Average theorem, Restriction theorem and Strichartz estimates

Transcription

1 Average theorem, Restriction theorem and trichartz estimates 2 August 27 Abstract We provide the details of the proof of the average theorem and the restriction theorem. Emphasis has been placed on the relation between the decay of the Fourier transform of the measure carried on a submanifold and the gain in regularity. The relation between the restriction theorem and a trichartz estimate is also explained. Theorem (Average theorem. Let be a smooth submanifold of, dσ be the induced Lebesgue measure on it, and dµ(x = η(xdσ(x where η Cc (. uppose for some α > dµ(ξ = O( ξ α as ξ. Define the average operator by Af(x = f(x ydµ(y. Then (a A maps L 2 ( to L 2 α(. (b A maps L 2α+2 2α+ ( to L 2α+2 (. Proof. The proof of (a is easy; just observe that if dµ(ξ C ξ α, then together with the trivial bound dµ(ξ C that holds by the finiteness of the measure dµ, we have dµ(ξ C( + ξ 2 α/2. Thus ( + ξ 2 α Âf(ξ 2 dξ = ( + ξ 2 α f(ξ 2 dµ(ξ 2 dξ C 2 f(ξ 2 dξ,

2 which implies that Af L 2 α C f L 2. To prove (b, we proceed in a number of steps. Localizing the average operator First we localize the average operator as follows. By means of a partition of unity, we may assume without loss of generality that the support of dµ lies in a coordinate chart U of (the average operator is a finite sum of such in any case. By a change of coordinate, restricting to a smaller coordinate patch if necessary, we may assume that is the graph of a function F (x there, i.e. is defined by Then = {(x, x n R: x n = F (x }. dµ(x = η (x + F (x 2 dx there, where η is a smooth function with compact support on. For simplicity, we write this as dµ(x = φ(x dx with φ C c (. In this local coordinate system, Af(x = f(x y, x n F (y φ(y dy. The idea is that we can embed it in an analytic family of operators and apply the complex interpolation theorem, which we state as follows: Theorem 2 (Interpolation of Operators. uppose α >, and that for each complex number s in the strip α R(s, T s is a linear operator defined on simple functions on such that for any fixed simple functions f and g, Φ(s = T s f(xg(xdx is analytic and bounded (as a function of s in the open strip α < R(s < and is continuous on the closure of the strip. uppose further that there are exponents p, p, q, q [, ] such that { T α+it f L q f L p uniformly for t R. Then where p = θ p + θ p, T +it f L q f L p T f L q f L p q = θ + θ, and θ = α q q α +. In particular, T extends uniquely to a bounded operator L p L q with norm at most. 2

3 Embedding in an analytic family of operators To embed A in an analytic family of operators, first observe that Af can be written as Af(x = f(x yδ (y n F (y φ(y dy. It is profitable to introduce a cut-off function ψ in the y n variable; indeed Af(x = f(x yδ (y n F (y ψ(y n F (y φ(y dy, where ψ Cc (R with ψ near and ψ = outside (,. The point is that there is a way to embed the distribution δ on R into an analytic family of distributions. Consider the distribution g (R j s (g := R Γ(s ts + g(tdt, where t + = max{t, } and s is a complex parameter. The integral converges if Re s >. It can be analytically continued to Re s > by integration by parts; indeed for Re s >, j s (g = R Γ(s ts + g(tdt = d sγ(s dt (ts g(tdt = Γ(s + ts g (tdt, the last of which converges if Re s >. Hence we can take the last integral as the definition of j s when Re s >. In particular, when s =, the distribution is j (g = Γ( t g (tdt = g (tdt = g(, hence j is just δ. More generally, we can define the distribution for Re s > k by the (convergent integral j s (g = ( k Γ(s + k ts+k g (k (tdt. R This gives us an analytic family of distributions, in the sense that j s (g is a holomorphic function of s for each chwartz function g on R. As a result, we are tempted to define an analytic family of operators by T s f(x = f(x yj s (y n F (y ψ(y n F (y φ(y dy, say for chwartz functions f, with j s acting on the y n variable. This can be written as T s f = f K s, where K s is the distribution defined by K s (y = j s (y n F (y ψ(y n F (y φ(y. ( 3

4 For Re s >, K s (y is just the function given by K s (y = Γ(s (y n F (y s + ψ(y n F (y φ(y. (Here we already see why it was good to incorporate ψ in our formula for A; it is only by introducing this harmless factor that K s has compact support in the y n variable, or at least has some decay as y n. Note also that the distribution K is just dµ. Boundedness of K s Now when Re s =, K s (y is a nice bounded function of y; indeed then K s (y C Γ(s, and hence T s maps L to L with norm at most CΓ(s. Note that so when Re s =, Γ(s sin πs = Γ( s, π Γ(s CeIm s, which grows exponentially with Im s as Im s. As a result, the maps T s does not quite satisfy the conditions of the interpolation that we stated above; this can be taken care of easily though, so let s not worry about it for now. Fourier transform of K s Next we would like to obtain an L 2 theory for T s for appropiate values of s. Thus we need to compute and estimate K s (ξ. First, suppose Re s >, so that K s (y is a nice integrable function of y. Then K s (ξ = (y n F (y s + Γ(s ψ(y n F (y φ(y e 2πiy ξ dy = ( (y n F (y s + Γ(s ψ(y n F (y e 2πiynξn dy n φ(y e 2πiy ξ dy R = Γ(s I(s, ξ n φ(y e 2πi(y ξ +F (y ξ n dy = Γ(s I(s, ξ n dµ(ξ (2 where I(s, ξ n = R t s + ψ(te 2πitξn dt; this follows from the change of variable y n F (y y n and that dµ(y = φ(y dy. Now when Re s, K s is a distribution with compact support, so K s (ξ is the 4

5 function defined by e 2πiy ξ, K s (y which depends analytically on s. Furthermore, Γ(s I(s, ξ n also has an analytic continuation to the right half plane by integration by parts, just as we analytically continued the distribution j s. ince both sides of (2 can be analytically continued to the whole complex plane, the identity continues to hold even when Re s, and we shall see that K s (ξ is a bounded function of ξ on the vertical line Re s = α, thereby giving us the desired L 2 theory of T s that allows us to apply complex interpolation. Estimating I(s, ξ n Now we estimate I(s, ξ n : first for Re s > we have the trivial bound Re s I(s, ξ n t+ ψ(t dt C t Re s dt C Re s. (3 R Actually via an integration by parts, we can obtain a better decay for large ξ n if Re s > : in that case I(s, ξ n = t s d ψ(t (e 2πitξn dt 2πiξ n dt = d 2πiξ n dt (ts ψ(te 2πitξn dt the boundary terms vanishing since Re s > and ψ vanish at infinity. This shows that then I(s, ξ n C d ξ n dt (ts ψ(t dt C ξ n. Indeed the larger Re s is, the more integration by parts one can perform and the better decay one obtains as ξ : if Re s > k for some positive integer k, then arguments analogous to above give I(s, ξ n C ξ n k. (4 It turns out that one can get a better estimate than the trivial one even if we just have Re s (, ]: the idea is that one can in effect perform an integration by parts Re s times even though Re s is not an integer. More precisely, one splits the integral I(s, ξ n = ε + ε <. The first term is estimated by A ε ε t s ψ(te 2πitξn dt = A + B, t Re s ψ(t dt C Re s ε Re s. The second term can be integrated by parts: B = t s d ψ(t (e 2πitξn dt ε 2πiξ n dt = ( ε s d ψ(εe 2πiεξn + 2πiξ n dt (ts ψ(te 2πitξn dt ε 5

6 so Hence B CεRe s ξ n + C ξ n ε d dt (ts ψ(t dt C Re sε ξ n I(s, ξ n A + B C Re s ( ε Re s + εre s ξ n. Re s Note that we are free to choose ε here. The first term is big when ε is small, while the second term is big when ε is large. One optimizes the expression by choosing ε such that the two terms are of the same order, say by setting. ε Re s = εre s ξ n, i.e. ε = ξ n. Then I(s, ξ n C Re s ξ n Re s, (5 a better decay than the trivial one when ξ n goes to infinity. (5 was proven just now for Re s (, ], but if Re s (k, k + ], then one can integrate by parts k times just as we did when we improved from estimate (3 to (4 and that allows one to see that (5 actually holds on the whole right half plane Re s >. Together with the trivial estimate, we see that when Re s >. I(s, ξ n C Re s ( + ξ n Re s (6 Next we estimate I(s, ξ n when Re s. Here we need a different integration by parts to reduce the estimate to the case Re s (, ]. By analytic continuation, if Re s ( k, k + ] for some positive integer k, then Γ(s I(s, ξ n = ( k Γ(s + k s+k dk t dt k (ψ(te 2πitξn dt where then Re (s + k (, ]. It follows from the proof of (6 that Γ(s I(s, ξ n C Re (s+k Γ(s + k (+ ξ n k (+ ξ n Re (s+k C Re s e Im s (+ ξ n Re s. (7 Back to L 2 theory Putting things back together, note that by the assumed decay dµ = O( ξ α and by the finiteness of the measure dµ, we have dµ(ξ C( + ξ α. From (7, we see that if we were to find a vertical line on the complex s plane on which K s (ξ = Γ(s I(s, ξ n dµ(ξ is bounded as a function of ξ, then the best that 6

7 we can do is to go to the straight line Re s = α; we could not hope to go beyond the left hand side of that. Indeed on this line K s (ξ C Re s e Im s ( + ξ n Re s ( + ξ α C α e Im s This proves that T s is bounded from L 2 to L 2 if Re s = α, with norm at most C α e Im s. Again, this does not quite satisfy the hypothesis of the complex interpolation theorem, but we shall take care of that now. Conclusion with complex interpolation Indeed it suffices to modify slightly the definition of T s to apply the complex interpolation theorem as stated. Let T s f(x = e s2 T s f(x. Then T is still the average operator A, but now one takes advantage of that e s2 = e (Re s 2 (Im s 2 which decays more rapidly then e Im s as Im s. Hence one has now { T α+it f L 2 C α f L 2 T +it f L C f L It follows from the complex interpolation theorem that T = A maps L 2α+2 2α+ boundedly to L 2α+2. Theorem 3 (Restriction theorem. Let, dµ and α be as in the average theorem. Then the Fourier transform maps L p ( to L 2 (, dµ, p = 2(α + α + 2. Proof. Let f be a chwartz function. R R duality lemma First, f(ξ 2 dµ(ξ = f(ξ f(ξdµ(ξ = f(ye 2πiy ξ f(xe 2πix ξ dµ(ξdydx ( = f(y dµ(y xdy f(xdx. This is just another way of saying that if Rf = f then Rf, Rf L 2 (,dµ = R Rf, f L 2 ( 7

8 with R Rf = f dµ(. (Recall that given an L 2 (dµ function g defined on, R g(x is the function defined on such that Rf, g L 2 (,dµ = f, R g L 2 ( say for all chwartz function f, where again Rf = f. But so Rf, g L 2 (,dµ = f(ξg(ξdµ(ξ = f(xe 2πix ξ g(ξdxdµ(ξ R n = f(x g(ξe 2πix ξ dµ(ξdx = f(x, g(ξe 2πix ξ dµ(ξ R g(x = g(ξe 2πix ξ dµ(ξ, L 2 ( for g L 2 (, dµ. (One can thus think of R as an extension operator. It follows that R Rf(x = f(ξe 2πix ξ dµ(ξ = f(y dµ(y xdy as claimed. To prove that R maps L p to L 2, it suffices to show that R R maps L p ( to L p (, i.e. f dµ( L p ( C f L p (, because Rf 2 L 2 (,dµ R Rf L p ( f L p (. Embedding into an analytic family of operators Now R Rf(x = f(x y dµ( ydy. Again we would like to embed this into an analytic family of operators. Let K s be an analytic family of distributions with K = dµ as in ( in the proof of the average theorem. Then it is natural to define, motivated by the proof of the average theorem, that T s f(x = e s2 f(x y K s ( ydy. This integral converges for all complex values of s, since we observed that K s (y is a function that grows at most polynomially in y for each fixed complex value of s, and f is chwartz. 8

9 L 2 theory Observe that T s f(ξ = e s2 f(ξks ( ξ and recall that e s 2 K s ( ξ C if Re s =. Hence when Re s =, T s maps L 2 to L 2 with norm at most C. L theory Next recall that e s 2 Ks (ξ C α if Re s = α. Hence when Re s = α, T s maps L to L with norm at most C α. Conclusion with complex interpolation It follows from the complex interpolation theorem now that R R = T maps L p ( boundedly to L p (, p = 2(α+ α+2. Hence the Fourier transform extends to a map from L p ( to L 2 (, dµ. Corollary. Let, dµ, α be as in the restriction theorem. Then the extension operator R g(x = g(ξe 2πix ξ dµ(ξ maps L 2 (, dµ boundedly to L q (, with q = 2(α +. α Proof. Note that the q in this corollary is the conjugate exponent to p in the restriction theorem. We have seen R maps L p ( to L 2 (, dµ, p = 2(α+ α+2. As a result, f, R g L 2 ( = Rf, g L 2 (,dµ so by duality Rf L 2 (,dµ g L 2 (,dµ C f L p ( g L 2 (,dµ R g L q ( C g L 2 (,dµ, q being the conjugate exponent to p, i.e. q = 2(α+ α. We shall now give some applications of the above theorems. Here we shall be dealing with non-compact submanifolds of an Euclidean space, and the measure involved will not be the one that is induced from the Lebesgue measure of the underlying Euclidean space. The proofs illustrate how one handles such non-compactness by introducing cut-off functions. 9

10 Corollary 2 (trichartz estimate for the linear chrodinger equation. Let f(x be a chwartz function on. If u(x, t = f(ξe 2πi(x ξ+t ξ 2 dξ so that u solves the linear chrodinger equation ( 2πi t x u(x, t = with initial value then with u(x, = f(x, u L q (dxdt C f L 2 (dx q = 2(n + 2. n Proof. The crucial submanifold here is the hypersurface = {(ξ, τ + : τ = ξ 2 }. The induced Lebesgue measure on the hypersurface is dσ = + 4 ξ 2 dξ, but the essential property for us would be its behaviour near the origin. We would like to apply the average theorem and the restriction theorem, but our surface-carried measure dσ does not have compact support. Therefore we need to introduce a cut-off function and let the support go to infinity. The non-isotropic dilations, (x, t (λx, λ 2 t that preserve the hypersurface, will also play a role. Let η Cc ( with η( =, and let dµ = η(ξdσ = η(ξ + 4 ξ 2 dξ. Then since has nowhere vanishing Gaussian curvature, dµ(x, t decays like dµ(x, t = O( (x, t n 2, so we can apply Corollary with α = n/2. The corollary to the restriction theorem says that if T f(x, t = f(ξe 2πi(x ξ+t ξ 2 η(ξ + 4 ξ 2 dξ then T f L q (dxdt C f L 2 (dξ = C f L 2 (dx,

11 where q = 2(n+2 n. Now pointwisely u(x, t = lim λ + = lim λ + ( ξ f(ξe 2πi(x ξ+t ξ 2 η + 4 ξ 2 λ λ dξ λ n f(λξe 2πi(λx ξ+λ2 t ξ 2 η(ξ + 4 ξ 2 dξ = lim λ + T (f λ(λx, λ 2 t where f λ (x = f(λ x. Hence for q = 2(n+2 n, since u L q (dxdt lim inf λ + T (f λ(λx, λ 2 t L q (dxdt (Fatou s lemma n+2 = lim inf λ + λ q T (f λ L q (dxdt (cale invariance n+2 lim inf λ + Cλ q f λ L 2 (dx (Restriction theorem n+2 = lim inf λ + Cλ q λ n 2 f L 2 (dx (cale invariance = C f L 2 (dx n + 2 q for this value of q. Hence we are done. + n 2 = Corollary 3 (Averaging along a paraboloid. Let f(x be a chwartz function on. If Af(x = f(x y, x n y 2 dy then Af L p ( C f L p ( with p = n + n. Proof. Again let dµ = η(y + 4 y 2 dy be a measure carried on the paraboloid {y n = y 2 }, where η Cc (. Then dµ(ξ = O( ξ n 2 because has nowhere vanishing Gaussian curvature. Hence the average theorem says that if T f(x = f(x y, x n y 2 η(y + 4 y 2 dy, then with T f L p ( C f L p ( p = n + n.

12 Now pointwisely Af(x = f(x y, x n y 2 dy ( y = lim f(x y, x n y 2 η + 4 y 2 λ + λ λ dy according to the dominated convergence theorem. However, where ( y f(x y, x n y 2 η + 4 y 2 λ λ dy =λ n f(x λy, x n λ 2 y 2 η(y + 4 y 2 dy =λ n T (f λ (λ x, λ 2 x n f λ (x = f(λx, λ 2 x n is the dilation of f by the relevant non-isotropic dilation. Hence since Af L p ( lim inf λ + λn T (f λ (λ x, λ 2 x n L p ( (Fatou s lemma = lim inf λ + λn λ n+ p T (f λ L p ( (cale invariance lim inf λ + Cλn λ n+ p f λ L p ( (Average theorem = lim inf λ + Cλn λ n+ p λ n+ p f L p ( (cale invariance = C f L p ( when p = n+ n. This completes the proof. n + n + p n + = p Corollary 4 (trichartz estimate for linearized KdV equation. Let f(x be a chwartz function on R. If u(x, t = f(ξe 2πi(xξ+tξ3 dξ so that u solves the linearized KdV equation 2 3 (4π t 3 u(x, t = x with initial value then u(x, = f(x, u L 8 (dxdt C f L 2 (dx. 2

13 Proof. The proof is the same as that of Corollary 2, except now that the relevant submanifold is the curve = {(ξ, ξ 3 : ξ R} in R 2 and the curvature of vanishes at the origin. The best possible decay for dµ is now dµ(ξ = O( ξ 3, so if we apply Corollary as in the proof of Corollary 2 with α = /3 we obtain u L 8 (dxdt C f L 2 (dx. Corollary 5 (Averaging along a curve of finite type. Let f(x be a chwartz function on R 2. If Af(x = f(x t, x 2 t k dt then with R Af L p (R 2 C f L p (R 2 p = 2k + 2 k + 2. Proof. Again the proof is the same as that of Corollary 3, except now that the relevant submanifold is the curve = {(t, t k : t R} in R 2. The curvature of vanishes at the origin. However, since it is a curve of finite type k, we still have a certain decay for dµ; indeed dµ(ξ = O( ξ k, so if we apply Theorem with α = /k as in the proof of Corollary 3 we obtain Af L p (R 2 C f L p (R 2 with p = 2k + 2 k