A NULL FORM ESTIMATE FROM STRICHARTZ ESTIMATES

TERENCE TAO

Abstract. I show how a certain null form estimate can be derived from dyadic decomposition techniques and an improvement to the standard Strichartz estimate.

1. Introduction

The purpose of this note is to prove the null form estimate

∥∇^{−1}Q(∂ϕ, ∂ψ)∥_{2} ≲ ∥f∥Ḣ^{1}∥g∥Ḣ^{1}  (1)

proved in [1], where ϕ is the solution to the free wave equation with initial data f in R^{3+1}, and similarly for ψ and g, and Q is one of the null forms

Q_{ij}(∂ϕ, ∂ψ) = ϕ_{i}ψ_{j} − ϕ_{i} · ψ_{j}.

A key tool is, of course, Strichartz’ estimate:

∥ϕ∥_{4} ≲ ∥f∥Ḣ^{1/2}.

In particular, if f has unit frequency (i.e. f̂ is supported on the annulus {|ξ| ∼ 1}) then

∥ϕ∥_{4} ≲ ∥f∥_{2}.  (2)

However, this estimate can be improved when f had even smaller frequency support. In fact, we have

Lemma 1.1. [2] If s is a positive integer, Q is a cube in the annulus {|ξ| ∼ 1} of sidelength 2^{−s}, and the Fourier transform of f is supported in Q, then we can improve (2) to

∥ϕ∥_{4} ≲ 2^{−εs}∥f∥_{2}

for some constant ε > 0. (Actually we can take ε = 1/2).

Proof Let’s begin by reviewing the proof of (2). We can write

ϕ(t) = U(t)f,

where

{ \atop U(t)f}(ξ) = β(ξ)e^{±t|ξ|}f̂(ξ)

and β is a cutoff to the annulus {|ξ| ∼ 1}. Then from the decay estimate

∥U(t)f∥_{∞} ≲ (1 + |t|)^{−1}∥f∥_{1}

and the energy estimate

∥U(t)f∥_{2} ≲ ∥f∥_{2}

we get (2) by the usual arguments.

1991 Mathematics Subject Classification. 42B15.

Now what’s different in this lemma is that we can replace U(t) by Ũ(t) defined by

{ \atop Ũ(t)f}(ξ) = ψ_{Q}(ξ)e^{±t|ξ|}f̂(ξ)

where ψ_{Q} is a narrower cutoff function, just supported on Q rather than the whole annulus. This doesn’t affect the energy estimate, but the decay estimate improves! We have

∥Ũ(t)f∥_{∞} ≲ 2^{−εs}(1 + |t|)^{−1}∥f∥_{1}

for some ε > 0. A good way to think about this is via the uncertainty principle. What does the kernel of U(t) look like for t ≫ 2^{s}? Well, it’s basically supported on the annulus of radius t and thickness 1 around the origin. (It’s the fundamental solution to the wave equation blurred out by an uncertainty of 1). It has a sup norm of O(1/t), hence the usual decay estimate. But since the Fourier transform of Ũ(t) is more pinched, the kernel of Ũ(t) is further blurred out; by the uncertainty principle it has a spatial uncertainty of 2^{s}. This smears the kernel around over a wider annulus - one of thickness 2^{s} instead of 1 - and this causes the sup norm to be lower. This is why we get an improved decay estimate. Of course, one can then go back and make this rigorous with stationary phase.

Anyway, the point is that we have an improved decay estimate, which when we plug it into the usual machinery gives an improved Strichartz estimate.

We now show how (1) follows from the above Lemma and a bit of dyadic decomposition. We want to estimate

∥∇^{−1}Q(∂ϕ, ∂ψ)∥_{2}.

Break up the data f and g into frequency pieces f = ∑_{k} f_{k}, g = ∑_{l} g_{l}, where f_{k} and g_{l} have frequency support on |ξ| ∼ 2^{k} and |ξ| ∼ 2^{l} respectively. Similarly decompose ϕ = ∑_{k} ϕ_{k}, ψ = ∑_{l} ψ_{l}. Now we simply use the triangle inequality^{1}, and estimate the above by

∑_{k} ∑_{l} ∥∇^{−1}Q(∂ϕ_{k}, ∂ψ_{l})∥_{2}.

On ϕ_{k}, ∂ essentially behaves like 2^{k}, and similarly for ψ_{l}, so this becomes

∑_{k} ∑_{l} 2^{k+l}∥∇^{−1}Q(ϕ_{k}, ψ_{l})∥_{2}.  (3)

Now we divide up into three cases: k > l + 10, l > k + 10, and l k.

Let’s first consider the contribution when k > l + 10. In this case Q(ϕ_{k}, ψ_{l}) has frequency about 2^{k}, so ∇^{−1} works like 2^{−k}. This will be quite beneficial since k is large We can thus estimate this portion of (3) by

∑_{k,l:k>l+10} 2^{l}∥Q(ϕ_{k}, ψ_{l})∥_{2}.

Now we simply estimate this by Hölder, ignoring the null structure:

∑_{k,l:k>l+10} 2^{l}∥ϕ_{k}∥_{4}∥ψ_{l}∥_{4}.

1One can of course be a bit more refined than the triangle inequality, and try to use some orthogonality. However, this null form estimate is quite far away from endpoint null form estimates, and one does not need such subtlety here.

Then we do Strichartz:

∑_{k,l:k>l+10} 2^{l}∥f_{k}∥Ḣ^{1/2}∥g_{l}∥Ḣ^{1/2}.

Now we eventually want Ḣ^{1} norms instead of Ḣ^{1/2} norms, so we pay 2^{k/2} and 2^{l/2} to upgrade these norms:

∑_{k,l:k>l+10} 2^{−(k−l)/2}∥f_{k}∥Ḣ^{1}∥g_{l}∥Ḣ^{1}.

Now we write l = k − s:

∑_{s>10} 2^{−s/2} ∑_{k} ∥f_{k}∥Ḣ^{1}∥g_{k−s}∥Ḣ^{1}.

Cauchy-Schwarz:

∑_{s>10} 2^{−s/2}(∑_{k} ∥f_{k}∥2 ; Ḣ^{1})^{1/2}(∑_{k} ∥g_{k−s}∥2 ; Ḣ^{1})^{1/2}.

By Plancherel, this is just

∑_{s>10} 2^{−s/2}∥f∥Ḣ^{1}∥g∥Ḣ^{1}

and we’re done for this part, since the s sum is very convergent.

The contribution for l > k + 10 is similar, so now let’s look at when k = l (say):

∑_{k} 2^{2k}∥∇^{−1}Q(ϕ_{k}, ψ_{k})∥_{2}.

In this case Q(ϕ_{k}, ψ_{k}) has frequency ≲ 2^{k}. To handle this, we introduce Littlewood-Paley projections Δ_{j} to frequency 2^{j}, and estimate the above by the triangle inequality as

∑_{k} ∑_{s>0} 2^{2k}∥Δ_{k−s}∇^{−1}Q(ϕ_{k}, ψ_{k})∥_{2}.

Now the ∇^{−1} contributes a 2^{−k+s}, so we get

∑_{s>0} 2^{s} ∑_{k} 2^{k}∥Δ_{k−s}Q(ϕ_{k}, ψ_{k})∥_{2}.

Note now that the ∇^{−1} has hurt us instead of helped us, as we have a 2^{s} factor instead of the very nice 2^{−s/2} we had earlier. So we can’t just use naive Hölder and Strichartz any more. Instead, we must exploit the null structure of Q.

We have two high frequencies (both ∼ 2^{k}) multiplying to a low frequency ∼ 2^{k−s}. This is only possible if the two original frequencies were almost diametrically opposite. If f_{k} had spatial frequency ξ and g_{k} had spatial frequency η, with |ξ| ∼ |η| ∼ 2^{k}, then we only get a contribution if ξ = −η + O(2^{k−s}). But in this case, the extra cancellation in the null-form kicks in. Take Q_{ij}, for instance: it is related to the multiplier

ξ_{i}η_{j} − ξ_{j}η_{i}.

We could crudely estimate this by |ξ||η|, and this roughly corresponds to treating Q_{ij} as just an ordinary product with no null structure. But when ξ = −η+O(2^{k−s}) the null form multiplier is a lot smaller than |ξ||η|, in fact it is only O(2^{−s}) times^{2} as large! (it has size O(2^{2k−s}) instead of O(2^{2k})). So it’s more appropriate to think

Q(ϕ_{k}, ψ_{k}) ≈ 2^{−s}ϕ_{k}ψ_{k}

than

Q(ϕ_{k}, ψ_{k}) ≈ ϕ_{k}ψ_{k}.

One can make this heuristic more precise with the Fourier transform and Plancherel, especially since we are estimating everything in L^{2}.

After this considerations we reduce ourselves to estimating

∑_{s>0} ∑_{k} 2^{k}∥Δ_{k−s}(ϕ_{k}ψ_{k})∥_{2}.

The divergence in s has now been reduced from exponential to polynomial. To finish the estimate we need an epsilon more decay in s. Ideally we’d like something like

∥Δ_{k−s}(ϕ_{k}ψ_{k})∥_{2} ≲ 2^{−εs}2^{−k}∥f_{k}∥Ḣ^{1}∥g_{k}∥Ḣ^{1},

which is an epsilon better than the Hölder-Strichartz argument above. If one puts this estimate into the previous one and crank, you get ∥f∥Ḣ^{1}∥g∥Ḣ^{1} as desired.

To obtain this epsilon improvement we use Lemma 1.1.

Everything in sight is scale-invariant (which isn’t surprising since the original estimate is), so we can make k=0:

∥Δ_{−s}(ϕ_{0}ψ_{0})∥_{2} ≲ 2^{−εs}∥f_{0}∥Ḣ^{1}∥g_{0}∥Ḣ^{1}.  (4)

Now f_{0} and g_{0} currently have frequency support on the unit annulus |ξ| ∼ 1. Now partition this annulus into a whole bunch of 2^{−s}-cubes Q; we’ll see in a minute why this decomposition is so natural. (There are about 2^{3s} of these cubes, but the exact number turns out to be irrelevant). We break up f_{0} = ∑_{Q} f_{Q} and g_{0} = ∑_{Q} g_{Q} accordingly, and similarly ϕ_{0} = ∑_{Q} ϕ_{Q} and ψ_{0} = ∑_{Q} ψ_{Q}.

By the triangle inequality, the left-hand side of (4) is bounded by

∑_{Q,Q′} ∥Δ_{−s}(ϕ_{Q}ψ′_{Q})∥_{2}.

The point of this decomposition is that Δ_{−s}(ϕ_{Q}ψ_{Q′}) is zero for most Q and Q′! In fact, the only way that ϕ_{Q} and ψ_{Q′} are going to multiply together to form something of spatial frequency 2^{−s} is if Q and Q′ are opposite each other, basically if Q′ = −Q. So the double sum (with 2^{6s} summands!) simplifies to a single sum:

∑_{Q} ∥Δ_{−s}(ϕ_{Q}ψ_{−Q})∥_{2}.

The Δ_{−s} has done its job of pruning the summation, and now we throw it away:

∑_{Q} ∥ϕ_{Q}ψ_{−Q}∥_{2}.

Hölder:

∑_{Q} ∥ϕ_{Q}∥_{4}∥ψ_{−Q}∥_{4}

2If we have the Q_{0} null form, then we may improve this to O(2^{−2s}) in the (+-) case, but in the (++) case we have no cancellation, and in fact the above estimate fails. In the Q_{ij} case it is in fact possible to prove these estimates more directly using the pointwise estimate |ξ_{i}η_{j} − ξ_{j}η_{i}| ≲ |ξ + η| min(|ξ|, |η|). Thanks to Sigmund Selberg for pointing these facts out.

Improved Strichartz:

2^{−εs} ∑_{Q} ∥f_{Q}∥_{2}∥g_{−Q}∥_{2}

Cauchy-Schwarz:

2^{−εs}(∑_{Q} ∥f_{Q}∥_{2}^{2})^{1/2}(∑_{Q} ∥g_{−Q}∥_{2}^{2})^{1/2}

and then by Plancherel we get (4) (since we’re at frequency 1, the L^{2} and Ḣ^{1} norms are the same), as desired.

References

[1] S. Klainerman, M. Machedon, Remark on Strichartz-type inequalities. With appendices by Jean Bourgain and Daniel Tataru. Internat. Math. Res. Notices 5 (1996), 201–220.

[2] S. Klainerman, D. Tatarum On the optimal regularity for Yang-Mills equations in R^{4+1}, preprint.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

  E-mail address: tao@math.ucla.edu