APRIORI ESTIMATES FOR THE 2-D WAVE EQUATION

Yung-Fu Fang and Manoussos G. Grillakis * Department of Mathematics

University of Maryland, College Park, MD 20742

Chapter 1, Introduction :

In the present work we will be concerned with certain regularity proper-ties of the linear Klein-Gordon equation. The motivation for this questions comes from certain nonlinear systems of wave equations, see [2]. In order to formulate the problem let us start by considering the Klein-Gordon Equation

*φ + m*2*φ = 0,* *(0.1a)*

*with initial data given at some fixed time, for example t = 0, by*

*φ(0, x) = 0 ,* *φ,t(0, x) = f (x).* *(0.1b)*

The conservation of energy, to be explained below, dictates as a natural
*hypothesis f ∈ L*2* _{(R}*2

_{). One can write the solution of the equation above in}

integral form as follows

*φ(t, x) =*
Z

*R*2

*R(t, x − y)f (y)dy ,* *(0.2a)*

* The second author is supported by a PYI, DMS-9157512 and a Sloan Fellowship

*where R(t, x) is the Riemann function, which in this case can be computed*
explicitly.
*R(t, x) :=* 1
*2π*
cos
h
*m*p*t*2* _{− |x|}*2
i
p

*t*2

*2*

_{− |x|}*H(t*2

*2*

_{− |x|}

_{) ,}

_{(0.2b)}*where H(t) is the Heaviside function. Notice that* p*t*2* _{− |x|}*2

_{is the }

*hyper-bolic distance between the point (t, x) ∈ R × R*2 _{and the origin with respect}

*to the Minkowski metric gµν* *:= diag(−1, 1, 1) . In the present paper we*

*would like to consider the solution φ(t, x) with rough initial data, namely*
*f ∈ L*2* _{(R}*2

_{). From the expression in (0.2a) it is obvious that φ(t, x) just}*fails to be in L∞ _{. Standard estimates however imply that φ is in the class}*

of bounded mean oscillation functions, an idea introduced by F. John and
*L. Nirenberg, [12], henceforth to be denoted by BM O. Functions in BM O*
have the property that, on every cube, they can be approximated in the
*L mean by their average, with an error independent of the cube, see [7].*
*Roughly speaking BM O functions can have logarithmic type of singularities*
*and there is no reason why the square of the function is still in BM O. Here*
*we will consider a square expression, namely φ*2* _{(t, x) and see that the L}*2

*norm in space of the first deivatives is bounded by a BM O function. It is*
*worthwhile to observe here that if one assumes that f ∈ B*_{2}*0,1(R*2_{), which is}

*the Besov space with indices (2, 1), then the solution is in L∞*_{.}

There are two standard estimates one can derive for the equation (0.1a), one is the energy estimate which follows from the conservation laws

*∂µTνµ* *= 0 .* *(0.3a)*

*Tµν* is the energy momentum tensor defined by

*Tµν* *= ∂µφ∂νφ + gµν*

¡

*∂αφ∂αφ + m*2*φ*2

¢

*.* *(0.3b)*

space-time we have the bound Z

*R*2

£

*|∇φ|*2*+ φ*2* _{,t}+ m*2

*φ*2¤

*dx = const.*

*(0.4)*

*The Sobolev inequality implies that for every fixed time φ(t, ·) is a BM O*
function.

The other estimate, due originally to Strichartz [4], reads

*kD1/2φkL*6* _{(R}*3

_{)}

*≤ Ckf k*2

_{L}*2*

_{(R}_{)}

*,*

*(0.5)*

*where D1/2* _{is half a derivative defined via the Fourier transform. Notice}

that the left hand side of the above inequality implies again ,via the Sobolev
*inequality, that φ is a BM O function but over space-time. Estimate (0.5) is*
derived using the Fourier transform and we will use these ideas in the present
work. As a matter of fact the estimate of Strichartz will follow from one of
the theorems that we will present later.

The solution of (0.1) can be written
*φ(t, x) =*

Z

*R*2

*sin(ω(ξ)t)eix·ξf (ξ)dµ(ξ)*ˆ *(0.6a)*

*ω(ξ) :=* p*|ξ|*2* _{+ m}*2

_{and}

_{dµ(ξ) :=}*dξ*

*ω(ξ)* *(0.6b)*

*Now squaring the above expression gives φ*2_{(t, x) as a combination of two}

terms
*A :=*
Z
*R*2* _{×R}*2

*cos [(ω(ξ*1

*) + ω(ξ*2

*))t] eix·(ξ*1

*+ξ*2)

*f (ξ*ˆ 1) ˆ

*f (ξ*2

*)dµ(ξ*1

*)dµ(ξ*2)

*B :=*Z

*R*2

*2*

_{×R}*cos [(ω(ξ*1

*) − ω(ξ*2

*))t] eix·(ξ*1

*+ξ*2)

*f (ξ*ˆ 1) ˆ

*f (ξ*2

*)dµ(ξ*1

*)dµ(ξ*2

*) .*

¿From the above considerations one can see that the square expression is built from the basic integral operator

*E(f )(t, x) :=*
Z
*R*2
*exp{i(ω(ξ)t + x · ξ)} ˆf (ξ)dµ(ξ)* *(0.7a)*
*ω(ξ) :=*p*|ξ|*2* _{+ m}*2

_{and}

_{dµ(ξ) :=}*dξ*

*ω(ξ)*

*.*

*(0.7b)*

*The measure dµ(ξ) is a natural object, it is the Leray measure on the *
Hy-perbola defined by the equation

n

*(τ, ξ) ∈ R × R*2 : *τ =*p*|ξ|*2* _{+ m}*2o

_{.}*¿From the operator E(f ) one can form two types of squares*

Φ*i(t, x) := E(f )E(g)(t, x)* and Φ*o(t, x) := E(f ) ¯E(g)(t, x) . (0.8)*

*We will see that in order to prove estimates for φ*2_{(t, x) we have to give a}

proof for both Φ*i,o* mentioned above.

Now let us state the first theorem.

*Theorem 0.1 : Consider the solution of equation (0.1a,b). Let Dt,x*

*denote any first order derivative and call*

*e(t) :=*
Z
*R*2
*|Dt,xφ*2*(t, x)|*2*dx*
1
2
*.* *(0.9a)*

*Then there is a positive function b(t) such that*

*e(t) ≤ b(t) ,* *(0.9b)*

*where*

*kbkbmo* *≤ Ckf k*2*L*2 *,* *(0.9c)*

*and the bmo norm is a local version of BM O.*

*Before we proceed we would like to explain the BM O and bmo norms*
as well as an important duality property of these spaces, see [6], [7] , [10].
Let us start by calling

*(g)I* :=
1
*|I|*
Z
*I*
*g(x)dx*

the average of a function over an arbitrary cube with edges parallel to the
*coordinate axis. The BM O norm is defined by*

*kgkBM O* := sup
*I*
1
*|I|*
Z
*I*
*|g(x) − (g)I|dx .* *(0.10)*

*The supremum above is taken over all cubes. The bmo norm is the following*
modification, see [6] , [10]
*kgkbmo* := sup
*|I|≤1*
1
*|I|*
Z
*I*
*|g − (g)I|dx + sup*
*|I|≥1*
1
*|I|*
Z
*I*
*|g|dx .* *(0.11)*

There is an important duality property for this space, namely ¡

*h*1¢*∗* *= bmo ,* *(0.12a)*

which in practical terms means the following
¯
¯
¯
¯
¯
¯
Z
*Rn*
*f gdx*
¯
¯
¯
¯
¯
¯*≤ kf kh*1*kgkbmo* *.* *(0.12b)*
*The space h*1 * _{is again a localized version of the Hardy space H}*1

_{. It can}

be described by giving the norm of the space, [9] , [10]. Consider first a
*smooth function η(x) supported in the unit ball of Rn* _{with the property}

R

*Rnη(x)dx = 1. Given an integrable function f , form the following average*

*M²[f ](x*0) := 1
*²n*
¯
¯
¯
¯
¯
¯
Z
*Rn*
*η*
µ
*x*0*− x*
*²*
¶
*f (x)dx*
¯
¯
¯
¯
¯
¯
and set
*f∗(x*0) := sup
*0<²<∞M²[f ](x*0*) .*

The usual Hardy space is defined by

*H*1 :=©*f ∈ L*1*(Rn) and f∗* *∈ L*1*(Rn*)ª
*Now form, for a function f ∈ L*1

*loc(Rn*)

*f∗∗ _{(x}*

0) := sup
*0<²<1*

*M²[f ](x*0*) .*

The local version of the Hardy space, introduced by Goldberg [6], is
*h*1 :=©*f ∈ L*1*(Rn) and f∗∗* *∈ L*1*(Rn*)ª *.*

*For a function f ∈ h*1_{(R}n_{) we set, see [9] , [10]}

*kf kh*1_{(R}n_{)} *:= kf∗∗k _{L}*1

_{(R}n_{)}

*.*

*There is a natural estimate for the operator E(f ) defined in (0.7). It*
reads

*kE(f )k _{L}*6

*3*

_{(R}_{)}

*≤ Ck ˆf k*2

_{L}

_{(dµ)}*.*

*(0.14a)*

*The right hand side of the above is the L*2_{norm defined via the Leray measure}

*on the Hyperbola dµ := dξ/ω(ξ). To be specific*

*k ˆf kL*2* _{(dµ)}* :=
Z

*R*2

*| ˆf (ξ)|*2

*dξ*

*ω(ξ)*

*1/2*

*.*

*(0.14b)*

This is essentially the same as estimate (0.5) and it is the estimate derived by
Strichartz. It is optimal as stated, however there is a smoothing phenomenon.
*One quarter of the derivative applied to E*2_{(f ) is bounded in an appropriate}

sense by the same right hand side. The theorem can be stated as follows.
*Theorem 0.2 : Consider the operator E(f )(t, x) defined in (0.7) and*
*the square expressions*

Φ*i(t, x) := E(f )E(g)* and Φ*o(t, x) := E(f ) ¯E(g) .* *(0.15a)*

*The following mixed space-time estimate holds*
Z
*R*
Z
*R*2
*|D1/4*Φ*i,o|*2*dx*
*4/3*
*dt*
*3/8*
*≤ Ck ˆf kL*2* _{(dµ)}kˆgk_{L}*2

_{(dµ)}*,*

*(0.15b)*

*where D1/4* _{denotes one quarter of the derivative taken via the Fourier }

*trans-form.*

*A final theorem is concerned with products E(f )E(g) or E(f ) ¯E(g) when*
*the smoothness of f and g is not the same, this is a refinement of Theorem*
0.1.

*Theorem 0.3 : Assume that φ and ψ are solutions of the equations*
*φ + m*2*φ = 0* *ψ + m*2*ψ = 0*

*with initial data*

*φ(0, x) := 0 φ,t(0, x) := f (x)* *,* *ψ(0, x) := 0 ψ,t(0, x) := g(x) .*
*Then*
°
°
*°(||τ | − |ξ|| + m)1/2φ ∗ ˆ*ˆ *ψ(τ, ξ)*
°
°
°
*L*2* _{(R}*3

_{)}

*≤ Ck ˆf /ωkL*2

*kˆg/kL*2

*,*

*(0.16a)*

*where ω :=*p*|ξ|*2* _{+ m}*2

_{. Moreover the quantity defined bellow}*²(t) :=*
Z
*R*2
*|φψ|*2*dx*
*1/2*
*(0.16b)*

*satisfies the estimate*

*²(t) ≤ b(t) ,* *(0.16c)*
*where*
*kbkbmo* *≤ C*
°
°
° ˆ*f /ω*
°
°
°
*L*2*kˆgkL*
2 *.* *(0.16d)*

*Notice that in the estimate above φ can only satisfy the estimate*
sup

*t* *kφ(t, ·)kL*

2* _{(R}*2

_{)}

*≤ k ˆf /ωk*2

_{L}*2*

_{(R}_{)}

*while ψ for each fixed time belongs to H*1_{. In what follows we will use ˆ to}

denote the Fourier transform of a function over the space-time variables. We will use ˜ to denote the partial Fourier transform over the space variables and we will always denote

Chapter 2, Demonstration :

We will start by considering the integral operator defined in (0.7),
name-ly
*E(f )(t, x) :=*
Z
*R*2
*exp[i(t*p*|ξ|*2* _{+ m}*2

_{+ x · ξ)] ˆ}_{f (ξ)}_{p}

*dξ*

*|ξ|*2

*2*

_{+ m}*.*

*(1.1a)*

In what follows we will denote

*ω(ξ) :=*p*|ξ|*2* _{+ m}*2

_{.}

_{(1.1b)}Notice that the Fourier transform of the expression above is ˆ

*E(f )(τ, ξ) = δ(τ*2*− |ξ|*2*− m*2*)H(τ ) ˆf (ξ)dµ(ξ)*

where

*dµ(ξ) := dξ/ω(ξ) = dξ/τ*

*is the Leray measure on the hyperbola defined by {τ = ω(ξ)}. The leray*
*measure can be explained as follows. Consider the function S := τ*2 * _{− |ξ|}*2

and write

*dτ dξ = dµSdS*

*the measure dµS* *is a natural object defined on the surface {S = s}. Consider*

the following quadratic expressions ˆ

Φ*i(τ, ζ) := ˆE(f ) ∗ ˆE(g)* *(1.2a)*

ˆ

Φ*o(τ, ζ) := ˆE(f ) ∗ ¯ˆE(g) .* *(1.2b)*

Notice that these are the Fourier transforms of the quantities

respectively. The idea of the estimate in Theorem 0.1 is very simple. It is based on the observation

°
°
*°(||τ | − |ζ|| + m)1/2*Φˆ*i,o(τ, ζ)*
°
°
°
*L*2* _{(R}*3

_{)}

*≤ Ck ˆf kL*2

*(R*2)

*kˆgkL*2

*(R*2)

*.*

*(A)*

To make this more precise and to prepare for Theorems 0.2,3 we need the following technical lemma.

*Lemma 1.1 : With the above notation we have the estimates*

*(|ζ| + |τ |) ˆ*Φ*i,o(τ, ζ) ∼ ˆKi,o(τ, ζ) ˆDi,o(τ, ζ)* *(1.3)*

*where*
ˆ
*Ki(τ, ζ) :=*
*τ1/2 _{H(τ}*2

*2*

_{− |ζ|}*2*

_{− 4m}

_{)H(τ )}*[(τ*2

*2*

_{− |ζ|}_{)}2

*2*

_{+ 4m}*2*

_{|ζ|}_{]}14

*(1.4a)*ˆ

*Ko(τ, ζ) :=*

*|ζ|*1 2

*(|ζ|*2

*− τ*2)14

*H(|ζ|*2

*− τ*2)

*{[(|ζ|*2

*2*

_{− τ}_{)}2

*2*

_{+ 4m}*2*

_{|ζ|}*2*

_{] (|ζ|}*2*

_{− τ}*2*

_{+ 4m}*14*

_{)}}*(1.4b)*

*and the quantities ˆDi,o* *satisfy the bounds*

*k ˆDi,okL*2* _{(R}*3

_{)}

*≤ Ckf k*2

_{L}*2*

_{(R}_{)}

*kgk*2

_{L}*2*

_{(R}_{)}

*,*

*(1.5)*

*where H(s) denotes the Heaviside function.*
Proof :

The proof is based on an explicit computation of the quantities ˆΦ*i,o* .

Let us compute first the quantity ˆΦ*o*

A) Computation of ˆΦ*o(τ, ζ) :*
*E(f ) ¯E(g)(t, x) =*
Z
*R*4
*exp(iΩ) ˆf (ξ)¯ˆg(−η)* *dξdη*
*ω(ξ)ω(η)* *(1.6a)*
where
*Ω := t(ω(ξ) − ω(η)) + x · (ξ + η) .* *(1.6b)*

*Call τ := ω(ξ) − ω(η) and set ζ := ξ + η and X := ξ − η. If we write*
*X = (X*1*, X*2*) where X*1 *is in the direction of ζ then X satisfies the equation*

*|ζ|*2 * _{− τ}*2

*|ζ|*2

*2*

_{− τ}*2*

_{+ 4m}*X*2 1

*τ*2

*−*

*X*2 2

*|ζ|*2

*2*

_{− τ}*2*

_{+ 4m}*= 1 .*

*(1.7a)*

*Notice that |ζ|*2* _{− τ}*2

_{≥ 0. Introducing natural coordinates on the hyperbola}defined by (1.7a)
*X*1 *:= a cosh α* *a :=*
*τ*p*|ζ|*2* _{− τ}*2

*2 p*

_{+ 4m}*|ζ|*2

*2*

_{− τ}*(1.7b)*

*X*2

*:= b sinh α*

*b :=*p

*|ζ|*2

*2*

_{− τ}*2*

_{+ 4m}

_{(1.7c)}*we can compute the Jacobian of the transformation dX*1*dX*2 *= Jdτ dα*

*J(τ, ζ, α) =* *|ζ|*

2* _{(|ζ|}*2

*2*

_{− τ}*2*

_{+ 4m}_{) cosh}2

*2*

_{α − τ}*2*

_{(|ζ|}*2*

_{− τ}_{)}

*(|ζ|*2* _{− τ}*2

_{)}32

*.* *(1.8a)*

On the other hand a straightforward calculation gives

*4ω(ξ)ω(η) =* *|ζ|*

2* _{(|ζ|}*2

*2*

_{− τ}*2*

_{+ 4m}_{) cosh}2

*2*

_{α − τ}*2*

_{(|ζ|}*2*

_{− τ}_{)}

*(|ζ|*2* _{− τ}*2

_{)}

*.*

*(1.8b)*

Combining (1.8a,b) gives the formula
ˆ
Φ*o(τ, ζ) =* p 1
*|ζ|*2* _{− τ}*2
Z

*R*ˆ

*f*µ

*ζ + X(α)*2 ¶

*¯ˆg*µ

*X(α) − ζ*2 ¶

*dα*

*(1.9a)*

*where the vector X(α) can be expressed in a simple manner using complex*
notation

*ζ := |ζ|eiφ* *and X(α) = eiφ[a(τ, ζ) cosh α + ib(τ, ζ) sinh α]* *(1.9b)*

*where a, b are defined in (1.7b,c) . We will need formula (1.9) later. Now the*
Lemma follows from the estimate

1
*|ζ|*2* _{− τ}*2
Z

*R*

*dα*

*J*

*∼*

*H(|ζ|*2

*2*

_{− τ}*2*

_{)(|ζ|}*2*

_{− τ}_{)}

*|ζ| [(|ζ|*2

*2*

_{− τ}*2*

_{+ 4m}*2*

_{) ((|ζ|}*2*

_{− τ}_{)}2

*2*

_{+ 4m}*2*

_{|ζ|}_{)]}

*1/2*

and the observation Z

*dτ dαdζJ| ˆf |*2* _{|ˆ}_{g|}*2

_{= k ˆ}_{f k}*L*2*kˆgk _{L}*2

*.*

B) Computation of ˆΦ*i(τ, ζ) :*

*The computation of E(f )E(g)(t, x) is similar. Let us mention what is*
*different. In this case we set τ := ω(ξ) + ω(η) , ζ := ξ + η and X := ξ − η ,*
*then dξdη = cdζdX and X = (X*1*, X*2*) with X*1 *in the direction of ζ satisfies*

the equation
*τ*2* _{− |ζ|}*2

*τ*2

*2*

_{− |ζ|}*2*

_{− 4m}*X*2 1

*τ*2 +

*X*2 2

*τ*2

*2*

_{− |ζ|}*2*

_{− 4m}*= 1 .*

*(1.10a)*

*Notice that τ*2* _{− |ζ|}*2

*2*

_{− 4m}

_{≥ 0. Choosing natural coordinates on the ellipse}defined by (1.10), namely
*X*1 *:= a cos α* *,* *a :=* *τ*
p
*τ*2* _{− |ζ|}*2

*2 p*

_{− 4m}*τ*2

*2*

_{− |ζ|}*(1.10b)*

*X*2

*:= b sin α*

*,*

*b :=*p

*τ*2

*2*

_{− |ζ|}*2*

_{− 4m}

_{(1.10c)}*we have for the Jacobian of the transformation dX*1*dX*2 *= Jdτ dα*

*J(τ, ζ, α) =* *τ*
2* _{(τ}*2

*2*

_{− |ζ|}*2¡*

_{) − |ζ|}*2*

_{τ}*2*

_{− |ζ|}*2¢*

_{− 4m}_{cos}2

_{α}*(τ*2

*2*

_{− |ζ|}_{)}

*3/2*

*,*

*(1.11a)*while

*4ω(ξ)ω(η) =*

*τ*2

*2*

_{(τ}*2*

_{− |ζ|}*2¡*

_{) − |ζ|}*2*

_{τ}*2*

_{− |ζ|}*2¢*

_{− 4m}_{cos}2

_{α}*τ*2

*2*

_{− |ζ|}*.*

*(1.11b)*

Finally we have the formula
ˆ
Φ*i(τ, ζ) =* p 1
*τ*2* _{− |ζ|}*2
Z

*S*1 ˆ

*f*µ

*ζ + X(α)*2 ¶ ˆ

*g*µ

*ζ − X(α)*2 ¶

*dα ,*

*(1.12a)*

*where again X(α) can be expressed using complex notation*

*and a, b are defined in (1.10b,c) . We will use formula (1.12) later. Again*
the Lemma follows from the estimate

1
p
*τ*2* _{− |ζ|}*2
Z

*S*1

*dα*

*J*

*∼*

*H(τ*2

*2*

_{− |ζ|}*2*

_{− 4m}_{)}p

*2*

_{τ}*2*

_{− |ζ|}*τ [(τ*2

*2*

_{− |ζ|}_{)}2

*2*

_{+ 4m}*2*

_{|ζ|}_{]}

*1/2*

and the computation Z

*dτ dαdζJ| ˆf |*2*|ˆg|*2 *= k ˆf kL*2*kˆgk _{L}*2

*.*

This completes the proof of the Lemma.

Remark : Notice that the previous computations can be combined in
*the case of the function φ(t, x) the solution of (0.1) as follows*

*φ*2*(t, x) =* * _{1/2}*1

*F (t, x)*

*where F ∈ H1/2 _{(R}*3

_{) is given via the Fourier transform as the even function}

defined by
ˆ
*F (τ, ζ) = i*
Z
*S*1
ˆ
*f*
µ
*ζ + X(α)*
2
¶
ˆ
*g*
µ
*ζ − X(α)*
2
¶
*dα if* *τ*2*− |ζ|*2 *> 0*
and
ˆ
*F (τ, ζ) =*
Z
*R*
ˆ
*f*
µ
*ζ + X(α)*
2
¶
ˆ
*g*
µ
*X(α) − ζ*
2
¶
*dα if* *|ζ|*2*− τ*2 *> 0 ,*
while *1/2* _{is defined by}
ˆ*1/2* :=
½ p
*|ζ|*2* _{− τ}*2

*2*

_{if |ζ|}*2*

_{− τ}

_{> 0}*i*p

*τ*2

*2*

_{− |ζ|}*2*

_{if τ}*2*

_{− |ζ|}

_{> 0}so that *1/2 1/2* _{=} _{. One can try to apply the energy estimate for the}

equation

*φ*2 = *1/2F*

Notice that ˆΦ*i* *is supported in the region {τ*2 *− |ζ|*2*− 4m*2 *≥ 0} while*

ˆ

Φ*o* *is supported in the region {|ζ|*2*− τ*2 *≥ 0}. Combining both cases we can*

write

*| ˆΦ(τ, ζ)| ≤ ˆQ(τ, ζ) ˆD(τ, ζ)* *(1.13a)*
where

*k ˆDkL*2 *≤ Ckf k _{L}*2

*kgk*2

_{L}*(1.13b)*

and ˆ*Q can be dominated using (1.4a,b), namely ˆQ can be chosen to be*
ˆ
*Q(τ, ζ) =* 1
*[(τ − |ζ|)*2* _{+ 4m}*2

_{]}

*1/4*+ 1

*[(τ + |ζ|)*2

*2*

_{+ 4m}_{]}

*1/4*

*.*

*(1.13c)*Call ˜

*Q(t, ζ) :=*Z

*R*

*exp(itτ ) ˆQ(τ, ζ)dτ*

*(1.13d)*the partial Fourier transform with respect to the dual time variable. Then a simple change of coordinates gives

˜
*Q(t, ζ) = 2 cos(t|ζ|)*
Z
*R*
*exp(itτ )dτ*
*[τ*2* _{+ 4m}*2

_{]}

*1/4*

*.*

*(1.13e)*

We want to isolate the singular part in the above expression, for this reason
call
*k(t) :=*
Z
*R*
*exp(itτ )dτ*
*[τ*2* _{+ 4m}*2

_{]}

*1/4*

*.*

*(1.14)*

*Now the function k(t) has the following properties stated in the lemma below.*
*Lemma 1.2 : The function k(t) defined in (1.14) satisfies the following*
*properties*

*|k(t)| ≤* *C*

*|t|1/2* if *|t| < 1* *(1.15a)*

*|k(t)| ≤* *C*

Proof :

To prove (1.15b) notice that by integration by parts we have
Z
*exp(itτ )dτ*
*[τ*2* _{+ 4m}*2

_{]}

*1/4*= 1

*2it*Z

*exp(itτ )τ dτ*

*[τ*2

*2*

_{+ 4m}_{]}

*5/4*and

_{Z}

*∞*0

*τ dτ*

*[τ*2

*2*

_{+ 4m}_{]}

*5/4*

*∼*

*C*

*√*

*m*

*In order to prove (1.15a) it is enough to notice that if |τ | is large then*
1

*[τ*2* _{+ 4m}*2

_{]}

*1/4*

*∼*

1
*|τ |1/2*

and this completes the proof.

We need one more technical lemma.

*Lemma 1.3 : Assume that k(t) ≥ 0 satisfies the conditions (1.15a,b)*
*from Lemma 1.2. Consider the operator defined by*

*I(f )(t) :=*
Z

*R*

*k(t − s)f (s)ds* *(1.16a)*

*where f (s) has compact support in the interval [−T, T ]. Then*

*kI(f )kL*2_{(R)}*≤ Ckf k _{h}*1

*(1.16b)*Proof : Compute Z

*R*

*I*2

*(f )(t)dt =*Z

*R*

*dt*Z

*[−T,T ]×[−T,T ]*

*k(t − s*1

*)k(t − s*2

*)f (s*1

*)f (s*2

*)ds*1

*ds*2 = Z

*[−T,T ]×[−T,T ]*

*h(s*1

*, s*2

*)f (s*1

*)f (s*2

*)ds*1

*ds*2 where

*h(s*1

*, s*2) := Z

*R*

*k(t − s*1

*)k(t − s*2

*)dt*

It is easy to see that because of (1.15a,b) we have

*h(s*1*, s*2*) ∼ | log |s*1*− s*2*||* if *|s*1*− s*2*| < 1*

*while h(s*1*, s*2*) is bounded if |s*1*− s*2*| ≥ 1. Hence for fixed s*1 we have

*kh(s*1*, .)kbmo* *≤ C*
*independent of s*1. Now
Z
*I*2*(f )(t)dt ≤ kf kL*1sup
*s*1
*kh(s*1*, .)kbmokf kh*1 *.*

This proves the Lemma.

Now we are in the position to prove
*Theorem 1.1 : Call*
*²(t) :=*
Z
*R*2
¯
*¯Dt,x*
¡
*E(f ) ¯E(g)*¢¯¯2*dx*
*1/2*
*(1.17a)*
*or*
*²(t) :=*
Z
*R*2
*|Dt,x(E(f )E(g))|*2*dx*
*1/2*
*(1.17b)*

*then there is a positive function b(t) defined almost everywhere such that*

*²(t) ≤ b(t)* *(1.18a)*

*and*

*kbkbmo* *≤ Ckf kL*2*kgk _{L}*2

*.*

*(1.18b)*

Proof :

Because of Lemma 1.1 the Fourier transform of both quantities

*Dt,x(E(f )E(g))* *and Dt,x*

¡

can be written as ˆ*G which satisfies*

*| ˆG(τ, ζ)| ≤ ˆQ(τ, ζ) ˆD(τ, ζ)*
where ˆ*D and ˜Q satisfy*

*k ˆDkL*2 *≤ Ckf k _{L}*2

*kgk*2

_{L}and

˜

*Q(t, ζ) = 2 cos(t|ζ|)k(t)*

*with k(t) defined in (1.14). Using tilde to denote the partial Fourier *
trans-form with respect to the dual time variable we have

˜

*G(t, ζ) =*
Z

*R*

*2 cos ((t − s)|ζ|) k(t − s) ˜D(s, ζ)ds*

and from this it follows that
°
°
° ˜*G(t, ·)*
°
°
°
*L*2 *≤*
Z
*R*
*k(t − s)*
°
°
° ˜*D(s, ·)*
°
°
°
*L*2*ds := b(t)*

*We want to use a duality argument. Pick an arbitrary function l(t) with*
*support in the time interval t ∈ [−T, T ] and compute*

*| (l, b) | ≤*
Z
*k(t − s)k ˜D(s, ·)kL*2*l(t, )dtds*
*≤ kDkL*2
°
°
°
°
Z
*k(t − s)l(t, )dt*
°
°
°
°
*L*2
*≤ kDkL*2*Ckl(·)k _{h}*1

*since k(t) satisfies (1.15a,b) in Lemma 1.2 . Hence b defines a bounded*
*functional on h*1 _{and}

*kb(·)k _{bmo}*

*≤ CkDkL*2

*≤ Ckf k*2

_{L}*kgk*2

_{L}Now we want to give the proof of the estimates in Theorems (0.2,3). Let us state the basic estimate from which all the others follow.

*Theorem 1.2 : Consider the integral operator of the type*
*E(f )(t, x) :=*
Z
*R*2
*exp{i(tω(ξ) + x · ξ)} ˆf (ξ)dµ(ξ)* *(1.19a)*
*ω(ξ) =*p*|ξ|*2* _{+ m}*2

_{with dµ(ξ) :=}*dξ*

*ω(ξ)*

*(1.19b)*

*Denote by ˆ*Φ*i,o* *the Fourier transform of E(f )E(g) and E(f ) ¯E(g) *

*respec-tively. The following estimates hold :*
°
°
*°|τ*2*− |ζ|*2*|1/4*Φˆ*i,o(τ, ζ)*
°
°
°
*L*2* _{(R}*3

_{)}

*≤ Ck ˆf (ξ)kL*2

*(dµ)kˆg(ξ)kL*2

*(dµ)*

*(1.20a)*

*and*° °

*°(||τ | − |ζ||)1/2*Φˆ

*i,o(τ, ζ)*° ° °

*L*2

*≤ C*° ° ° ˆ

*f /ω*° ° °

*L*2

*2*

_{(R}_{)}

*kˆgkL*2

*2*

_{(R}_{)}

*.*

*(1.20b)*Proof :

Consider first the quantity ˆΦ*i* which is the integral given in (1.12a),

recall that ˆΦ*i* *is supported in the region τ ≥ 0 and τ*2*− |ζ|*2 *≥ 4m*2 . Both

estimates follow from the same computation so we will prove only (1.20b). Set

*ξ1,2*:= *ζ ± X(α)*

2 *.*

Cauchy-Schwartz inequality gives
*(τ − |ζ|)*
¯
¯
¯ ˆΦ*i*
¯
¯
¯2 *≤*
_{p}*τ − |ζ|*
*τ*2* _{− |ζ|}*2
Z

*S*1

*ω(ξ*1)

*ω(ξ*2)

*dα* Z

*S*1

*| ˆf (ξ*1

*)|*2

*|ˆg(ξ*2

*)|*2

*ω(ξ*2)

*ω(ξ*1)

*dα*p

*τ*2

*2*

_{− |ζ|}*.*

Observe first that, see (1.11a,b)
*dτ dαdζ*
p

*τ*2* _{− |ζ|}*2 =

*dξ*1*dξ*2

*and the estimate will follow after the inequality with respect to τ , ζ, provided*
that the quantity

*τ − |ζ|*
p
*τ*2* _{− |ζ|}*2
Z

*S*1

*ω(ξ*1)

*ω(ξ*2)

*dα*

*is bounded for every τ , ζ. It is straightforward to calculate*

*ω(ξ1,2) = τ ± q|ζ| cos α* with *q :=*
s
*1 −* *4m*2
*τ*2* _{− |ζ|}*2

*.*

*Call λ := τ /|ζ|q. We have*Z

*S*1

*ω(ξ*1)

*ω(ξ*2)

*dα =*Z

*S*1

*λ + cos α*

*λ − cos αdα ∼*1

*√*

*λ*2

_{− 1}and the quantity

*τ − |ζ|*
p

*(τ*2* _{− |ζ|}*2

*2*

_{)(λ}

_{− 1)}*is bounded for every τ and ζ.*

In order to prove the estimates for the term ˆΦ*o* , see (1.9a), we need a

new idea. Consider a dyadic decomposition of the functions ˆ*f and ˆg. This*
*can be achieved as follows, we start with a smooth function ˆa supported in*
*the set {1/2 ≤ |ξ| ≤ 1} and form the dyadic dilations ˆaj(ξ) := ˆa(ξ/2j*). It is

*easy to see that ˆa can be chosen so that*

*+∞*
X
*j=−∞*
*ˆaj(ξ) = 1* if *ξ 6= 0 .*
Now set
*ˆa*0*(ξ) := 1 −*
*+∞*
X
*j=1*
*ˆaj(ξ)*
and define
ˆ
*fj(ξ) := ˆaj(ξ) ˆf (ξ)* where *j = 0, 1, · · · + ∞ .*

*Denote the dyadic components fj* *and gk* and consider the coresponding

expression ˆΦ*o,jk*. The proof (1.20a) will follow from the estimate

°
°
*°(|ζ|*2*− τ*2)*1/4*Φˆ*o,jk*
°
°
°
*L*2 *≤*
*C*
2*|j−k|/4k ˆfjkL*2*(dµ)kˆgkkL*2*(dµ)* *.* *(1.21)*
Observe that
*ω(ξ1,2) = |ζ|q cosh α ± τ* with *q :=*
s
1 + *4m*2
*|ζ|*2* _{− τ}*2

and, see (1.8a,b),

*dτ dαdζ*
p

*|ζ|*2* _{− τ}*2 =

*dξ*1*dξ*2

*ω(ξ*1*)ω(ξ*2) *.*

*Case A : Assume that j ∼ k and both j and k large, otherwise there*
is nothing to prove. In this case

*ω(ξ*1)
*ω(ξ*2) =
*cosh α + λ*
*cosh α − λ* *∼ 1 with λ :=*
*τ*
*|ζ|q* *< 1 .*

*It is easy to see that for any λ the variable α has to belong to a bounded*
*interval. The most interesting case is when λ is close to one, in which case*
*α has to be large and belong to an interval of the form*

*eα* *∈*
·
2*j*
*|ζ|q,*
2*j+1*
*|ζ|q*
¸
*.*
Hence using Cauchy-Schwartz again

¯
¯
¯ ˆΦ*o,jk*
¯
¯
¯2 *≤*
µZ
*dα*
¶ Z
*| ˆf (ξ*1*)|*2*|ˆg(ξ*2*)|*2
*dα*
*|ζ|*2* _{− τ}*2

*.*

Multiplying the above inequality with p*|ζ|*2* _{− τ}*2

_{and integrating with }

*re-spect to τ and ζ gives (1.21).*

*Case B : Assume k << j. In this case |ζ| ∼ 2j* _{and τ ∼ 2}j_{. Without}

*loss of generality we can assume that τ > 0. Now*
*ω(ξ*1)
*ω(ξ*2)
= *cosh α + λ*
*cosh α − λ* *∼*
2*j*
2*k* *>> 1 ,*

*where λ := τ /|ζ|q ∼ 1. For this to be satisfied α must be close to zero and*
in particular
*α*2 *≤* 2
*k*
2*j* hence
Z
*dα ∼* 2
*k/2*
2*j/2* *.*

Now we have after using Cauchy-Schwartz again
¯
¯
¯ ˆΦ*o,jk*
¯
¯
¯2 *≤* 2
*k/2*
2*j/2*
Z
*| ˆf (ξ*1*)|*2*|ˆg(ξ*2*)|*2 *dα*
*|ζ|*2* _{− τ}*2

*,*

*from which the estimate follows after integration in τ and ζ.*
Finally we want to prove the estimate

°
°
*°(|ζ| − |τ |)1/2*Φˆ*o*
°
°
°
*L*2 *≤ Ck ˆf /ωkL*2*kˆgkL*2 *.*
Write
ˆ
Φ*o* =
X
*j,k*
ˆ
Φ*o,jk*=
X
*|j−k|>2*
ˆ
Φ*o,jk*+
X
*|j−k|≤2*
ˆ
Φ*o,jk* = ˆΦ1*o*+
X
*|j−k|≤2*
ˆ
Φ*o,jk* *. (1.22)*

The first term in the above expression can be estimated as follows. Notice
first that
ˆ
Φ1* _{o}* = p 1

*|ζ|*2

*2 Z ˆ*

_{− τ}*f*µ

*ζ + X(α)*2 ¶

*¯ˆg*µ

*X(α) − ζ*2 ¶

*χA(τ, ζ, α)dα ,*

*where χA(τ, α, ζ) is the characteristic function of a set contained in*

*A ⊂*
½
*(τ, α, ζ)* : *ω(ξ*1)
*ω(ξ*2)
*≤* 1
2 or
*ω(ξ*1)
*ω(ξ*2)
*≥ 2*
¾
*.*
The relation
*ω(ξ*1)
*ω(ξ*2)
= *cosh α + λ*
*cosh α − λ* with *λ :=*
*τ*
*|ζ|q* *< 1*

*implies that for every fixed (τ, ζ) the range of possible values of α belong to*
*a bounded set i.e {α ∈ R : |α| ≤ C}. Using Cauchy-Schwartz one more time*
we have
*(|ζ| − |τ |)*
¯
¯
¯ ˆΦ1* _{o}*
¯
¯
¯2

*≤* p

*|ζ| − |τ |*

*|ζ|*2

*2 Z*

_{− τ}*|α|<C*

*ω(ξ*1)

*ω(ξ*2)

*dα* Z

*| ˆf (ξ*1

*)|*2

*|ˆg(ξ*2

*)|*2

*ω(ξ*2)

*ω(ξ*1)

*dα*p

*|ζ|*2

*2*

_{− τ}*.*

¿From the observation
Z
*|α|<C*
*ω(ξ*1)
*ω(ξ*2)*dα =*
Z
*|α|<C*
*cosh α + λ*
*cosh α − λdα ∼*
1
*√*
*1 − λ*2 *where λ :=*
*τ*
*|ζ|q* *< 1*

*we have after integration with respect to τ, ζ the estimate*
°
°
*°(|ζ| − |τ |)1/2*Φˆ1
*o*
°
°
°
*L*2 *≤ Ck ˆf /ωkL*2*kˆgkL*2 *.*

To handle the last term in the expansion (1.22) observe that
*(|ζ| − |τ |)*
¯
¯
¯ ˆΦ*o,jk*
¯
¯
¯2 *≤* p*|ζ| − |τ |*
*|ζ|*2* _{− τ}*2
µZ

*dα*¶ Z

*| ˆfj|*2

*|ˆgk|*2

*dα*p

*|ζ|*2

*2*

_{− τ}*.*

*Since |j−k| ≤ 2 the variable α must belong to a bounded interval. Integrating*
*with respect to τ and ζ we obtain*

°
°
°p*|ζ| − |τ | ˆ*Φ*o,jk*
°
°
°
*L*2 *≤ Ck ˆfj/ωkL*2*kˆgkkL*2 *.*

*Summing over j, k and taking into account the fact that j ∼ k gives the*
desired estimate.

The estimate of Strichartz follows from the multiplier estimate
*kf kL*3* _{(R}*3

_{)}

*≤ k*

¯

*¯τ*2* _{− |ζ|}*2¯

_{¯}

*1/4*

_{f (τ, ζ)k}_{ˆ}

*L*2

*3*

_{(R}_{)}

see [3].

Remark : A more careful analysis taking into account the fact that ˆΦ*i*

*is supported in the region where τ*2 * _{− |ζ|}*2

*2*

_{− 4m}

_{≥ 0 and the observation}that ˆΦ*o* *is the convolution over a hyperbola that vanishes when |ζ| = |τ | can*

prove the following slight improvement.
°
°
*°(||ζ| − |τ || + m)1/2*Φˆ*i,o*
°
°
*° ≤ Ck ˆf /ωkL*2*kˆgk _{L}*2

*.*

*(1.23)*

Now we are in the position to prove Theorem (0.2). Proof of Theorem 0.2 :

Observe that from Theorem 1.2 we have
*(|τ | + |ζ|)1/4*Φˆ*i,o* *∼*
ˆ
*Di,o*
*||τ | − |ζ||1/4* *,*
where
*k ˆDi,okL*2 *≤ Ck ˆf k _{L}*2

*2*

_{(dµ)}kˆgk_{L}

_{(dµ)}*.*

Denote by ˜ the partial Fourier transform with respect to the time variable and observe that

Z
*R*
*eiτ t _{dτ}*

*||τ | − |ζ||1/4*

*∼*

*eit|ζ|*

*|t|3/4*for

*|t|*

*small .*

*Consider an arbitrary function l(t, x). A duality argument gives*
¯
¯
¯
³
*ˆl, D1/4*_{Φ}_{ˆ}
*i,o*
´¯_{¯}
*¯ ≤*
Z
*R×R*
1
*|t − s|3/4k˜l(t, ·)kL*2
°
°
° e*Di,o(s, ·)*
°
°
°
*L*2*dtds*
*≤ kDi,okL*2
°
°
°
°
°
°
Z
*R*
*kl(t, ·)kL*2
*|t − s|3/4* *dt*
°
°
°
°
°
°
*L*2_{(R)}*≤ CkDi,okL*2
°
*°kl(t, ·)k _{L}*2

*2*

_{(R}_{)}° °

*L8/3*

_{(R)}*,*

where we used the Littlewood Hardy Polya inequality in the last step. This concludes the proof.

Proof of Theorem 0.3 :

We will give only an outline. The proof of Theorem 0.3 follows exactly the same steps as in the proof of Theorem 1.1, starting from the observation that because of (1.23) we have

ˆ
Φ*i,o* *∼*
1
*(||τ | − |ζ|| + m)1/2*
ˆ
*Di,o* *,*
where
°
°
° ˆ*Di,o*
°
°
°
*L*2* _{(R}*3

_{)}

*≤ C*° ° ° ˆ

*f /ω*° ° °

*L*2

*2*

_{(R}_{)}

*kˆgkL*2

*(R*2)

*.*.

REFERENCES

[1] Barcelo-Taberner, B. : On the Restriction of Fourier Transform to a Conical Surface. Trans. Amer. Math. Soc. 292 321-333 (1985).

[2] Grillakis, M.G. : Apriori estimates and Regularity of Nonlinear Waves. Proceedings of ICM 1187-1194 (1994)

[3] Harmse, J. : On Lebesque space Estimates for the Wave Equation. Ind. Univ. Math. J. 39 (1990), 229-248.

[4] Strichartz, S.R. : Restrictions of Fourier Transforms to Quadratic Surfaces and Decay of Solutions of Wave Equations. Duke Math. J. 44 No. 3 , 705-714 (1977).

[5] Stein, E.M. : Oscillatory integrals in Fourier Analysis. Beijing Lec-tures in Harmonic Analysis. Princeton U. Press (1986) 307-355.

[6] Goldberg, D. : A local version of real Hardy spaces. Duke Math. J. 46 (1979), 27-42.

[7] Torchinsky, A. : Real Variable Methods in Harmonic Analysis. Aca-demic Press, 1986.

[8] Klainerman, S. and Machedon, M. : Smoothing Estimates for Null Forms and Applications. Duke Math. J. 81 (1995) 99-133.

[9] Evans, L.C. and Muller, S. : Hardy Spaces and the Two-Dimensional Euler Equations with Nonnegative Vorticity. J. Amer. Math. Soc. 7 (1994) 199-219.

*[10] Fefferman, C. Stein, E.M. : Hp* _{Spaces of several variables. Acta}

Math. 129, 137-193.

[11] Kapitanski, L. V. : Global and Unique Weak Solutions of Nonlinear Wave Equations. Math. Res. Lett. 1 (1994), no 2, 211-223.

[12] John, F. Nirenberg, L. : On functions of bounded mean oscillation. Comm. Pure Appl. Math. 14, 415-426 (1961).