LECTURE NOTES 3 FOR 254B

TERENCE TAO

1. Bochner-Riesz summation

We all know that given the Fourier transform

f̂(ξ) = ∫_{R_{n}} e^{−2πix·ξ}f(x) dx

of a function, one can reconstruct it via the Fourier inversion formula

f(x) = ∫_{R_{n}} e^{2πix·ξ}f̂(ξ) dξ,

at least if f is smooth. But if f is not smooth - for instance, if f is merely in L^{p} - it is not clear whether the Fourier inversion formula is true, or even make sense.

To make things more precise, let us define the partial Fourier integrals S_{R}f(x) by

S_{R}f(x) = ∫_{|ξ|≤R} e^{2πix·ξ}f̂(ξ) dξ,

and ask whether

S_{R}f → f in L^{p} norm

for arbitrary f ∈ L^{p}. There are other types of convergence, such as pointwise convergence, but they are harder to deal with. (Note that weak convergence is not a problem here).

The operators S_{R} are known as disc multipliers (because they are Fourier multipliers corresponding to the characteristic function of the disc |ξ| ≤ R), and the problem was known as the disc multiplier problem (at least, until the problem was solved).

The answer to the disc multiplier problem is completely known. In one dimension n = 1, one has convergence in L^{p} norm for all 1 < p < ∞. However in higher dimensions one only has convergence in L^{2}; for p ≠ 2 convergence fails! This phenomenon was a great shock at the time it was discovered (by Fefferman in the late 70s), and the reason has to do with Besicovitch sets, which we will return to later in this course.

To begin with, we first recall that L^{p} convergence questions are basically equivalent to an L^{p} norm estimate. (This is one reason why this type of convergence question is easier than other types).

Lemma 1.1 (Uniform boundedness principle). Let 1 < p < ∞, and suppose that T_{R} is a sequence of linear operators such that T_{R}f → f in L^{p} norm as R → ∞ for test functions f. Then in order for T_{R}f to converge to f in L^{p} norm for all L^{p} functions f (i.e. not just for test functions) it is a necessary and sufficient condition that we have the estimate

∥T_{R}f∥_{p} ≲ ∥f∥_{p} (1)

for all sufficiently large R, where the constants are independent of R.

Proof This should be in any decent real analysis book. (Me, I was brought up on Rudin’s ”real and complex analysis”.)

Thus to settle the question of whether S_{R}f converges in L^{p} to f, it suffices to show the bound

∥S_{R}f∥_{p} ≲ ∥f∥_{p}

uniformly in R. Actually, it suffices to prove this for R = 1, since a simple scaling argument shows that the R = 1 case automatically implies the general case. More generally, if a multiplier operator of the form

T̂ f(ξ) = m(ξ)f̂(ξ)

is bounded on L^{p}, and L is any affine transformation whatsoever, then

T̂_{L}f(ξ) = m(Lξ)f̂(ξ)

is also bounded on L^{p} with exactly the same operator norm. (Exercise).

So we have to see whether

∥S_{1}f∥_{p} ≲ ∥f∥_{p}.

Let’s first consider the one-dimensional case. In this case we see that

S_{1}f = F^{−1}(χ_{[−1,1]}f̂)

where F^{−1} is the inverse Fourier transform. Using the identity

χ_{[−1,1]}(x) = \frac{1}{2}(sgn(x + 1) − sgn(x − 1))

we see that it suffices to prove the L^{p} boundedness of the operator

f ↦ F^{−1}(sgn(x ± 1)f̂).

By the invariance of multipliers under affine transformations remarked upon earlier, it thus suffices to prove

f ↦ F^{−1}(sgn(x)f̂).

But this operator is just \frac{i}{π} H, where H is the Hilbert transform

Hf(x) = p.v. ∫ f(x − y)\frac{dy}{y} .

The result now follows from the classical L^{p} boundedness result of the Hilbert transform.

Now let’s look at the n-dimensional case. When p = 2 we have boundedness directly from Plancherel, since the multiplier corresponding to S_{1} is bounded. What about p ≠ 2? It turns out that we do not have boundedness here!

Proposition 1.2 (Fefferman). For any N > 0 and p > 2 there exists a non-zero test function f such that

∥S_{1}f∥_{p} ≥ N∥f∥_{p}. (2)

We’ll delay the proof of Fefferman’s result for a while, and ask how to get around the lack of convergence. One way to do this is to make the frequency cutoff more gentle. For any δ ≥ 0, define the Bochner-Riesz mean S_{R}^{δ} by

S_{R}f(x) = ∫_{|ξ|≤R}(1 − \frac{|ξ|^{2}}{R^{2}} )^{δ}e^{2πix·ξ}f̂(ξ) dξ.

When δ = 0 this is just the disc multiplier S_{R}. When δ = 1 these operators are essentially Fejer means. The point is that the larger δ gets, the smoother the multiplier gets at |ξ| = R, which helps convergence.

These operators were first systematically studied in one dimension by Riesz, and in higher dimensions by Bochner.

It’s easy to see that S_{R}^{δ}f converges to f uniformly if f is a test function. Thus, we may pose the same question as before, namely whether S_{R}^{δ}f converges to f in L^{p} norm for all f in L^{p}. The question reduces as before to verifying whether

∥S_{1}^{δ}f∥_{p} ≲ ∥f∥_{p}. (3)

A necessary condition was found by Herz:

Theorem 1.3 (Herz). In order for (3) to hold, one must have

|\frac{1}{p} − \frac{1}{2}| < \frac{2δ + 1}{2n}. (4)

In particular, we see that the larger δ gets, the larger the interval on which we have a shot at convergence. This doesn’t quite give Fefferman’s theorem, by the way, but it does give the weaker observation that (2) holds whenever p ≥ 2n/(n − 1) or p ≤ 2n/(n + 1).

Let’s see how the proof goes. We first observe that S_{1}^{δ} is a Fourier multiplier operator with multiplier (1 − |ξ|^{2})_{+}^{δ}. This multiplier is real, so the operator is selfadjoint. Thus boundedness on L^{p} is equivalent to boundedness on L^{p′}. (Actually, even if the multiplier was not real, this last statement would still be true - exercise!). So it suffices to consider the p < 2 case.

Let f be a Schwarz function whose Fourier transform is identically 1 on the unit ball. (To find this function, first choose f̂). Testing (3) with this value of f, we see that we must have

∥m̂_{δ}∥_{p} ≲ 1.

where

m_{δ}(ξ) = (1 − |ξ|^{2})_{+}^{δ}.

We claim that

m̂_{δ}(x) = \frac{Ce^{2πi|x|} + Ce^{−2πi|x|} + o(1)}{|x|^{\frac{n+1}{2} +δ}} (5)

as |x| → ∞; inserting this estimate into the previous we obtain a contradiction unless

\frac{n}{p} < \frac{n + 1}{2} + δ;

and this is just a re-arrangement of (4).

We’ll defer the proof of (5) to an appendix. Note however that it is very similar to the formula for d̂σ. Morally speaking, d̂σ is comparable to m_{−1}; this is akin to the heuristic that the delta function is “of the same strength as” the distribution 1/x. Note that every time δ is lowered by 1, (5) predicts that m̂_{δ} is multiplied by roughly |x|; this is consistent with the heuristic observation that the derivative of m_{δ} is roughly comparable to m_{δ−1}. Also, this formula matches up with the ones for the classical Dirichlet and Fejer kernels, which correspond to the cases (n, δ) = (1, 0) and (1, 1) respectively. As one last heuristic, note that m_{δ} is in L^{2} when δ > −1/2, and the same goes for the right-hand side of (5). None of this adds up to a proof, but it can help in remembering the numerology of (5).

Given the necessary conditions of Herz and Fefferman, it is natural to conjecture

Conjecture 1.4 (Bochner-Riesz conjecture). Let δ > 0 and 1 ≤ p ≤ ∞ be such that (4) holds. Then S_{R}^{δ}f converges to f in L^{p} norm for all f ∈ L^{p}.

This conjecture has only been completely settled for n ≤ 2. There are some easy cases we can dispose of first. When p = 2 the conjecture is true, since S_{R}^{δ} is given by a bounded multiplier. When δ > (n − 1)/2 the formula (5) implies that the convolution kernel of S_{R}^{δ} is integrable (note that there is no singularity near zero, in fact m̂_{δ} must be analytic since m_{δ} is compactly supported), so the claim follows from Young’s inequality.

It is possible to interpolate between these results using (for instance) complex interpolation. However, using restriction theory one can do better.

Theorem 1.5 (Fefferman). Let S be the unit sphere. Suppose that p is such that R_{S}(2 → p) holds. Then the Bochner-Riesz conjecture is true for this exponent p.

Proof Fix δ such that (4) holds. We have to show that

∥f ∗ K_{δ}∥_{p} ≲ ∥f∥_{p}

where K_{δ} is the Fourier transform of m_{δ}.

This looks like the estimate we used to prove the Tomas-Stein theorem, except that it’s (L^{p}, L^{p}) instead of (L^{p}, L^{p′}). As before, we break up K as

K_{δ} = ϕK_{δ} + ∑_{k>0} ψ_{k}K_{δ};

we won’t need any fancy moment conditions on ϕ or ψ this time. This is because we have strict inequality in (4) and don’t have to do anything subtle. (People have studied the endpoint, because one still gets weak-type (p, p) estimates there even though strong-type estimates fail. For the endpoint one does need to be more delicate. See [3], [6]).

The function ϕK_{δ} is a bump function, and so convolving with this guy is clearly bounded on L^{p} (e.g. by Young’s inequality). So as before we are left with showing that

∥ ∑_{k>0} f ∗ ψ_{k}K_{δ}∥_{p} ≲ ∥f∥_{p}.

Since we have a bit of room in (4), we will just use the triangle inequality. Actually, we’re going to show

∥f ∗ ψ_{k}K_{δ}∥_{p} ≲ 2^{(n( \frac{1}{p} − \frac{1}{2} )− \frac{1}{2} −δ)k}∥f∥_{p}. (6)

When (4) holds, the constant happens to decay exponentially in k, so we get what we want from the triangle inequality.

The kernel ψ_{k}K_{δ} is supported on an annulus {x : |x| ∼ 2^{k}}. So this operator is somewhat localized; the values of f at a point x only influence points which are close by (within O(2^{k})). Now we invoke a useful lemma that reduces the study of such ”local” operators to a compact set.

Lemma 1.6. Suppose T is a linear operator taking functions on R^{n} to functions of R^{n}. Suppose T is local in the sense that the support of T f always remains within R of the support of f for some R > 0. Then for any 1 ≤ p ≤ q ≤ ∞, the bound

∥T f∥_{q} ≲ ∥f∥_{p} for all f ∈ L^{p}(R^{n}) (7)

is equivalent to the bound

∥T f∥_{L^{q}(B(x,2R)} ≲ ∥f∥_{p} for all f ∈ L^{p}(B(x, R)) (8)

holding uniformly in x. In other words, to show (7) it suffices to test it for functions supported on an R-ball.

Intuitively, the idea is that functions on distinct R-balls basically do not interfere with each other.

Proof Clearly (7) implies (8). Now suppose that (8) holds uniformly in x. Let f be a function in L^{p}(R^{n}). We cover R^{n} by a finitely overlapping collection {B} of R-balls, and make a partition of unity 1 = ∑_{B} ψ_{B} subordinate to this partition. We write

∥T f∥_{q}^{q} = ∫ |T (∑_{B} ψ_{B}f)|^{q} = ∫ | ∑_{B} T (ψ_{B}f)|^{q}.

Since T is local, the functions T (ψ_{B}f) are just supported on the double 2B of B. These balls are finitely overlapping, so we have the pointwise estimate

| ∑_{B} T (ψ_{B}f)|^{q} ≲ ∑_{B} |T (ψ_{B}f)|^{q}.

Putting this back into the previous estimate and simplifying, we obtain

∥T f∥_{q} ≲ (∑_{B} ∥T (ψ_{B}f)∥_{q}^{q})^{1/q}.

Applying (8) we get

∥T f∥_{q} ≲ (∑_{B} ∥ψ_{B}f∥_{p}^{q})^{1/q}.

We now observe the elementary inequality

(∑_{B} a_{B}^{q})^{1/q} ≤ (∑_{B} a_{B}^{p})^{1/p}

which follows from the assumption q ≥ p. This is easily established by checking the easy cases q = p and q = ∞ and then using Hölder’s inequality to interpolate. Thus we have

∥T f∥_{q} ≲ (∑_{B} ∥ψ_{B}f)∥_{p}^{p})^{1/p}.

Since the balls B are finitely overlapping, it is easy to see that the right-hand side is essentially ∥f∥_{p}.

The condition that q ≥ p is absolutely necessary. This is an example of Littlewood’s principle: ”the higher exponents are always on the left”. More precisely:

Lemma 1.7. Let T be a non-zero translation invariant operator. Then the estimate ∥T f∥_{q} ≲ ∥f∥_{p} is only possible if q ≥ p.

Proof Let ϕ be any bump function such that T ϕ is non-zero. Let N > 0 be a large number, and let x_{1}, … x_{N} be N very widely separated points. Define f by

f(x) = ∑_{i=1}^{N} ϕ(x − x_{i}).

If the above estimate held for f, we would have

∥ ∑_{i=1}^{N} T ϕ(x − x_{i})∥_{q} ≲ ∥ ∑_{i=1}^{N} ϕ(x − x_{i})∥_{p}

since T is translation invariant. Now if the x_{i} are sufficiently far apart, the summands are essentially disjointly supported. The right-hand side grows in N like N^{1/p}, and the left-hand side grows like N^{1/q}. Thus we must have N^{1/q} ≲ N^{1/p} for all large N, which implies q ≥ p as desired.

Let’s return to the proof of (6). From the above discussion we just have to prove

∥f ∗ ψ_{k}K_{δ}∥_{L^{p}(B(x,C2^{k}))} ≲ 2^{(n( \frac{1}{p} − \frac{1}{2} )− \frac{1}{2} −δ)k}∥f∥_{p}

for all f supported on a ball B(x, \frac{1}{2}C2^{k}). By translation invariance we may take x = 0.

We’re supposed to be using the Tomas-Stein restriction theorem, which is an (L^{p}, L^{2}) estimate. But we don’t have an L^{2} norm on the right, just an L^{p} norm. But we can fix that easily enough, using Hölder’s inequality and the fact that our L^{p} norm is on a compact set. More precisely, we have

∥F ∥_{L^{p}(B(0,C2^{k}))} = ∥F χ_{B(0,C2^{k})}∥_{p} ≤ |B(0, C2^{k})|^{\frac{1}{p} − \frac{1}{2}} ∥F ∥_{2} ≲ 2^{nk( \frac{1}{p} − \frac{1}{2} )}∥F ∥_{2};

note that the hypothesis R_{S}(2 → p) implies that p < 2. So we only need to show that

∥f ∗ ψ_{k}K_{δ}∥_{2} ≲ 2^{−k/2}2^{−δk}∥f∥_{p}

for all f supported on B(0, C2^{k}).

Okay, this is looking more like Tomas-Stein, but we don’t have a Fourier transform on the right-hand side like we do in a restriction estimate. But that’s okay, we just use Plancherel:

∥f̂{ \atop ψ_{k}K_{δ}}∥_{2} ≲ 2^{−k/2}2^{−δk}∥f∥_{p}.

Since K_{δ} is the Fourier transform of m_{δ}, this is

∥f̂(p̂si_{k} ∗ m_{δ})∥_{2} ≲ 2^{−k/2}2^{−δk}∥f∥_{p}.

We will shortly show the key estimate

|ψ̂_{k} ∗ m_{δ}(ξ)| ≲ 2^{−δk}(1 + 2^{k}d(ξ, S))^{−N} (9)

for any N. Assuming this for the moment, we see that we just need to show that

∥f̂(1 + 2^{k}d(ξ, S))^{−N}∥_{2} ≲ 2^{−k/2}∥f∥_{p}

to finish the proof. We square this as

∫ \frac{|f̂|^{2}}{(1 + 2^{k}d(ξ, S))^{2N}} dξ ≲ 2^{−k}∥f∥_{p}^{2}.

We first deal with the portion when d(ξ, S) > 1/2. This is an error term, and we shall be very crude. We may estimate f̂ pointwise by 2^{Ck}∥f∥_{p} for some C, just from the definition of f̂, Hölder’s inequality, and the fact that f is compactly supported. On the other hand, the denominator in the integral is 2^{−Nk} for any N, and rapidly decreasing as ξ → ∞. This decay beats all other factors, and the bound is easy to prove. So we just need to show

∫_{1/2≤|ξ|≤3/2} \frac{|f̂|^{2}}{(1 + 2^{k}d(ξ, S))^{2N}} dξ ≲ 2^{−k}∥f∥_{p}^{2}.

We use polar co-ordinates, discarding the bounded Jacobian, and rewrite this as

∫_{1/2}^{3/2}(1 + 2^{k}|r − 1|)^{−2N}(∫_{rS^{n−1}} |f̂|^{2} dω) dr ≲ 2^{−k}∥f∥_{p}^{2}

Now we recall the hypothesis R_{S^{n−1}}(p → 2). By earlier remarks we therefore have R_{rS^{n−1}}(p → 2) uniformly for 1/2 ≤ r ≤ 3/2. So we can estimate the left-hand side of this as

∫_{1/2}^{3/2}(1 + 2^{k}|r − 1|)^{−2N}∥f∥_{p}^{2} dr

and the claim follows. (To evaluate the r integral, do the interval where |r−1| ≤ 2^{−k} and |r − 1| > 2^{−k} separately.)

We now have to show (9). The idea is as follows. Since ψ_{k} vanishes at the origin to infinite order, ψ̂_{k} has vanishing moments to infinite order. On the other hand, m_{δ} is quite smooth (except on the unit sphere). The combination of the two should lead to much cancellation.

We first smooth out m_{δ} a bit to remove this annoying singularity at the sphere. Write

m_{δ}(ξ) = m_{δ}(ξ)ϕ(2^{k}(|ξ| − 1)) + m_{δ}(ξ)(1 − ϕ(2^{k}(|ξ| − 1))) = m_{δ}^{0} + m_{δ}^{1}

where ϕ is a bump function which equals 1 near the origin (it doesn’t matter whether this is the same ϕ we used to define ψ_{k}, but it may as well be). The second term here is nice and smooth, but let’s look at the first term for the moment. It’s supported in the annulus of thickness C2^{−k} around the unit sphere, and is quite small there, only about 2^{−δk}. In particular, we see that

m_{δ}^{0} ≲ ∫_{|r−1|≲2^{−k}} 2^{−δk}dσ_{r}dr

where dσ_{r} is surface measure on the sphere of radius r. Thus we can estimate the contribution of m_{δ}^{0} to (9) as

∫_{|r−1|≲2^{−k}} 2^{−δk}dσ_{r} ∗ |ψ̂_{k}|dr

However, from our previous work we know that

dσ_{r} ∗ |ψ̂_{k}|(x) ≲ 2^{k}(1 + 2^{k}d(x, S))^{−N};

(r is so close to 1 that it may as well be 1). Thus we see that the contribution of this case is acceptable.

Now we turn to the main term, which is

ψ̂_{k} ∗ m_{δ}^{1}(ξ).

To handle this we will (implicitly) take advantage of the moment conditions of ψ̂_{k} by writing it as a derivative of a similar function.

Since ψ_{0} vanishes near the origin, we may write ψ_{0}(x) = |x|^{2N}η(x) for some bump function η and some large N. This implies that

ψ_{k}(x) = 2^{−2Nk}|x|^{2N}η(2^{−k}x)

so that

ψ̂_{k}(ξ) = 2^{−2Nk}Δ^{N}φ_{k}(ξ)

where

φ_{k}(ξ) = 2^{nk}η̂(2^{k}ξ).

Since derivatives can be passed between the two factors of a convolution via integration by parts, we thus see that

ψ̂_{k} ∗ m_{δ}^{1} = 2^{−2Nk}φ_{k} ∗ Δ^{N}m_{δ}^{1}.

We now observe that φ_{k} is an approximation to the identity of thickness 2^{−k}, so that

|φ_{k}(x)| ≲ 2^{nk}(1 + 2^{k}|x|)^{−N}.

Also, repeated applications of the product rule show that

|2^{−2Nk}Δ^{N}m_{δ}^{1}(ξ)| ≲ 2^{−δk}(1 + 2^{k}d(x, S))^{−N+δ}.

So we are basically back to the same situation we had for the m_{δ}^{0} term, except that the analogue of m_{δ}^{0} is not perfectly supported on the annulus of thickness C2^{−k} but has a rapidly decreasing tail. But this tail causes no extra difficulty, as can be easily verified (it is convenient to divide into the cases d(ξ, S) ≤ 1/2 and d(ξ, S) > 1/2 respectively). This completes the proof of (9) and thus of the theorem.

Thus we have proven the Bochner-Riesz conjecture for all 1 ≤ p ≤ 2(n + 1)/(n + 3). The Bochner-Riesz problem is self-adjoint, so we also have proven it for p ≥ 2(n + 1)/(n − 1).

The region 2(n + 1)/(n + 3) < p < 2(n + 1)/(n − 1) is much more difficult, except when p = 2. By interpolation one can get some results, but they are not optimal (we do not get the sharp value of δ. Further progress has been made on this problem, but surprisingly little. When n ≤ 2 the whole conjecture is completely proven. When n = 3 we know a little bit more than the above results, which give the conjecture for 1 ≤ p ≤ 4/3 or p ≥ 4; we actually know the conjecture to be true for p ≤ 26/19 and p ≥ 26/7. As one can see we are still a ways from settling the whole thing.

2. Fefferman’s disc multiplier counterexample

We now prove Proposition 1.2. By duality we may restrict ourselves to the case p > 2.

We first observe a nice observation about multiplier operators, which lets us start with a boundedness result for a high-dimensional operator and deduce the corresponding result for a low dimensional “slice” of the operator. (It would be very nice if we could somehow reverse this direction of implication and use the lowdimensional theory to understand the high-dimensional theory!)

Lemma 2.1 (De Leeuw). Suppose that m is a smooth function on R^{n}, and that the operator T defined by

T̂ f(ξ) = m(ξ)f̂(ξ)

is bounded on L^{p}(R^{n}). Then the operator T_{0} defined by

T̂_{0}g(\underline{ξ}) = m(\underline{ξ}, 0)ĝ(\underline{ξ})

for \underline{ξ} ∈ R^{n−1}, is bounded on L^{p}(R^{n−1}).

Proof From the invariance of L^{p} multiplier bounds under affine transformations, we see that we may replace m(ξ) by

m_{R}(\underline{ξ}, ξ_{n}) = m(\underline{ξ}, ξ_{n}/R)

in the definition of T without affecting the L^{p} boundedness property. Letting R → ∞ and taking limits, we may replace m by

m_{∞}(\underline{ξ}, ξ_{n}) = m(\underline{ξ}, 0).

In other words, the operator

{ \atop T_{∞}f}(\underline{ξ}, ξ_{n}) = m(\underline{ξ}, 0)f̂(\underline{ξ}, ξ_{n})

is bounded on L^{p}(R^{n}). If we now apply this fact to a function of the form

f(\underline{ξ}, ξ_{n}) = g(\underline{ξ})ψ(ξ_{n})

and observe that

f̂(\underline{ξ}, ξ_{n}) = ĝ(\underline{ξ})ψ̂(ξ_{n})

we obtain the desired result.

Because of De Leeuw’s principle, it suffices to prove Fefferman’s result for n = 2.

Let 1 ≪ a ≪ R be large numbers to be chosen later. (a is going to end up at about 100, and R is going to be sent off to infinity). Let ψ be a non-negative bump function on the square [0, 1] × [0, 1] which is strictly positive on the sub-square [1/3, 2/3] × [1/3, 2/3], and consider the function

f(x_{1}, x_{2}) = e^{2πix_{1}}ψ(\frac{x_{1}}{R^{2}} , \frac{x_{2}}{aR}).

Thus this function is supported on the rectangle [0, R^{2}] × [0, aR], and has a certain oscillation. Its Fourier transform is given by

f̂(ξ_{1}, ξ_{2}) = aR^{3}ψ̂(R^{2}(ξ_{1} − 1), aRξ_{2}).

Since ψ̂ is rapidly decreasing, we see that f̂ is concentrated near the point (1, 0). Since

Ŝ_{1}f(ξ) = χ_{|ξ|≤1}(ξ)f̂(ξ),

we expect that it will not be too wasteful to approximate χ_{|ξ|≤1} by χ_{ξ_{1}≤1}:

Ŝ_{1}f(ξ) = χ_{ξ_{1}≤1}(ξ)f̂(ξ) − χ_{ξ_{1}≤1,|ξ|≥1}f̂(ξ).

We think of the first term as the main term, and the second term as an error term.

Let’s first deal with the error term. The contribution to S_{1}f(x) is

− ∫ e^{2πixξ}χ_{ξ_{1}≤1,|ξ|≥1}aR^{3}ψ̂(R^{2}(ξ_{1} − 1), aRξ_{2}) dξ.

We’ll estimate this crudely^{1} by absolute values:

∫_{ξ_{1}≤1,|ξ|≥1} aR^{3}|ψ̂(R^{2}(ξ_{1} − 1), aRξ_{2})| dξ_{1}dξ_{2}.

1Getting estimated by absolute values is a common fate for error terms. The subtle analysis methods tend to get reserved for the main terms.

We first deal with the case where ξ_{1} ≤ 0. In this case the ψ̂ term is incredibly small: it is O(R^{−N}) for any N, and decays rapidly at infinity. So it’s easy to see that the total contribution of this part is O(R^{−N}). As for the remainder, we observe that

ξ_{1} ≥ 0, |ξ| ≥ 1 ⟹ ξ_{1} > 1 − \frac{ξ_{2}^{2}}{2} .

Thus it suffices to estimate

∫_{1−^{\frac{ξ^{2}_{2}}{2}} ≤ξ_{1}≤1} aR^{3}|ψ̂(R^{2}(ξ_{1} − 1), aRξ_{2})| dξ_{1}dξ_{2}.

We make the change of variables x = R^{2}(ξ_{1} − 1), y = aRξ_{2}:

∫_{− \frac{y^{2}}{2a} ≤x≤0} |ψ̂(x, y)| dxdy.

Since ψ̂ is rapidly decreasing, we certainly have

|ψ̂(x, y)| ≲ \frac{1}{(1 + y^{2})^{1}0}

and so we can estimate this as

∫_{−∞}^{∞} \frac{y^{2}}{2a} \frac{1}{(1 + y^{2})^{10}} dy = O(1/a).

Now let’s look at the main term. We rewrite it on the Fourier side as

χ_{ξ_{1}≤1}(ξ)f̂(ξ) = aR^{3}(ψ̂χ_{H})(R^{2}(ξ_{1} − 1), aRξ_{2})

where H = {(ξ_{1}, ξ_{2}) : ξ_{1} ≤ 0}. The inverse Fourier transform of this is

e^{2πix_{1}}(ψ ∗ F^{−1}χ_{H})(\frac{x_{1}}{R^{2}} , \frac{x_{2}}{aR}).

Since

χ_{H}(ξ_{1}, ξ_{2}) = \frac{1}{2} − \frac{1}{2}sgn(ξ_{1})

we have (in the sense of distributions)

F^{−1}χ_{H}(x_{1}, x_{2}) = \frac{1}{2}δ(x_{1})δ(x_{2}) − \frac{i}{2π} \frac{1}{x_{1}} δ(x_{2})

where the distribution \frac{1}{x_{1}} is interpreted in the principal value sense. Thus

ψ ∗ F^{−1}χ_{H}(x_{1}, x_{2}) = \frac{1}{2}ψ(x_{1}, x_{2}) + \frac{1}{2πi}p.v. ∫ ψ(x_{1} − y, x_{2})\frac{dy}{y} .

In particular, since ψ is non-negative on [1/3, 2/3] × [1/3, 2/3] we see that

|ψ ∗ F^{−1}χ_{H}(x_{1}, x_{2})| ∼ 1

on the rectangle [10, 11] × [1/3, 2/3]. Thus the main term is comparable to 1 in magnitude on the rectangle [10R^{2}, 11R^{2}] × [R/3, 2R/3]. Since this is much bigger than the error term if a is big enough, we see that S_{1}f is comparable to 1 on this rectangle.

By rotating and translating this example, we obtain

Lemma 2.2. Let T be a R^{2} × aR rectangle in the plane with arbitrary position and orientation, and let T′ be the rectangle of the same length and one third of the width which is shifted over by 10R^{2} in the direction of the long axis of T. Then there exists a function f_{T} supported on T such that |f_{T} | ≤ 1 on T, and |S_{1}f_{T} | ∼ 1 on T′.

In other words, we can make a function supported on T, such that the disc multiplier essentially shifts this function by ten units to a slightly different rectangle T′.

So far we have not obtained (2), except for small N. The idea is to get several rectangles T and pile them together. The key geometrical fact is:

Lemma 2.3. For any M > 0, there exists an R > 0 and a collection of R^{2} × aR rectangles T, such that the rectangles T are all disjoint, and

| ⋃ T′| ≤ M^{−1}| ⋃ T |. (10)

Let’s see how this implies Fefferman’s result. Let M be a large number to be chosen later, and let {T } be the rectangles given by the above lemma. Define the corresponding functions f_{T} from Lemma 2.2. We define f by

f = ∑_{T} ε_{T} f_{T}

where ε_{T} are iid random signs. Since the f_{T} have disjoint supports, we have

∥f∥_{p} = (∑_{T} ∥f_{T} ∥_{p}^{p})^{1/p}

regardless of the choice of signs ε_{t}. Each f_{T} has size about 1 on T, so we have

∥f∥_{p} ≲ (∑_{T} |T |)^{1/p} = | ⋃_{T} T |^{1/p}.

What about S_{1}f? We have

S_{1}f = ∑_{T} ε_{T} S_{1}f_{T}

so by Khinchin’s inequality we have

E(∥S_{1}f∥_{p}^{p}) ∼ ∥(∑_{T} |S_{1}f_{T} |^{2})^{1/2}∥_{p}^{p}.

On the other hand, |S_{1}f_{T} | ∼ 1 on T′. Thus we have

E(∥S_{1}f∥_{p}^{p}) ≳ ∥(∑_{T} |χ_{T′}|^{2})^{1/2}∥_{p}^{p}.

The right-hand side simplifies to

∥ ∑_{T} χ_{T′}∥_{p/2}^{p/2}.

From Hölder’s inequality we have

∥ ∑_{T} χ_{T′}∥_{1} ≤ ∥ ∑_{T} χ_{T′}∥_{p/2}| ⋃_{T} T |^{1−2/p}.

Since the left-hand side of this is ∑_{T} |T′| ∼ | ⋃_{T} T |, this can be re-arranged using (10) as

∥ ∑_{T} χ_{T′}∥_{p/2}^{p/2} ≳ M^{p/2−1}| ⋃_{T} T |.

Combining all our estimates together, we see that

∥S_{1}f∥_{p} ≳ M^{p/2−1}∥f∥_{p};

since p > 2, Fefferman’s result is thus proved by taking M sufficiently large.

3. The Besicovitch set construction

We now prove Lemma 2.3.

Consider a the right-angled triangle T with vertices (0, 0), (1, 0), (0, 1). Divide the base into N intervals from (i/N, 0) to ((i + 1)/N), where i = 0, 1, …, N − 1. This divides the big triangle T into N triangles T_{i} of base 1/N and height 1. Currently the union of these triangles has measure 1/2.

The key lemma is

Lemma 3.1. Let N be a power of 2. There exists a decreasing sequence of numbers a_{0} ≥ … ≥ a_{N−1} such that the union of the translated triangles T_{i}+a_{i}e_{1} has measure O(log log N/ log N).

Assuming this lemma for the moment, let R_{i} be a 1/10x1/(10N) rectangle inscribed inside T_{i}+a_{i}e_{1}, and let R′_{i} be defined as in the previous section, with the orientation chosen so that R′_{i} is below the x-axis. Then it is easy to see that the R′_{i} are all disjoint (because of the monotonicity of the a_{i}), so that | ⋃ R′_{i}| ∼ 1. Lemma 2.3 then follows by setting N roughly equal to R/a and rescaling by R^{2}.

It remains to prove Lemma 3.1. Let us first try to understand the case N = 2. Suppose we try the sequence a_{0} = 0, a_{1} = −δ, where 0 < δ ≪ 1 is some number we shall choose later. This pushes the triangles in a little bit. It’s easy to see that the new union of triangles consists of a triangle similar to T but 1 − δ times as small, together with a bowtie shaped object whose area is only O(delta^{2}) the size of the original triangle T.

Now let’s go back to the general case. Suppose we take the sequence a_{i} = −⌊\frac{i+1}{2}⌋δ/N. Look at what this does to each pair (T_{2n−1}, T_{2n}) of adjacent triangles. By similar considerations to the above, these triangles become squished together to form a triangle T′_{n} similar to T_{2n−1} ∪ T_{2n} but 1 − δ times as small, together with a bowtie of area about O(δ^{2}) times the size of T_{2n−1} ∪ T_{2n}. Also, the triangles T′_{n} as n = 0, …, N/2 − 1 are arranged to form a single triangle which is similar to T but is 1 − δ times as small.

In short, we have adjusted the original configuration T_{0} ⋃ … ⋃ T_{N−1} to become a new configuration T′_{0} ⋃ … ⋃ T′_{N/2−1} of 1 − δ the size, and half as many slices, together with a bunch of bow-ties whose total area is at most O(δ^{2}).

Now take the new triangles T′_{0}, …, T′_{N/2−1} and repeat the same procedure, translating each triangle T′_{i} by −⌊\frac{i+1}{2}⌋(1−δ)δ/(N/2). This induces a further translation on the component triangles T_{2i−1}, T_{2i} which make up T′_{i}, and also move the old bow-ties around. By doing this we create some new triangles T′′_{0}, …, T′′_{N/4}, which are arranged in a triangle which is similar to but (1−δ)^{2} times the size of the original triangle T, plus some new bow-ties whose total area is at most O(δ^{2}(1 − δ)^{2}).

We keep repeating this procedure until the number of slices goes down to 1. We then end up with a small triangle which is similar to but (1 − δ)^{log N} times the size of T, plus a large number of bow-ties whose total area is at most

δ^{2} + δ^{2}(1 − δ)^{2} + δ^{2}(1 − δ)^{4} + … = O(δ).

Thus we have achieved a total area of

O((1 − δ)^{log N} + δ),

and the result follows by choosing δ so that log(1 − δ) = log log N/ log(N).

There are other nice constructions which work too. One of the simplest is the following example of Kahane. Let E be the middle-halves Cantor set, that is the set of all reals in [0, 1] whose base four expansion only consists of 0s and 1s. Let E_{N} be the N^{th} stage of this set, i.e. the set whose first N digits in the base four expansion consists of only 0s and 1s. Now consider the set formed by taking all line segments between {0} × E_{N} and {1} × E_{N}. This set contains about N 1/10xN/10 rectangles, whose translated counterparts are all disjoint, but has total measure about log log N/ log N.

The above construction is closely related to the Kakeya problem. Define a Besicovitch set in R^{n} to be a set which contains a line segment in every direction. The above construction can be adapted to show that Besicovitch sets can have arbitrarily small measure, or even zero measure for n = 2, and a slight modification gives the same result for all n ≥ 2. The Kakeya conjecture asserts that even though Besicovitch sets can have zero measure, they must still have Hausdorff dimension n. In a later set of notes we shall study this conjecture and see why it is so closely related to restriction and Bochner-Riesz type problems. For now, we shall digress to non-linear Schrodinger equations.

4. Appendix: the Bochner-Riesz kernel

We now prove the bound (5). The classical way to do this would be to do some clever changes of variables and compute (5) exactly as an expression involving a Bessel function. In fact, the exact formula is

{ \atop m_{δ}(x)} = π^{−δ}Γ(1 + δ)|x|^{−(n/2)−δ}J_{n/2+δ}(2π|x|).

We’re going to do this in a more modern way - fuzzier, with not as many exact integrals or special functions - but more robust. I’ll be sketchy with the details; those of you who remember the previous quarter’s class will find much of this familiar (indeed, this stuff is very close to the computation of the Helmholtz kernel we did last quarter).

Since m_{δ} is radial, we may take ξ = λe_{n} for some large λ. We need to evaluate the integral

∫_{|ξ|≤1}(1 − |ξ|^{2})^{δ}e^{2πiλξ_{n}} dξ.

We smoothly cut this into three pieces: the piece where |ξ − e_{n}| ≪ 1 (the north pole), the piece where |ξ + e_{n}| ≪ 1 (the south pole), and everything else (the equator), where |ξ_{n}| ≤ 1 − ε for some ε > 0.

Let’s deal with the equator first. The core part where |ξ| ≪ 1 is extremely small by the principle of non-stationary phase, so let’s just look at the surface part |ξ| ∼ 1. In this case we can use polar co-ordinates and reduce to

∫_{r∼1}(1 − r^{2})_{+}^{δ}r^{n−1} ∫_{S^{n−1}:|ω_{n}|≤1−ε} e^{2πiλrω_{n}} dω.

The inner integral, and hence the entire integral, is also rapidly decreasing in λ by the principle of non-stationary phase. So this part is also negligible.

Now let’s look at the north pole part. We can approximate this part by

(1 − |ξ|^{2})_{+}^{δ} = fdω ∗ dμ + error

where the error is smoother than the main term, f is a smooth function on the sphere supported on the northern polar cap, and

dμ(\underline{ξ}, ξ_{n}) = δ(\underline{ξ})η(ξ_{n})(−ξ_{n})_{+}^{δ}

is a measure supported on the vertical axis, with η a bump function which equals 1 near the origin. Indeed, we can easily work out that

fdω ∗ dμ(\underline{ξ}, ξ_{n}) = f(\underline{ξ}, Φ(\underline{ξ}))η(ξ_{n} − Φ(\underline{ξ})J(\underline{ξ})(Φ(\underline{ξ}) − ξ_{n})_{+}^{δ}

where J is some Jacobian factor; by choosing f properly one can make this a good approximation to the kernel m^{δ} near the north pole. The error vanishes to order δ + 1 or more at the sphere; one can do a similar decomposition with this error, with a new error term now vanishing to order δ + 2. We can continue until the error is so smooth that we can easily absorb it into the error term of (5).

We now look at the treatment of the main term. The contribution to (5) of the main term is

f̂dω(λe_{n})d̂μ(λe_{n}).

The first factor is Ce^{2πiλ}λ^{−(n−1)/2} + o(λ^{−(n−1)/2}) by the discussion used to derive the Fourier transform of surface measure. We now claim that the other factor is (C + o(1))λ^{−1−δ}. Since the \underline{ξ} variable is pretty much irrelevant here, this claim is equivalent to

{ \atop η(−ξ_{n})_{+}^{δ}}(λ) = (C + o(1))λ^{−1−δ}.

If it were not for the η, this would be true, since (−ξ_{n})_{+}^{δ} is homogeneous of degree δ, and in k dimensions, the homogeneity of a function and the homogeneity of its Fourier transform always add up to −k. The η has the effect of convolving the Fourier transform with a Schwartz function, which is quite harmless and only perturbs the homogeneity slightly.

Anyway, putting the terms together we get the desired estimate (5).

References

1. M. Christ, Weak type endpoint bounds for Bochner-Riesz multipliers, Revista Mat. Iberoamericana 3 (1987), 25–31.

2. M. Christ, Weak type (1,1) bounds for rough operators, Annals of Mathematics, 128 (1988): 19–42.

3. C. Fefferman, Inequalities for strongly singular convolution operators, Acta Math. 124 (1970), 9–36.

4. C. Fefferman, The multiplier problem for the ball, Ann. of Math. 94 (1971): 330–336.

5. E. M. Stein, M. H. Taibleson, and G. Weiss, Weak-type estimates for maximal operators on certain H^{p} spaces, Rend. Circ. Mat. Palermo suppl. 1 (1981): 81–97.

6. T. Tao, Weak-type endpoint bounds for Riesz means, Proc. Amer. Math. Soc. 124 (1996), 2797–2805.

7. T. Tao, The Bochner-Riesz conjecture implies the Restriction conjecture, to appear, Duke Math J.

8. T. Tao, The weak-type endpoint Bochner-Riesz conjecture and related topics , to appear, Indiana Math. J.

18 TERENCE TAO

9. P. Tomas, A restriction theorem for the Fourier transform, Bull. Amer. Math. Soc. 81 (1975), 477–478.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555 E-mail address: tao@math.ucla.edu