Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 25H (Differential Geometry)

For a regular simple closed plane curve of length $L$ enclosing area $A$, the planar isoperimetric inequality is $L^2\ge 4\pi A$, with equality precisely for a circle.

The periodic Wirtinger inequality states that an absolutely continuous $L$-periodic function $h$ of mean zero with square-integrable derivative satisfieswith equality only for a linear combination of $\sin (2\pi s/L)$ and $\cos (2\pi s/L)$. Parametrize the curve by arc length and translate its position vector $r=(x,y)$ so that both coordinate means vanish. Then $\int |r{|}^2\le L^3/(4{\pi }^2)$, since $|r^{\prime }|=1$. Green's theorem and the Cauchy-Schwarz inequality giveEquality requires a first-harmonic parametrization and equality in Cauchy-Schwarz; its radius vector is a constant multiple of the rotated unit tangent. It is therefore a circle. Conversely, a circle attains equality.

The Fenchel theorem states that every closed regular space curve has total curvature at least $2\pi$:Equality holds precisely for a convex plane curve, traversed once.

For the anticlockwise boundary of the region, the inward normal is the left normal, so the signed curvature obeys the turning tangent theorem ${\int }_0^{L}k(s)ds=2\pi$. Therefore some $s_0$ has $k(s_0)\le 2\pi /L$. The enclosed area satisfies $A\ge d_1{d}_2$, and the planar isoperimetric inequality gives $L\ge 2\sqrt{\pi A}$. ConsequentlyThis uses the interpretation of the curve as a simple boundary traversed once; the inward normal and enclosed region in the question have that meaning.