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Parametrize a positively oriented closed plane curve at constant arc length speed on $[0,2\pi ]$, and let $c_n$ be its complex-position Fourier coefficients. Green's theorem and Parseval's identity give $A=\pi {\sum }_{n}n|c_n{|}^2$ and $L^2/(4{\pi }^2)={\sum }_n{n}^2|c_n{|}^2$. Since $n\le n^2$ for integers, the planar isoperimetric inequality follows. Equality leaves only the constant and positive first harmonic, a translated circle.

Fix two attachments a distance $2a$ apart on a straight line and a simple fence of length $2l$ in one half-plane, with $a<l<\pi a/2$. There is a unique $\theta \in (0,\pi /2)$ satisfying $l/a=\theta /\sin \theta$: its derivative has numerator $\sin \theta -\theta \cos \theta >0$, since that numerator starts at zero and has derivative $\theta \sin \theta >0$. A radius-$R$ minor circular arc with half-angle $\theta$ then has the required chord and length. Its graph is $y=\sqrt{R^2-x^2}-\sqrt{R^2-a^2}$, for $-a\le x\le a$, and sector-minus-triangle geometry gives the displayed area.

To prove global maximality, attach the complementary major arc of this circle, lying in the opposite half-plane, to any competing fence. The resulting closed curve has perimeter $2l+(2\pi R-2l)=2\pi R$. Its enclosed area is the competitor's area plus a fixed major-segment area. The planar isoperimetric inequality bounds the total by $\pi R^2$, with equality only for a circle. Subtracting that fixed added area proves the minor arc is the unique maximizing shape on the chosen side. The same arc follows from the Beltrami identity for the area-minus-length functional.

The plane planar isoperimetric inequality says that a simple closed curve of length $L$ enclosing area $A$ satisfies $L^2\ge 4\pi A$, with equality for circles.

For sufficiently small $r$, below the injectivity radius at $p$, geodesic polar coordinates give the surface metric $d{r}^2+j(r,\theta {)}^{2}d{\theta }^2$. The radial variation field is a Jacobi field, soIts integral equation is $j=r-{\int }_0^r(r-u)K({\gamma }_{\theta }(u))j(u,\theta )du$. Continuity of curvature and $j(u,\theta )=u+o(u)$, uniformly in direction, giveThe Gauss lemma makes the angular metric factor both the circle line element and the polar area factor. ThereforeSquaring the length expansion and subtracting givesThis identifies the requested remainder and proves its ratio to $r^4$ tends to zero. For $K(p)>0$, sufficiently small geodesic disks have $L^2<4\pi A$, the opposite strict inequality to the planar bound. Their positive curvature allows more area for the same perimeter; the Euclidean inequality is not a universal inequality for curved surfaces.

The planar planar isoperimetric inequality is $4\pi |\Omega |\le |\partial \Omega {|}^2$, with equality exactly for a disk. For one boundary component of length $L$, take its unit-speed periodic parametrization $\gamma =(x,y)$ and subtract its mean position. Green's area formula and the Cauchy-Schwarz inequality giveApply the periodic Wirtinger inequality to each mean-zero coordinate: $\int |\gamma {|}^2\le (L/2\pi {)}^2\int |{\gamma }^{\prime }{|}^2$. Since unit speed gives the final integral $L$, it follows that $|A_{\mathrm{signed}}|\le L^2/(4\pi )$. For several boundary components, orient each with the domain on its left. Green's formula, followed by the triangle inequality, givesThis proves the stated inequality, including disconnected domains or holes. Equality permits only one component; equality in the two integral inequalities makes its coordinates first sine-cosine harmonics with the corresponding orthogonal rotation relation. Unit speed then makes the curve a circle, so its domain is a disk. Conversely disks attain equality.

For the positively oriented simple closed boundary, the turning tangent theorem gives $\int k(s)ds=2\pi$. If $L$ is its length, continuity gives a point with $k(s_0)\le 2\pi /L$. The planar isoperimetric inequality gives $L^2\ge 4\pi A(K)$, henceFor the reversed orientation the curvature integral is $-2\pi$, so a point of nonpositive curvature already proves the stated positive upper bound. The region-bounding interpretation is the ordinary Jordan-curve one; self-intersections would require specifying a different notion of enclosed area.

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.

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 planar planar isoperimetric inequality iswith equality precisely for a disc. Let the positively oriented boundary be parametrized by arclength as $r(s)=(x(s),y(s))$, $0\le s\le L$. Translate its mean to zero. Then $|r^{\prime }|=1$, and Green's theorem gives $2A={\int }_0^L(x{y}^{\prime }-y{x}^{\prime })ds$.

The periodic Wirtinger inequality says that for a real absolutely continuous $L$-periodic function $h$ of mean zero and square-integrable derivative,with equality exactly for $h(s)=a\cos (2\pi s/L)+b\sin (2\pi s/L)$. Apply it to both coordinates and use Cauchy-Schwarz inequality:If equality holds, the Wirtinger equalities give $r=p\cos (2\pi s/L)+q\sin (2\pi s/L)$. The arclength condition forces $|p|=|q|=L/(2\pi )$ and $p\cdot q=0$, so the boundary is a circle. A circle attains equality, making the constant optimal.

The inequality and equality characterization persist for finitely many corners. A rectifiable piecewise smooth simple closed curve has a Lipschitz arclength parametrization, with $|r^{\prime }|=1$ almost everywhere. Green's theorem and the absolutely continuous version of Wirtinger's inequality used above still apply. More generally one may approximate a rectifiable Jordan curve by smooth curves while preserving the limiting length and area. If the allowed isolated singularities give infinite length, the inequality is trivially satisfied in the extended sense.

The periodic Wirtinger inequality says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, thenEquality holds exactly for a linear combination of $\cos (2\pi s/L)$ and $\sin (2\pi s/L)$.

First suppose $\partial \Omega$ is one positively oriented simple closed curve, parametrized by arc length as $\gamma (s)=(x(s),y(s))$ for $0\le s\le L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. Green theorem and the Cauchy-Schwarz inequality giveApplying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x^{\prime }{)}^2+(y^{\prime }{)}^2=1$ yieldsand therefore the planar isoperimetric inequalityFor several boundary components, apply the simple-curve result to the relevant enclosed regions and use ${\sum }_j{L}_j^2\le ({\sum }_j{L}_j{)}^2$; holes only decrease the area.

Equality in both inequalities forcesfor constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

Along an extremal, differentiate $f-y^{\prime }{f}_{y^{\prime }}$. The chain rule givesWhen $f_x=0$, the Euler-Lagrange equation makes the right side zero. This proves the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=k$.

Place the wall on $y=0$ with attachments at $x=\pm a$, and write the fence above it as $y(x)\ge 0$. Maximize area $A={\int }_{-a}^{a}ydx$ subject to length ${\int }_{-a}^a\sqrt{1+y^{\prime 2}}dx=2l$. The isoperimetric constraint gives the augmented integrand $f=y-R\sqrt{1+y^{\prime 2}}$, where $R$ is a constant multiplier. Its Euler-Lagrange equation is $y^{\prime \prime }/(1+y^{\prime 2}{)}^{3/2}=-1/R$, so the fence has constant negative curvature. Equivalently the Beltrami identity is $y-R/\sqrt{1+y^{\prime 2}}=k$.

The resulting maximum-area circular segment with fixed arc length and chord hasHere $0<\theta <\pi /2$ is the half-angle subtended by the arc. Its half-length is $R\theta =l$ and half-chord is $R\sin \theta =a$. The function $\theta /\sin \theta$ increases from one to $\pi /2$, so the given range $a<l<\pi a/2$ determines a unique minor arc. Its enclosed area is $A_{\max }=R^2(\theta -\sin \theta \cos \theta )$.

There is also a global maximality proof for simple fences in the chosen half-plane, without assuming competitors are graphs. Complete the candidate minor arc with the complementary major arc of the same circle, which lies below the wall. Attach this same fixed major arc to any competing fence of length $2l$. The completed closed curve always has perimeter $2\pi R$, and its area is the competitor's fenced area plus the same fixed area below the wall. The planar isoperimetric inequality bounds that total by $\pi R^2$, attained by the candidate circle. Hence no competitor encloses more area, and equality requires the circular arc.

Write the speed as $v=\sqrt{{\dot{x}}^2+{\dot{y}}^2}>0$ and takeThe two Euler-Lagrange equations areFor example ${\mathcal{L}}_x=\dot{y}/2$ and ${\mathcal{L}}_{\dot{x}}=-y/2+\lambda \dot{x}/v$, yielding the first equation. Integrating once gives constants $c_1,c_2$ with $\lambda \dot{x}/v=y+c_1$ and $\lambda \dot{y}/v=-x+c_2$. The case $\lambda =0$ would force both velocities to vanish, contrary to a regular curve. Squaring and adding therefore givesThus every regular stationary curve is a circle, whose center is set by the integration constants. A simple closed traversal of length $L$ has radius $L/(2\pi )$ and positive enclosed area $L^2/(4\pi )$. The planar isoperimetric inequality proves that this stationary candidate is the global maximum; the necessary variational equations alone would not prove that inequality. The stationary curves of the planar isoperimetric problem give this argument independently of the chosen parametrization.

For a regular closed planar curve, maximize signed enclosed area at fixed length using the variational integrand $\mathcal{L}=(x\dot{y}-y\dot{x})/2+\lambda \sqrt{{\dot{x}}^2+{\dot{y}}^2}$. Write the nonzero speed as $v$. The Euler-Lagrange equations integrate to $\lambda \dot{x}/v=y+c_1$ and $\lambda \dot{y}/v=-x+c_2$. A zero multiplier would force a zero velocity, so squaring and adding yields the displayed circle. The argument uses only regularity and is independent of the speed used to trace the curve. A positively oriented simple traversal of prescribed length $L$ encloses area $L^2/(4\pi )$; the planar isoperimetric inequality supplies the separate global maximality statement.