Lemma 1. **Assume that** H is C
2 on IR2n\{0} and that H′′(x) **is non-degenerate**
for any x 6= 0. Then any local minimizer xe of ψ **has minimal period** T .

Proof. We know that xe, or xe + ξ for some constant ξ ∈ IR2n, is a T **-periodic**
solution of the Hamiltonian system:

$$k-1\geq k-1\geq1\ .$$
$$(222)$$

x˙ = JH′(x) . **(20)**
There is no loss of generality in taking ξ = 0. So ψ(x) ≥ ψ(xe**) for all** xe in some neighbourhood of x in W1,2IR/T ZZ**; IR**2n.

But this index is precisely the index iT (xe) of the T -periodic solution xe **over**
the interval (0, T **), as defined in Sect. 2.6. So**
iT (xe) = 0 . **(21)**
Now if xe has a lower period, T /k **say, we would have, by Corollary 31:**
iT (xe) = ikT /k(xe) ≥ kiT /k(xe) + k − 1 ≥ k − 1 ≥ 1 . **(22)**
This would contradict (21), and thus cannot happen. ⊓⊔
Notes and Comments. **The results in this section are a refined version of [1]; the**
minimality result of Proposition 14 was the first of its kind.

To understand the nontriviality conditions, such as the one in formula (16),
one may think of a one-parameter family xT , T ∈
2πω−1, 2πb−1
∞
**of periodic**
solutions, xT (0) = xT (T ), with xT **going away to infinity when** T → 2πω−1, which is the period of the linearized system at 0.

Table 1. This is the example table taken out of The TEXbook, **p. 246**

| Year      | World population   |
|-----------|--------------------|
| 8000 B.C. | 5,000,000          |
| 50 A.D.   | 200,000,000        |
| 1650 A.D. | 500,000,000        |
| 1945 A.D. | 2,300,000,000      |
| 1980 A.D. | 4,400,000,000      |

Theorem 1 (Ghoussoub-Preiss). Assume H(t, x) is (0, ε)**-subquadratic at** infinity for all ε > 0, and T **-periodic in** t

$$\mathbf{\partial}J$$
$\mathbf{v}$
H(t, ·) is convex ∀t **(23)**

H(·, x) is T−periodic ∀x (24) H(t, x) ≥ n (kxk) with n(s)s −1 → ∞ as s → ∞ (25)
4