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1 MATH 205C: STATIONARY PHASE LEMMA For ω, consider an integral of the form I(ω) = e iωf(x) u(x) dx, where u Cc (R n ) complex valued, with support in a compact set K, and f C (R n ) real valued. Thus, I(ω) C(K) sup u ; we are interested in the asymptotic behavior as ω. If f has no critical points, then as ω, the exponential becomes highly oscillatory, and one expects that I(ω) 0 rapidly. Indeed, I(ω) C ω for all in this case. To see this, we note that if f never vanishes then, with f (x) 2 = j f 2, e iωf(x) = iω Leiωf(x), L = f (x) 2 j f, and so, with L t the transpose of L, i.e. L t = f j f (x) 2, with the relevant factors acting as multiplication operators, I(ω) = (Le iωf(.) )(x) u(x) dx = e iωf(x) (L t u)(x) dx, iω iω so by induction I(ω) = (iω) e iωf(x) ((L t ) u)(x) dx, leading to the conclusion that () I(ω) C(K, f, )ω sup D α u. α Of course, here we only needed u C with support in K for this estimate. For a moment, also consider non-real valued f with Im f 0 so e iωf(x) = e ω Im f(x) for ω. Then, as long as f 0, the above calculation goes through if we replace f by its complex conjugate in the definition of L, so () also holds. Moreover, if f does have some critical points, but at these Im f > 0, then a partition of unity argument (noting that the set of critical points is closed, as is the set of points where f is real, while supp u is compact) allows one to reduce consideration of the integral the two cases where either Im f 0 and f 0, which we just analyzed, or instead Im f > 0 (hence bounded below by a positive constant). In the latter case, one actually gets I(ω) e ω inf Im f u L, i.e. one has exponential decay. Thus, if Im f 0, and in addition f = 0 implies Im f > 0, then () still holds. Returning to real valued f, the interesting case is if f has some critical points, and the simplest setting is if these are non-degenerate, i.e. if f (x 0 ) = 0 implies that the Hessian f is invertible at x 0. Thus, one assumes that if f (x 0 ) = 0 then (2) f(x) = f(x 0 ) + 2 Q(x x 0, x x 0 ) + R(x, x 0 ),

2 2 where R(x, x 0 ) C x x 0 3, with Q a non-degenerate symmetric bilinear quadratic form; writing it as Q(x x 0, x x 0 ) = A(x x 0 ), x x 0 for the standard inner product on R n, A is a symmetric invertible linear operator on R n. The signature of Q, and of A, can be defined as the pair (, n ) where is the maximal dimension of a subspace on which Q is positive definite, or the number of positive eigenvalues of A. It is also useful to have a single number sign Q = sign A = (n ), the difference of the number of positive and negative eigenvalues of A. Note that non-degenerate critical points are isolated, for f (x) A(x x 0 ) vanishes quadratically at x 0, and A is invertible. Thus, by using a partition unity, when the critical points are non-degenerate, one can easily arrange that there is a single critical point x 0 over supp u, which one may arrange to be at 0 by a translation of the coordinates. In this case, one has f (x) C x x 0 for a suitable C > 0, so f (x) C x x 0. It is useful to observe that even if f has a non-degenerate critical point at 0, thus the above integration by parts argument breas down in general, it continues to wor if u vanishes to sufficiently high order at 0. It is convenient to be somewhat more general (due to the singularities the integration by parts argument induces), so suppose that u C (R n \ {0}), with support in K, with D α u C x 2 α for α. Then, by the divergence theorem in the last step, I(ω) = lim e x ρ iωf(x) u(x) dx = iω lim = ( iω lim e iωf(x) (L t u)(x) dx + x ρ x ρ x =ρ (Le iωf(.) )(x) u(x) dx f 2 ( j f ) ν j )e iωf(x) u(x) ds(x), where ν is the outward unit normal of R n \ {x : x ρ}. Now, by the assumptions on u, the estimate on f and the vanishing of f at 0, the integrand in the surface integral is bounded by Cρ, and thus the surface integral has limit 0 as ρ 0. Correspondingly, I(ω) = iω lim x ρ e iωf(x) (L t u)(x) dx = iω as L t u is bounded by the assumptions. Thus, I(ω) C(K, f)ω sup (D α u) x 2+ α. α e iωf(x) (L t u)(x) dx, Further, if u C (R n \ {0}), with support in K, with D α u C x 2 α for α, then L t u C (R n \ {0}), with support in K, with D α L t u C x 2( ) α for α. Thus, the argument can be applied iteratively to conclude that (3) I(ω) C(K, f)ω sup (D α u) x 2+ α. α We now show how to use this to reduce the case of general f to that of a quadratic form, 2 Q(x x 0, x x 0 ). With the notation of (2), let f s (x) = f(x 0 ) + 2 Q(x x 0, x x 0 ) + sr(x, x 0 ), s [0, ], so f 0 is quadratic (plus a constant), and f = f. Let I(ω, s) = e iωfs(x) u(x) dx.

3 3 Then differentiating in s 2 times yields I (2) (ω, s) = (iω) 2 e iωfs(x) R(x, x 0 ) 2 u(x) dx. Since R is C and vanishes cubically at x = x 0, v x0 (x) = R(x, x 0 ) 2 u(x) satisfies (D α v x0 )(x) C x x 0 6 α when u C 3 (R n ) with compact support in K and α 3. Thus, applying (3) with 3 in place of, we deduce that By Taylor s theorem, (4) I(ω, ) j<2 I (2) (ω, s) Cω. j! I(j) (ω, 0) sup s [0,] (2)! I(2) (ω, s), so modulo ω decay, I(ω) can be calculated merely by calculating I (j) (ω, 0) = (iω) j e iωf0(x) R(x, x 0 ) j u(x) dx (5) = (iω) j e iωf(x0) e iωq(x x0,x x0)/2 R(x, x 0 ) j u(x) dx for j < 2, i.e. we are reduced to the case of purely quadratic phase. We now want to compute integrals of the form I 0 (ω) = e iω 2 Ax,x u(x) dx, with u being C. Note that we could replace u by any function that has the same Taylor series to order 2 in view of our preceding estimates, and one way to proceed would be to compute explicitly such integrals. Instead, we rewrite using Parseval s formula I 0 (ω) = (F e iω 2 A.,. )(ξ)(fu)(ξ) dξ We mae a general definition: if P C (R n ) with D α P (ξ) C α ξ Mα, i.e. if P is polynomially bounded with all derivatives, let P (D)u = F (P (.)(Fu)(.)), so P (D) maps S to itself and S to itself. Thus, we have (6) I 0 (ω) = (2π) n ((F e iω 2 A.,. )(D)u)(0). Our first tas is to compute the inverse Fourier transform of the imaginary Gaussian. First, recall that and (F(t e t2 /2 ))(τ) = 2πe τ 2 /2, (F (t e t2 /2 ))(τ) = 2π e τ 2 /2. A change of variables gives that for α > 0 (F (t e αt2 /2 ))(τ) = 2πα e τ 2 /(2α). Now, both sides are analytic functions of α with values in S(R) in Re α > 0, and both sides are continuous functions of α with values in S (R) in Re α 0, α 0, so the formula continues to hold there (with the square root being the standard one

4 4 for positive reals, with a branch cut along the negative reals). In particular, when α = βi, β > 0, we deduce that (F (t e ±iβt2 /2 ))(τ) = 2πβ e ±iπ/4 e iτ 2 /(2β). Now, if A is a non-degenerate symmetric linear map on R n, then there is an orthogonal transformation O diagonalizing A as Λ = OAO t = diag(λ j ), with rows of O given by orthonormal eigenvectors e j of A, and with λ j 0 the eigenvalues of A, so Ax, x = j λ j e j, x 2, i.e. in coordinates y j = e j, x, y = Ox, eeping in mind that O has determinant of absolute value, (2π) n e ix ξ e i Ax,x /2 dx = (2π) n e iy Oξ e i j λjy2 j /2 dy Thus, (7) (F e iω 2 A.,. )(ξ) = = Now, by Taylor s formula, f(y) = (2π) n/2 λ /2 ei(π/4)... λ n /2 j sign λj e i j (Oξ)2 j /(2λj) = (2π) n/2 det A /2 ei(π/4) sign A e i (OAO t ) Oξ,Oξ /2 = (2π) n/2 det A /2 ei(π/4) sign A e i A ξ,ξ /2. (2πω) n/2 det A /2 ei(π/4) sign A e i ω A ξ,ξ /2 f (j) (x) (y x) j (y x) + j! ( )! we have for w R (or indeed for w with Im w 0) e iw ( iw) j j! ( )! w 0 ( t) f () (ty + ( t)x) dt, Thus, with P (.) = B.,., using Plancherel s theorem, 0 ( t) dt = w! e ip (D) u ( ip (D)) j u j! L 2! P (D) u L 2.. Now, we would lie to replace the L 2 norm on the left by just the absolute value of the value of the argument at 0 in order to obtain an expansion for (6) (while strengthening the norm on the right, of course); this goal is reached if we can replace the L 2 norm by the L norm, for the argument is Schwartz when u is in C c (R n ). But by the Sobolev embedding, for s > n/2 integer, there is C > 0 such that v L C( v L 2 + D α v L 2), α =s

5 5 so as P (D) and e ip (D) commute with D α (for they are all Fourier multipliers) e ip (D) u ( ip (D)) j u j! L ( C e ip (D) u ( ip (D)) j u j! L 2 + D α( e ip (D) u ( ip (D)) j u ) L 2 j!) α =s ( C e ip (D) u ( ip (D)) j u j! L 2 + e ip (D) (D α u) ( ip (D)) j Dα u j! α =s C ( P (D) u L 2 + ) P (D) D α u L 2! α =s C! B ( u L 2 + ) D α u L 2 C B α =s β 2+s D β u L 2, where in the penultimate step we used that P (ξ) B ξ 2 for ξ R n so P (ξ) Fu L 2 B ξ 2 Fu L 2 and thus P (D) u L 2 B u L 2 by Plancherel, and then in the last step we expanded = n j= D2 j. Here C depends on the Sobolev inequality parameters C, s and on only. Applying this with B = ω A /2, (6) and (7) give (8) (2π) n/2 I 0 (ω) ω n/2 det A /2 ei(π/4) sign A ( iω A D, D ) j u j! (0) π n/2 det A /2 (2ω) n/2 C A D β u L 2, β 2+s Combining this with (4) and (5), we deduce that for L j differential operators of order 2j, and L 0 the identity, (9) I(ω) e iωf(x0) (2π) n/2 sign f (x ω n/2 det f ei(π/4) 0) ω j L (x 0 ) /2 j u(x 0 ) ω n/2 C f (x 0 ) D β u L 2, β 2+s which is the stationary phase lemma. Notice that if (D α u)(x 0 ) vanishes for α 2l +, l <, then the first l terms in the sum on the left hand side vanish, so the expansion starts with ω n/2 l. Assuming f(x 0 ) = 0, since ω l ( ωi)(ω) l = ω l e iωf(x) f(x) l u(x) dx, and f vanishes quadratically at x 0, derivatives of I in ω possess a similar expansion, starting with a multiple if ω n/2. Factoring out e iωf(x0) in general, (0) ω l ( ω(e l if(x0)ω (2π) n/2 I))(ω) sign f (x ω n/2 det f ei(π/4) 0) ω j L (x 0 ) /2 l,j u(x 0 ) ω n/2 C l f (x 0 ) D β u L 2, β 2+2l+s L 2 )

6 6 where L l,j are differential operators of the form L l+j f(x) l. One can integrate this expression to obtain the already-nown expansion for I, which shows that the coefficients in the expansion of ω(e l if(x0)ω I) are the term-by-term differentiated coefficients of e if(x0)ω I. Finally, ( ωi)(ω) l = ( ω(e l if(x0)ω (e if(x0)ω I))(ω) can be computed by the product rule. One can also allow parametric dependence on another variable y R m, i.e. for f C (R n+m ) with Im f 0, consider the integral I(ω, y) = e iωf(x,y) u(x, y) dx, u Cc (R n+m ). R n As before I decays rapidly in ω, uniformly in y unless f has critical points in x at which it is real, i.e. unless there exists (x 0, y 0 ) such that f x(x 0, y 0 ) = 0 and Im f(x 0, y 0 ) = 0. In addition, differentiation under the integral sign shows that in this case I is C in (ω, y), with Dy α DωI l still rapidly decreasing. If f is real, and has some critical point (x 0, y 0 ) in x which is non-degenerate, i.e. f xx is invertible, so d x f x,... d x f x n are linearly independent at (x 0, y 0 ), then the implicit function theorem guarantees that the joint zero set of f x,..., f (x n ) is a C graph over a neighborhood of y 0 in R m, i.e. there is a C function X defined on a neighborhood of y 0 such that in a neighborhood of (x 0, y 0 ), the critical points of f are given by (X(y), y), and these are non-degenerate. Thus, the previous arguments are applicable, uniformly in y, so the stationary phase expansion (0) holds with the y dependence added in. Writing I(ω, y) = e iωf(x(y),y) R n e iωf (x,y) u(x, y) dx, F (x, y) = f(x, y) f(x(y), y), notice that, with yj hand side, standing for the jth component of the second slot on the right ( yj F )(x, y) = ( yj f)(x, y) ( yj f)(x(y), y) (( x f)(x(y), y)) X (y), y j and thus vanishes at x = X(y) since that is a critical point of f in x. Since x f, and thus x F, have a non-degenerate zero at x = X(y), we actually obtain ( yj F )(x, y) = G j (x, y)( x F )(x, y). Differentiation under the integral sign gives, with Ĩ(ω, y) = e iωf (x,y) u(x, y) dx, R n yj Ĩ(ω, y) = (iω yj F (x, y))e iωf (x,y) u(x, y) dx + e iωf (x,y) yj u(x, y) dx. R n R n Now, the second term has a stationary phase expansion as before. On the other hand, the first term is G j (x, y)(iω x F (x, y))e iωf (x,y) u(x, y) dx R n = G j (x, y)( x e iωf (x,y) ) u(x, y) dx = e iωf (x,y) x (G j (x, y)u(x, y)) dx, R n R n so the stationary phase expansion is again applicable. Thus, yj I possesses a stationary phase expansion. Since the expansion can be integrated in y j, this proves

7 7 that the coefficients of the expansion of the derivative are the differentiated coefficients of the expansion of I (including the statement that the latter are differentiable). Iterating the argument shows that ω l y α ωi l still possesses an expansion as in (0), with the expansion given by term-by-term differentiation. We finally discuss an interpretation of the stationary phase result by partially compactifying (or bordifying) the (ω, y)-space. Thus, consider the new variable h = ω (0, ]. Then the stationary phase lemma, together with ω ω = h h and that ω l ω l can be rewritten as a linear combination of (ω ω ) j for j l, and similarly for h l h l, shows that I(h, y) = e if(x,y)/h u(x, y) dx = e if(x(y),y)/h h n/2 J(h, y), with h l l h α y J j h j Lj,l,α u, where means that if one sums over j <, then one obtains an error terms bounded by C l,α, h, and the series on the right is the term-by-term derivative of h j Lj,0,0 u. But notice that h l j<l hj Lj,0,0 u = 0, so in particular, taing = l, h l h l α y J C l,α,l h l, i.e. h l α y J C l,α,l for all l and α, i.e. all partial derivatives of the C function J on (0, ] h R m y are bounded. This implies that J has a unique C extension to [0, ] R m : uniqueness is automatic as (0, ] R m is dense in [0, ] R m. On the other hand, h l α y J extends continuously to h = 0 since l h α y J(h 0, y) = l h α y J(, y) l+ h h 0 α y J(h, y) dh, and l+ h y α J(h, y) is bounded, and then repeated integration shows that for l l, α α, h l α y J extends as well (which of course would follow from the same argument with (l, α) replaced by (l, α )), with extension given by the appropriate integral of l+ h y α J(h, y), so in particular this extension is differentiable (l l, α α ) times. As l and α are arbitrary, we deduce that the extension of J we defined is C ; in view of the uniqueness, one usually simply writes J for the extension as well. A slight generalization is then obtained by allowing u to depend on h as well, i.e. consider u Cc (R n R m [0, ]), and consider J(h, y) = e if(x,y)/h u(x, y, h) dx. Regarding the h in u as one of the parameters y, we have then J(h, y) = Ĵ(h, y, h), where Ĵ(h, y, h) = e if(x,y)/h u(x, y, h) dx = e if(x(y),y)/h h n/2 Ĵ (h, y, h) with Ĵ C ([0, ] R m [0, ]), so restricting to h = h yields with J C ([0, ] R m ). J(h, y) = e if(x(y),y)/h h n/2 J(h, y),