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My Proof:

Let the arc-length parametrization of the curve be $\gamma(s) = \langle x(s),y(s)\rangle$. By Green's Theorem, the area $\mathcal{A}$ is

$$ \mathcal{A} = \int_{0}^{2\pi} -y(s)x'(s) dx$$

Then, consider the following:

$$\pi-\mathcal{A}=\int_{0}^{2\pi} \frac{1}{2}ds - \int_{0}^{2\pi} -y(x) x'(s) ds $$ $$=\frac{1}{2}\int_{0}^{2\pi} 1+2y(x)x'(s) ds$$ $$=\frac{1}{2}\int_{0}^{2\pi} (x' + y)^2 ds + \frac{1}{2}\int_{0}^{2\pi} -y^2 + (y'(s))^2 ds$$

The first integral is obviously greater than zero. The second integral is not so obvious, but since we have, by Parsevel's identity,

$$\int_{0}^{2\pi} (y'(s))^2 ds - \int_0^{2\pi} y^2 ds=\sum|na_n|^2 - \sum|a_n|^2 = \sum|a_n|^2(n^2-1)$$

Since, by assumption, $\int_{0}^{2\pi} y(s) ds = 0$, as a result, $a_0 = 0$. So the above sum is always greater than $0$ since $|n|\ge 1$. So we have showed that $\pi \ge \mathcal{A}$.

Then we study when we have equality. (I am not very confident about the following part). Suppose we do have the equality, then we must have both integrals equal to zero. That is, we must have

$$x'(s) = -y(s)$$ and the only values $n$ would take are $1$ and $-1$. That implies that

$$y(s) = a_1 e^{is} + a_{-1}e^{-is} = a_1\cos(s) + a_1\sin(s)i + a_{-1}\cos(s) - a_{-1}\sin(s)i$$

Since we know that $y(s)$ is actually real, we must have $a_1 = a_{-1}$ and thus

$$y(s) = 2a_1\cos(x)$$

This would imply that

$$x(t) = 2a_1\sin(t)$$

This finishes the proof.

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  • $\begingroup$ $n$ might be zero too, so saying that $(n^2-1)\ge 0$ after using Parseval (on the fifth line) is not valid. $\endgroup$ – Chee Han Jun 5 '16 at 8:07
  • $\begingroup$ Also, it would be great if you could justify why the only values for $n$ are $\pm 1$ when you are considering the equality part. $\endgroup$ – Chee Han Jun 5 '16 at 8:09

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