By HAMILTON R.S.

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**Example text**

For our solutions E will be a smooth function of the parameters x and y. Since a change of phase 8 -> 0 + a rotates x and y but does not change the energy, we see that £ is a smooth function of x2 + y2. Indeed in our expansion 5 = [xcos vt + y sin vt]/c + and hence E = (mg/2c){x2 + y2) + • • • . By the inverse function theorem we can solve for x2 + y2 as a smooth function of the energy E. Now z is a smooth function of x2 + y2, v is a smooth function of z near z = 0, and T = 2TT/V. Hence the period Tis a smooth function of the energy E for E > 0.

X-axis. Suppose / has a Taylor series = "a + bx + cy + dx1 + • • • . We must estimate the normal derivative c. We can find a constant C, depending only on the derivatives of degree at most 2 of f\ dD, such that on 3D f(Xy)-a-bx*ZC{y 116 R. S. HAMILTON If dD is strictly convex, then y > ex2 for some e > 0. If we let g be the affine function bx + C 1 + - g v then / < g on dD and hence on all of D. Then c < C(l + 7 ) . The same trick estimates c from below also. Finally we must estimate the Holder norm of fx and fy.

The same must be true of any diffeomorphism conjugate to a rotation. Thus one way to find a diffeomorphism / not conjugate to a rotation is to make/*(x) = x for some point x but fk(y) ¥= y for some other point}'. The rotation t -> t + 2w/k by an angle 2ir/k has fk(t) s t (mod27r) for all r. If we modify it by an extra little push just in the interval 0 < x < 2ir/k, so that /(0) = 2ir/k but f(ir/k) > 1-n/k, then fk(0) S 0 but fk{ir/k) s it/k (mod27j-). Then this diffeomorphism cannot be the exponential of a vector field.

### THE INVERSE FUNCTION THEOREM OF NASH AND MOSER by HAMILTON R.S.

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