A remark on the Strichartz inequality in one dimension
aa r X i v : . [ m a t h . A P ] J a n A REMARK ON THE STRICHARTZ INEQUALITY INONE DIMENSION
RYAN FRIER AND SHUANGLIN SHAO
Abstract.
In this paper, we study the extremal problem for the Strichartzinequality for the Schr¨odinger equation on R . We show that the solu-tions to the associated Euler-Lagrange equation are exponentially decay-ing in the Fourier space and thus can be extended to be complex analytic.Consequently we provide a new proof to the characterization of the ex-tremal functions: the only extremals are Gaussian functions, which wasinvestigated previously by Foschi [7] and Hundertmark-Zharnitsky [11]. Introduction
To begin, we note that the Strichartz inequality for an arbitrary dimension d is k e it ∆ f k /d ≤ C d k f k , where k f k p = (cid:0)R R d | f ( x ) | p dx (cid:1) /p , and e it ∆ f = π ) d R R d e ix · ξ + i | ξ | t b f ( ξ ) dξ ,see e.g., [12, 18]. Strichartz’s Inequality has long been studied. The originalproof of Strichartz inequality is due to Robert Strichartz in [17] in 1977.Define the Fourier transform as b f ( ξ ) = R R d e − ix · ξ f ( x ) dx and the space-time Fourier transform as ˜ F ( τ, ξ ) = R R × R n e i ( τt + x · ξ ) F ( t, x ) dt dx . Note thatin the case of d = 1 we have (cid:13)(cid:13) e it ∆ f (cid:13)(cid:13) ≤ C k f k ; e it ∆ f = 12 π Z R e ixξ + iξ t b f ( ξ ) dξ. (1)Define C = sup ( (cid:13)(cid:13) e it ∆ f (cid:13)(cid:13) k f k : f ∈ L , f = 0 ) . (2)We say that f is an extremizer or a maximzer of the Strichartz inequalityif f = 0 and k e it ∆ f k = C k f k . The extremal problem for the Strichartzinequality (1) is (a) Whether there exists an extremzier for (1)? (b) If itexists, what are the characterizations of extremizers, e.g., continuity anddifferentiability? What is the explicit formulation of extremizers? are they Date : January 5, 2021. unique up to the symmetries of the inequality? In this note, we are mainlyconcerned with question (b).By D. Foschi’s 2007 paper [7] we know that the maximizers are Gaussianfunctions of the form f ( x ) = e Ax + Bx + C , where A, B, C ∈ C , and ℜ{ A } < f ( x ) = e −| x | is a maximizer in dimension 1. Thus f must satisfy(1), and obtains equality with C .Hundertmark and Zharnitsky in [11] showed a new representation usingan orthogonal projection operator for dimension 1 and 2. The representationthat was found is Z R Z R (cid:12)(cid:12) e it ∆ f ( x ) (cid:12)(cid:12) dxdt = 12 √ h f ⊗ f ⊗ f, P ( f ⊗ f ⊗ f ) i L ( R ) , Z R Z R (cid:12)(cid:12) e it ∆ f ( x ) (cid:12)(cid:12) dxdt = 14 h f ⊗ f, P ( f ⊗ f ) i L ( R ) for dimensions d = 1 and d = 2, respectively, where P , P are certainprojection operators. Using this, they were able to obtain the same results.In [13] Kunze showed that such a maximizer exists in dimension 1. In [14],the second author showed the existence of a maximizer in all dimensions forthe Strichartz inequalities for the Schr¨odinger equation. Likewise, in [16],Brocchi, Silva, and Quilodr´an investigated sharp Strichartz inequalities forfractional and higher order Schr¨odinger equations. There they discussedthe rapid L decay of extremizers, which we will also discuss and use it toestablishing a characterization of extremizers.We will take inspiration from [15] to show a different method of provingthat extremizers are Gaussians. More precisely, in this note, we are inter-ested in the problem of how to characterize extremals for (1) via the studyof the associated Euler-Lagrange equation. We show that the solutions ofthis generalized Euler-Lagrange equation enjoy a fast decay in the Fourierspace and thus can be extended to be complex analytic, see Theorem 1.1.Then as an easy consequence, we give an alternative proof that all extremalfunctions to (1) are Gaussians based on solving a functional equation ofextremizers derived in Foschi [7], see (5) and Theorem 1.2. The functionalequality (5) is a key ingredient in Foschi’s proof in [7]. To prove f in (5) tobe a Gaussian function, local integrability of f is assumed in [7], which isfurther reduced to measurable functions in Charalambides [2].Let f be an extremal function to (1) with the constant C . Then f satisfies the following generalized Euler-Lagrange equation,(3) ω h g, f i = Q ( g, f, f, f, f, f ) , for all g ∈ L , REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 3 where ω = Q ( f, f, f, f, f, f ) / k f k L > Q ( f , f , f , f , f , f ) is theintegral Z R b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) δ ( ξ + ξ + ξ − ξ − ξ − ξ ) × δ ( ξ + ξ + ξ − ξ − ξ − ξ ) dξ dξ dξ dξ dξ dξ , (4)for f i ∈ L ( R ), 1 ≤ i ≤ δ ( ξ ) = (2 π ) − d R R d e iξ · x dx in the distribution sense, d = 1 ,
2. The proof of (3) is standard; see e.g. [6, p. 489] or [9, Section 2]for similar derivations of Euler-Lagrange equations.
Theorem 1.1. If f solves the generalized Euler-Lagrange equation (3) forsome ω > , then there exists µ > such that e µ | ξ | b f ∈ L ( R ) . Furthermore f can be extended to be complex analytic on C . To prove this theorem, we follow the argument in [10]. Similar reason-ing has appeared previously in [5, 8]. It relies on a multilinear weightedStrichartz estimate and a continuity argument. See Lemma 3.2 and Lemma3.3, respectively.Next we prove that the extremals to (1) are Gaussian functions. We startwith the study of the functional equation derived in [7]. In [7], the functionalequation reads(5) f ( x ) f ( y ) f ( z ) = f ( a ) f ( b ) f ( c ) , for any x, y, z, a, b, c ∈ R such that(6) x + y + z = a + b + c, x + y + y = a + b + c , In [7], it is proven that f ∈ L satisfies (5) if and only if f is an extremalfunction to (1). Basically, this comes from two aspects. One is that in theFoschi’s proof of the sharp Strichartz inequality only the Cauchy-Schwarzinequality is used at one place besides equality. So the equality in theStrichartz inequality (1), or equivalently the equality in Cauchy-Schwarz,yields the same functional equation as (5) where f is replaced by ˆ f . Theother one is that the Strichartz norm for the Schr¨odinger equation enjoysan identity that(7) k e it ∆ f k L ( R ) = C k e it ∆ f ∨ k L ( R ) for some C > f is a locally integrable function. In [15], Jiangand the second author studied the two dimensional case of (5) and provedthat the solutions are Gaussian functions. These can be viewed as investiga-tions of the Cauchy functional equations (5) for functions supported on theparaboloids. To characterize the extremals for the Tomas-Stein inequalityfor the sphere in R , in [4], Christ and the second author study the func-tional equation of similar type for functions supported on the sphere and RYAN FRIER AND SHUANGLIN SHAO prove that they are exponentially affine functions. In [2], Charalambidesgeneralizes the analysis in [4] to some general hyper-surfaces in R n that in-clude the sphere, paraboloids and cones as special examples and proves thatthe solutions are exponentially affine functions. In [2, 4], the functions areassumed to be measurable functions.By the analyticity established in Theorem 1.1, Equations (5) and (6) havethe following easy consequence, which recovers the result in [7, 11]. Theorem 1.2.
Suppose that f is an extremal function to (1) . Then (8) f ( x ) = e A | x | + B · x + C , where A, C ∈ C , B ∈ C and ℜ ( A ) < . Developing the Extremizer
We want to show that if f solves the generalized Euler-Lagrange equation(3), then there exists some µ > e µ | ξ | b f ∈ L . Furthermore, we can extend f to be entire. To begin, we note that byFoshi’s paper [7], we have for a maximizer f , f ( x ) f ( y ) f ( z ) = F ( x + y + z , x + y + z ). Thus for any ( x, y, z ) , ( a, b, c ) ∈ R such that x + y + z = a + b + c (9)and x + y + z = a + b + c , (10)then f ( x ) f ( y ) f ( z ) = F ( x + y + z , x + y + z )= F ( a + b + c , a + b + c )= f ( a ) f ( b ) f ( c ) . That is, f ( x ) f ( y ) f ( z ) = f ( a ) f ( b ) f ( c ) . (11)Let us assume that f has an entire extension. Then f restricted to R isreal analytic. By [7, Lemma 7.9], such nontrival f ∈ L is also nonzero. Weprove the following theorem. Theorem 2.1. If f is a maximizer for (1) , then f ( x ) = e Ax + Bx + C , where A, B, C ∈ C . REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 5
Proof.
Consider ϕ ( x ) = log( f ( x )). We know from [7, Lemma 7.9] that f isnowhere 0, so ϕ is well defined. Since f is analytic, then so is ϕ . Hence bythe power series expansion we have ϕ ( x ) = ϕ (0) + ϕ ′ (0) x + ϕ ′′ (0)2 x + X k ≥ ϕ ( k ) (0) k ! x k . Hence it is true for a, b, c, d, e, g such that ( a, b, c ) , ( d, e, g ) satisfy equations(9) and (10). That is, ϕ ( a ) = ϕ (0) + ϕ ′ (0) a + ϕ ′′ (0)2 a + X k ≥ ϕ ( k ) (0) k ! a k ,ϕ ( b ) = ϕ (0) + ϕ ′ (0) b + ϕ ′′ (0)2 b + X k ≥ ϕ ( k ) (0) k ! b k ,ϕ ( c ) = ϕ (0) + ϕ ′ (0) c + ϕ ′′ (0)2 c + X k ≥ ϕ ( k ) (0) k ! c k ,ϕ ( d ) = ϕ (0) + ϕ ′ (0) d + ϕ ′′ (0)2 d + X k ≥ ϕ ( k ) (0) k ! d k ,ϕ ( e ) = ϕ (0) + ϕ ′ (0) e + ϕ ′′ (0)2 e + X k ≥ ϕ ( k ) (0) k ! e k ,ϕ ( g ) = ϕ (0) + ϕ ′ (0) g + ϕ ′′ (0)2 g + X k ≥ ϕ ( k ) (0) k ! g k . By equation (11) we know that ϕ ( a ) + ϕ ( b ) + ϕ ( c ) = ϕ ( d ) + ϕ ( e ) + ϕ ( g ).Thus by using the power series expansions and equations (9), (10) and (11) RYAN FRIER AND SHUANGLIN SHAO we have that0 = ϕ ( a ) + ϕ ( b ) + ϕ ( c ) − ϕ ( d ) − ϕ ( e ) − ϕ ( g )= ϕ (0) + ϕ ′ (0) a + ϕ ′′ (0)2 a + X k ≥ ϕ ( k ) (0) k ! a k + ϕ (0) + ϕ ′ (0) b + ϕ ′′ (0)2 b + X k ≥ ϕ ( k ) (0) k ! b k + ϕ (0) + ϕ ′ (0) c + ϕ ′′ (0)2 c + X k ≥ ϕ ( k ) (0) k ! c k − ϕ (0) − ϕ ′ (0) d − ϕ ′′ (0)2 d − X k ≥ ϕ ( k ) (0) k ! d k − ϕ (0) − ϕ ′ (0) e − ϕ ′′ (0)2 e − X k ≥ ϕ ( k ) (0) k ! e k − ϕ (0) − ϕ ′ (0) g − ϕ ′′ (0)2 g − X k ≥ ϕ ( k ) (0) k ! g k = ϕ ′ (0)( a + b + c − d − e − g ) + ϕ ′′ (0)2 ( a + b + c − d − e − g )+ X k ≥ ϕ ( k ) (0) k ! ( a k + b k + c k − d k − e k − g k )= X k ≥ ϕ ( k ) (0) k ! ( a k + b k + c k − d k − e k − g k ) . That is, X k ≥ ϕ ( k ) (0) k ! ( a k + b k + c k − d k − e k − g k ) = 0 , (12)where ( a, b, c ) , ( d, e, g ) ∈ R satisfy equations (9) and (10). Consider a = x, b = − x, c = x, g = 0, by solving the equations d + e = x,d + e = 3 x , we obtain that d = √ x, e = −√ x . Then a k + b k + c k − e k − f k − g k = 2 x k + ( − x ) k − √ x ! k − − √ x ! k . REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 7
When k is even, − − ( 1 + √
52 ) k − ( 1 − √
52 ) k ! ≥ ( 32 ) k − > k ≥
3. When k is odd, − − ( 1 + √
52 ) k + ( √ −
12 ) k ! ≥ ( 32 ) k − > k ≥
3. This shows that ϕ k (0) = 0 when k ≥
3. Hence ϕ ( k ) (0) = 0 for all k . Thus ϕ ( x ) = Ax + Bx + C . Therefore f ( x ) = e Ax + Bx + C . (cid:3) Establishing the Exponential Decay in Fourier Space
Consider the integral Q ( f , f , f , f , f , f ) = Z R b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) b f ( ξ ) δ ( ξ + ξ + ξ − ξ − ξ − ξ ) × δ ( ξ + ξ + ξ − ξ − ξ − ξ ) dξ dξ dξ dξ dξ dξ . Notice that if f is an extremal function to Strichartz estimate, then f must satisfy the generalized Euler-Lagrange equation ω h g, f i = Q ( g, f, f, f, f, f )(13)for all g ∈ L , where ω = || f || Q ( f, f, f, f, f, f ) >
0, and δ ( ξ ) = π R R e iξx dx in the distribution sense.Define η := ( η , η , η , η , η , η ) ∈ R ,a ( η ) := η + η + η − η − η − η ,b ( η ) := η + η + η − η − η − η . The choice of a ( η ) and b ( η ) are useful, since when a ( η ) = 0 = b ( η ), then η ,more specifically, ( η , η , η ) , ( η , η , η ) ∈ R , satisfy equations (9) and (10).For ε ≥ , µ ≥ , and ξ ∈ R define F ( ξ ) := F µ,ε ( ξ ) = µξ εξ . For h i ∈ L ( R ) , ≤ i ≤
6, define the weighted multilinear integral M F as M F ( h , h , h , h , h , h ) = Z R e F ( η ) − P k =2 F ( η k ) Π k =1 | h ( η k ) | δ ( a ( η )) δ ( b ( η )) dη. It is easy to see that M F ( h , h , h , h , h , h ) ≤ Z R Π k =1 | h ( η k ) | δ ( a ( η )) δ ( b ( η )) dη. (14) RYAN FRIER AND SHUANGLIN SHAO
Indeed, on the support of a and b , η ≤ X i =2 η i . We also note that F ( ξ ) is an increasing function, and F ( ξ ) ≥ ξ, µ ,and ε . So equation (14) can be derived by | M F ( h , h , h , h , h , h ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z R e F ( η ) − P k =2 F ( η k ) Π k =1 | h ( η k ) | δ ( a ( η )) δ ( b ( η )) dη (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z R (cid:12)(cid:12)(cid:12) e F ( η ) − P k =2 F ( η k ) (cid:12)(cid:12)(cid:12) Π k =1 | h ( η k ) | δ ( a ( η )) δ ( b ( η )) dη ≤ Z R Π k =1 | h ( η k ) | δ ( a ( η )) δ ( b ( η )) dη. We state the following key lemma, which is established by using the Hausdorff-Young inequality. The two dimensional such estimate is due to Bourgain [1]that is much harder.
Lemma 3.1. (cid:13)(cid:13) e it ∆ h e it ∆ h (cid:13)(cid:13) L t,x ≤ CN − / k h k L k h k L , (15) where h ∈ L is supported on | ξ | ≤ s and h ∈ L is supported on | η | ≥ N s ,for N ≫ and s ≫ . Equation (15) has been established in [16]. We provide a proof for com-pleteness. Let b f be supported on | ξ | ≤ s and b g be supported on | η | ≥ N s where N ≫ s ≫
1, and note that e it ∆ f e it ∆ g = 1(2 π ) Z R Z R e ixξ + itξ b f ( ξ ) e ixη + itη b g ( η ) dηdξ = 1(2 π ) Z R Z R e ix ( ξ + η )+ it ( ξ + η ) b f ( ξ ) b g ( η ) dηdξ. Consider the change of variables γ = ξ + η and τ = ξ + η . Let | J | = √ τ − γ be the corresponding Jacobian. Thus 2 τ − γ = ( ξ − η ) . Let | ξ | ≤ s and | η | ≥ N s , where
N >
1. If η >
0, then η > ξ . If η <
0, then η < ξ . In either case, the Jacobian is well defined. Likewise, by considering2 k N s ≤ | η | ≤ k +1 N s , we see that | J | . (2 k N s ) − . Let G ( γ, τ ) := b f γ + p τ − γ ! b g γ − p τ − γ ! | J | . Then we have e it ∆ f e it ∆ g = 1(2 π ) Z e ixγ + itτ G ( γ, τ ) d ( γ × τ )= 1(2 π ) e G ( x, t ) . REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 9
Thus by the Hausdorff-Young inequality and change of variables, we have (cid:13)(cid:13) e it ∆ f e it ∆ g (cid:13)(cid:13) L x,t = (cid:13)(cid:13)(cid:13) ˜ G ( x, t ) (cid:13)(cid:13)(cid:13) L x,t ≤ k G ( γ, τ ) k L / γ,τ = (cid:18)Z | G ( γ, τ ) | / d ( γ × τ ) (cid:19) = Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b f γ ± p τ − γ !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)b g γ ∓ p τ − γ !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | J | d ( γ × τ ) = (cid:18)Z (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | b g ( η ) | | J | | J | − d ( ξ × η ) (cid:19) . The above continues to equal ∞ X k =0 Z | ξ |≤ s, k Ns ≤| η |≤ k +1 Ns (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | b g ( η ) | | J | d ( ξ × η ) ! ≤ ∞ X k =0 Z | ξ |≤ s, k Ns ≤| η |≤ k +1 Ns (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | b g ( η ) | | J | d ( ξ × η ) ! . ∞ X k =0 (cid:18)Z (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | b g ( η ) | (2 k sN ) − d ( ξ × η ) (cid:19) = ( sN ) − ∞ X k =0 − k Z | ξ |≤ s, k Ns ≤| η |≤ k +1 Ns (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) | b g ( η ) | d ( ξ × η ) ! = ( sN ) − Z | ξ |≤ s | b f ( ξ ) | dξ ! ∞ X k =0 − k Z k Ns ≤| η |≤ k +1 Ns | b g ( η ) | dη ! . For (cid:16)R | ξ |≤ s | b f ( ξ ) | / (cid:17) / , we wish to show that R | ξ |≤ s (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) / . s / k f k / .Consider H¨older’s inequality, for p = 4 and q = . Then Z | ξ |≤ s (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) / = Z | ξ |≤ s · (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) / ≤ Z | ξ |≤ s ! / Z | ξ |≤ s (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) ! / = (2 s ) / Z | ξ |≤ s (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) ! / / ≤ (2 s ) / k b f k / = (2 s ) / k f k / , where the final step is a consequence of Plancherel’s theorem. Hence Z | ξ |≤ s (cid:12)(cid:12)(cid:12) b f ( ξ ) (cid:12)(cid:12)(cid:12) / ! / ≤ (2 s ) / k f k . As for (cid:16)R k Ns ≤| η |≤ k +1 Ns | b g ( η ) | / (cid:17) / , we use a similar technique as wedid for b f . Specifically, Z k Ns ≤| η |≤ k +1 Ns | b g ( η ) | / ≤ Z kNs ≤| η |≤ k +1 Ns ! / Z k Ns ≤| η |≤ k +1 Ns | b g ( η ) | ! / = (cid:16) k N s (cid:17) / Z k Ns ≤| η |≤ k +1 Ns | b g ( η ) | ! / / ≤ (cid:16) k N s (cid:17) / k b g k / = (cid:16) k N s (cid:17) / k g k / . Hence (cid:16)R k Ns ≤| η |≤ k +1 Ns | b g ( η ) | / (cid:17) / ≤ (2 k N s ) / k g k . By pairing thiswith the above we have (cid:13)(cid:13) e it ∆ f e it ∆ g (cid:13)(cid:13) L x,t . ( sN ) − Z | ξ |≤ s | b f ( ξ ) | dξ ! × ∞ X k =0 − k Z k Ns ≤| η |≤ k +1 Ns | b g ( η ) | dη ! ≤ ( sN ) − / (2 s ) / k f k ∞ X k =0 − k/ (2 k N s ) / k g k = 2 / N − / k f k k g k ∞ X k =0 − k/ = CN − / k f k k g k . If we pair the estimate in Lemma 3.1 with H¨older’s inequality and the L → L Strichartz inequality, we get the following lemma.
Lemma 3.2.
Let h k ∈ L ( R ) , ≤ k ≤ , and s ≫ , N ≫ . Suppose thatthe Fourier transform of h is supported on { ξ : | ξ | ≤ s } and the Fouriertransform of h is supported on {| ξ | ≥ N s } . Then M F ( h , h , h , h , h , h ) ≤ CN − / Π k =1 k h k k . REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 11
Next we focus on establishing Theorem 1.1. If it can be shown that e µξ b f ∈ L for some µ >
0, then e λ | ξ | b f ∈ L for some 0 < λ < µ . Then bythe Fourier inversion equation we have that f ( z ) = π R R e izξ − λ | ξ | e λ | ξ | b f ( ξ ) dξ .Thus ∂ z f ( z ) = ∂ z (cid:18) π Z R e izξ − λ | ξ | e λ | ξ | b f ( ξ ) dξ (cid:19) = Z R ∂ z (cid:16) e izξ − λ | ξ | (cid:17) e λ | ξ | b f ( ξ ) dξ = 0 . So f can be extended to complex analytic on C . To prove Theorem 1.1, weestablish Lemma 3.3.
Let f solve the generalized Euler-Lagrange equation (7) for ω as defined just below equation (7) , k f k = 1 , and define b f > := b f | ξ |≥ s for s > . Then there exists some s ≫ such that for µ = s − , ω (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) ≤ o (1) (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + o (1) , (16) where lim s →∞ o i (1) = 0 uniformly for all ε > , i = 1 , , and the constant C is independent of ε and s .Proof. For this proof we follow the proof of lemma 2 . (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) = h e F ( · ) b f > , e F ( · ) b f > i = h e F ( · ) b f > , b f i = h e F ( · ) f > , f i . So by equa-tion (7) we have that ω (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) = Q ( e F ( · ) f > , f, f, f, f, f )= Z R e F ( ξ ) − P k =2 F ( ξ k ) h > ( ξ ) h ( ξ ) h ( ξ ) h ( ξ ) h ( ξ ) h ( ξ ) δ ( a ( ξ )) δ ( b ( ξ )) dξ, where ξ = ( ξ , ξ , ξ , ξ , ξ , ξ ), and h ( ξ i ) := e F ( ξ i ) b f ( ξ i ) and h > ( ξ i ) := e F ( ξ i ) b f > ( ξ i ) for 1 ≤ i ≤
6. Thus ω (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) ≤ M F ( h > , h, h, h, h, h ). Define h ≪ := h | ξ | , h, h, h, h, h ) = M F ( h > , h < , ..., h < ) + X j ,...,j M F ( h > , h j , ..., h j ):= A + B, where j i is either < or > , and at least one of the subscripts is > . We breakup A into M F ( h > , h ≪ , h < , ..., h < ) + M F ( h > , h ∼ , h < , ..., h < ) := A + A . By lemma 3 . A . s − / k h > k k h ≪ k k h < k . Note that k h < k = Z (cid:12)(cid:12)(cid:12) e F ( ξ ) b f ( ξ )1 | ξ | k k h ≪ k k h < k + k h > k k h ∼ k k h < k ≤ s − / k h > k e µs e µs + k h > k e µs k f ∼ k e µs = e µs k h > k (cid:16) s − / e µs − µs + k f ∼ k (cid:17) = o (1) k h > k = o (1) (cid:13)(cid:13)(cid:13) e F ( · ) b f (cid:13)(cid:13)(cid:13) , where o (1) = e µs (cid:16) s − / e µs − µs + k f ∼ k (cid:17) . Thus o (1) → s → ∞ .As for B , let B := P j ,...,j M F ( h > , h j , ..., h j ) containing precisely 1 h > ∈ { h j , ..., h j } , B k := P j ,...,j M F ( h > , h j , ..., h j ) containing precisely k h > ∈ { h j , ..., h j } . For example, B = M F ( h > , h > , h < , h < , h < , h < ) + M F ( h > , h < , h > , h < , h < , h < )+ M F ( h > , h < , h < , h > , h < , h < ) + M F ( h > , h < , h < , h < , h > , h < )+ M F ( h > , h < , h < , h < , h < , h > )= CM F ( h > , h < , h > , h < , h < , h < )= CM F ( h > , h ≪ , h > , h < , h < , h < ) + CM F ( h > , h ∼ , h > , h < , h < , h < ) . By the same argument as for A , we see that B . s − / k h > k k h ≪ k k h < k + k h > k k h ∼ k k h < k = o (1) k h > k . Hence B . o (1) (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) where o (1) → s → ∞ . Following asimilar process, we find that B k . (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) k +12 . By setting µ = s − we REMARK ON THE STRICHARTZ INEQUALITY IN ONE DIMENSION 13 have e µs = e . Thus we have ω (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) ≤ o (1) (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + o (1) (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + C (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) + (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) . (17)Dividing both sides of inequality (17) by (cid:13)(cid:13)(cid:13) e F ( · ) b f > (cid:13)(cid:13)(cid:13) we obtain the desiredresult. (cid:3) To see that e µξ b f ∈ L as required in Theorem 1.1, we also follow adiscussion in [15]. Define H ( ε ) = Z | ξ |≥ s (cid:12)(cid:12)(cid:12) e F s − ,ε ( ξ ) b f (cid:12)(cid:12)(cid:12) dξ ! / . Note here that s is fixed, but we have control over that term. By thedominated convergence theorem we have that H ( ε ) is continuous on (0 , ∞ ),and is therefore connected on (0 , ∞ ). To see that H is bounded uniformlyon (0 , ∞ ), consider the function G ( x ) = ω x − Cx − Cx − Cx − Cx on(0 , ∞ ) (refer to lemma 3 . s large enough such that o (1) ≤ ω ).This is very clearly bounded above by lemma 3 .
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