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The proof I'm referring to is to the following theorem.

Assume $S$ is a compact regular and strictly convex energy surface for the Hamiltonian field $X_H$ in $\mathbb{R}^{2n}$. Then $S$ carries a closed orbit.

Steps before my passage:

  1. We build a Hamiltonian $H$ with Hessian everywhere positive definite, $H(0)=0$, $\mathcal{C}^2$, positively homogeneous of degree 2 and having $S$ as the level set where it is 1;
  2. We call $G$ its Fourier transform;

  3. We consider the functional:

    $$I(z)=\int_0^{2\pi}G(\dot z)dt,$$

    defined on $H_1(S^1)$, and look for a minimum of it in $\mathcal{A}$, the subset of $\mathcal{F}$ where $\frac{1}{2}\int_0^{2\pi}\langle Jz,\dot z\rangle dt=1$, $J$ being the standard symplectic matrix $(\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix})$ and $\langle\cdot,\cdot\rangle$ being the Euclidean inner product, and $\mathcal{F}$ being the subspace of $H_1(S^1)$ where $\int_0^{2\pi}z(t)dt=0$;

  4. We establish $I$ has a minimum $z_\ast$ on $\mathcal{A}$, through a minimizing sequence $z_j$ which converges to it weakly (and its derivatives seem to weakly converge to $\dot z_\ast$, but for that see here.

The the book says:

Ad (iii): Since $z_*$ is a minimum, we have $$\tag{1.68}\int_0^{2\pi}\langle\nabla G(\dot {z}_*),\dot\zeta\rangle=0$$ for every test function $\zeta\in\cal F$ satisfying $$\tag{1.69}\int_0^{2\pi}\langle Jz_*,\dot\zeta\rangle=0.$$

Now, according to this, to find the minimum I should compute the EL equations for:

$$I(z)=\int_0^{2\pi}[G(\dot z)-\lambda(\langle Jz,\dot z\rangle-2)]dt.$$

But if I put $f(\epsilon)=I(z_\ast+\epsilon\zeta)$ and compute $f'(0)$ I get:

\begin{align*} f'(0)={}&\lim_{\epsilon\to0}\frac1\epsilon\int[G(\dot z_\ast+\epsilon\dot\zeta)-G(z_\ast)-\lambda(\langle J(z_\ast+\epsilon\zeta),\dot z_\ast+\epsilon\dot\zeta\rangle-\langle Jz_\ast,\dot z_\ast\rangle)]dt={} \\ {}={}&\int_0^{2\pi}[\langle\nabla G(\dot z_\ast),\dot\zeta\rangle+\lambda(\langle J\zeta,\dot z_\ast\rangle+\langle Jz_\ast,\dot\zeta\rangle)]dt. \end{align*}

Assuming (1.69), the third term goes away, and the rest sums to zero, but how do I deduce the first term must be zero?

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MickG
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1 Answers1

1

Let us just integrate the second term by parts:

$$\int_0^{2\pi}\langle J\zeta,\dot z_\ast\rangle dt=\langle J\zeta,z_\ast\rangle|_0^{2\pi}-\int_0^{2\pi}\langle J\dot\zeta,z_\ast\rangle.$$

Both $\zeta$ and $z_\ast$ are periodic, hence term 1 vanishes. Moving $J$ to the right of the inner product yields a minus by antisymmetry of $J$, so the second term is precisely the LHS of (1.69), hence 0.

By the way way, one can see the condition (1.69) is the condition for $\zeta\in T_{z_\ast}\mathcal{A}$. To see this, one simply expands the inner product in the integral:

$$\int_0^{2\pi}\langle J\dot z_\ast+J\dot\zeta\epsilon,z_\ast+\epsilon\zeta\rangle=\int\langle J\dot z_\ast,z_\ast\rangle+\epsilon\int\langle J\dot\zeta,z_\ast\rangle+\epsilon\int\langle J\dot z_\ast,\zeta\rangle+\epsilon^2\int\langle J\dot\zeta,\zeta\rangle.$$

So at the first order in $\epsilon$, the constraint is fixed, thus the vector $\zeta$ is tangent to the constrained subset of the curve space.

MickG
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