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Expand a real function with zero endpoint values in its Fourier sine series, $y={\sum }_{n\ge 1}{a}_n\sin (n\pi x/L)$. The Parseval identity gives $\int y^2=(L/2)\sum a_n^2$. For its weak derivative, the Fourier cosine series has coefficients $(n\pi /L)a_n$ and zero constant term, soThe derivative coefficient identity follows by integration by parts, rather than by assuming pointwise differentiation of an arbitrary series. This proves the Wirtinger inequality and its equality case. Under unit L2 norm, the only real minimizers are $y=\pm \sqrt{2/L}\sin (\pi x/L)$.

Fix an admissible $u$. The preceding result gives ${\int }_0^b{u}^{\prime 2}dx\ge {\lambda }^2{\int }_0^b{u}^{2}dx$ for every ${\lambda }^2<{\pi }^2/b^2$. Letting ${\lambda }^2$ increase to ${\pi }^2/b^2$ provesEquality is attained byIndeed the derivative and function squared integrals are $A^2{\pi }^2/(2b)$ and $A^{2}b/2$. These are all equality cases: an equality function minimizes the nonnegative threshold quadratic form, so its first variation vanishes. The resulting weak equation $u^{\prime \prime }+(\pi /b{)}^{2}u=0$ makes $u$ smooth and, with the endpoint conditions, yields precisely these multiples. This is the Wirtinger inequality.

For fixed terminal time and fixed endpoint, use the maximizing form of the Pontryagin maximum principle. For dynamics $\dot{q}=f(q,u,t)$ and running cost $L$, there are multipliers $(p,p_0)$, not all zero, with $p_0\ge 0$, and Hamiltonian of an optimal-control problem $\mathcal{H}=p\cdot f-p_{0}L$. Along an optimum, $\dot{q}={\mathcal{H}}_p$, $\dot{p}=-{\mathcal{H}}_q$, and the chosen control maximizes $\mathcal{H}$ over admissible controls almost everywhere. Fixed terminal state imposes no prescribed terminal value of $p$; fixed time imposes no free-time condition $\mathcal{H}(T)=0$. The normal case scales $p_0=1$; abnormal extremals have $p_0=0$.

For this problem the normal Hamiltonian of an optimal-control problem isMaximization gives $u=p_x+ky$, $v=p_y-kx$. The costate equations give ${\dot{p}}_x=kv$, ${\dot{p}}_y=-ku$, hence $\dot{u}=2kv$, $\dot{v}=-2ku$. Nonzero velocities rotate at constant angular rate, and rotational symmetry permits $u(0)=A$, $v(0)=0$. For a closed path, $2k=2\pi n$, with integer $n\ne 0$. Then $z(1)=A^2/(2\pi n)$, so the required positive endpoint has $n>0$, $A^2=2\pi n$ and cost $\pi n$.

To prove a global minimum rather than only classify stationary extremals, write $r=(x,y)$ and subtract its time average; this changes neither its derivative nor $\int (y\dot{x}-x\dot{y})dt$. The resulting closed mean-zero path satisfies the periodic Wirtinger inequality $\int |r{|}^2\le (4{\pi }^2{)}^{-1}\int |\dot{r}{|}^2$. This follows immediately by its Fourier expansion: each nonzero mode of the derivative has squared multiplier at least $4{\pi }^2$. The Cauchy-Schwarz inequality now givesThe argument extends to square-integrable controls by the periodic Sobolev/Fourier interpretation. Thus the cost is at least $\pi$ for every feasible control. Equality is attained byThese satisfy every endpoint and have cost $\pi$. Therefore the minimum is $\pi$, an area-constrained quadratic control cost. The global inequality also prevents an overlooked abnormal candidate from lowering it; in fact an abnormal finite maximizing Hamiltonian of an optimal-control problem would force both linear control coefficients to vanish and then force zero velocity, incompatible with $z(1)=1$.

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 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.

Assume the regular Sturm–Liouville setting: real coefficients with $p>0$, $w>0$ and sufficient continuity and differentiability for the integrations below. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation in the direction $\eta$ isVariations vanishing at both endpoints and the supplied fundamental lemma giveAt a free endpoint the value of $\eta$ is arbitrary, so the natural condition is $p{y}^{\prime }=0$ there. At a fixed endpoint $\eta =0$, so no such derivative condition is imposed; the prescribed endpoint value is imposed instead. This distinction also allows one endpoint of each type.

For eigenfunctions $u,v$ of distinct eigenvalues $\lambda ,\mu$ under the same homogeneous self-adjoint endpoint conditions, multiply their equations by $v,u$ and subtract. Integration givesThe boundary expression vanishes for both natural conditions and homogeneous fixed-value conditions. Thus $\boxed{{\int }_a^{b}wuv=0}$ for different eigenvalues.

In the specified fixed-endpoint case the eigenproblem is $-y^{\prime \prime }-y=\lambda y$, $y(0)=y(1)=0$. There are no nonzero solutions when $\lambda +1\le 0$; the elementary exponential or affine solutions and the two endpoint conditions prove this. For $\lambda +1>0$, the first condition leaves a sine and the second requires frequency $n\pi$. The normalized stationary functions and eigenvalues areDistinct sine modes are orthogonal by the product-to-sum formula, or by the preceding Green second identity. Although they need not satisfy the natural derivative conditions, their fixed zero values also annihilate the boundary expression, so orthogonality is expected.

The Wirtinger inequality ${\int }_0^1{y}^{\prime 2}\ge {\pi }^2{\int }_0^1{y}^2$ gives $J[y]\ge {\pi }^2-1$. For completeness it follows by writing $y=v\sin \pi x$ for a smooth function supported away from the endpoints and integrating the cross term: $\int (y^{\prime 2}-{\pi }^2{y}^2)=\int {\sin }^2(\pi x)v^{\prime 2}\ge 0$. Approximation extends it to the Dirichlet Sobolev class. Equality is attained by the first sine mode; the eigenproblem or equality condition gives only its scalar multiples. Hence the two normalized minimizers are $\boxed{y=\pm \sqrt{2}\sin \pi x}$ and the minimum is $\boxed{{\pi }^2-1={\lambda }_1}$.

There is a wording distinction in the source: the two sign choices of the minimum are not orthogonal to each other, since their inner product is $-1$. Orthogonality applies to the stationary modes with different eigenvalues, not to both signs of the same minimizer.

For $y=\pm \sqrt{2/\pi }\sin x$, both the constraint and endpoint values hold, the extremal equation has $\lambda =1$, and direct integration gives $J[y]=1$.

For any admissible function, its derivative has Fourier cosine series coefficients $n{a}_n$ and zero constant coefficient. This follows by integration by parts: ${\int }_0^{\pi }{y}^{\prime }\cos (nx)dx=n{\int }_0^{\pi }y\sin (nx)dx$, and ${\int }_0^{\pi }{y}^{\prime }=0$. Thus the Parseval identity givesFor finite sums the same identity follows directly from the cosine product formula in the hint; the $L^2$ identity justifies the general case. Equality holds only if $a_n=0$ for all $n\ge 2$, and normalization then fixes $a_1=\pm \sqrt{2/\pi }$. ThereforeThis is the minimum Dirichlet energy at fixed L2 norm on an interval, equivalently the sharp Wirtinger inequality for this interval.