}{L},{}\]
\[{}\frac{1}{L} \int_0^L \left| \frac{d \gamma}{dt}(t) \right|^2 dt{}\]
\[{}= \sum_{n=-\infty}^\infty \frac{4 \pi^2 n^2}{L^2} |\hat{\gamma}(n)|^2,{}\]
so the desired inequality to prove is
\[{}\left| 2 \pi \sum_{n=-\infty}^\infty n |\hat{\gamma}(n)|^2 \right|{}\]
\[{}\leq \frac{L}{2 \pi} \sum_{n=-\infty}^\infty \frac{4 \pi^2 n^2}{L} |\hat{\gamma}(n)|^2,{}\]
i.e.,
\[ \left| \sum_{n=-\infty}^\infty n |\hat{\gamma}(n)|^2 \right| \leq \sum_{n=-\infty}^\infty n^2 |\hat{\gamma}(n)|^2. \]
[Step 3.] The key idea to finish up is that \(|n| \leq n^2\) with equality exactly when \(n=-1,0,1\). Because of cancellation effects, the inequality also won't be sharp if both the \(n=1\) and \(n=-1\) coefficients are nonzero. If the only nonzero Fourier coefficients are \(n = -1,0\) or \(n = 0,1\), knowing what we know about complex exponentials, \(\gamma\) must parametrize a circle.