Kovač’s solution of the maximal Fourier restriction problem

About 2 years ago, Müller Ricci and Wright published a paper that opened a new line of investigation in the field of Fourier restriction: that is, the study of the pointwise meaning of the Fourier restriction operators. Here is an account of a recent contribution to this problem that largely sorts it out.

1. Maximal Fourier Restriction
Recall that, given a smooth submanifold \Sigma of \mathbb{R}^d with surface measure d\sigma , the restriction operator {R} is defined (initially) for Schwartz functions as

\displaystyle f \mapsto Rf:= \widehat{f}\Big|_{\Sigma};

it is only after having proven an a-priori estimate such as \|Rf\|_{L^q(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} that we can extend {R} to an operator over the whole of L^p(\mathbb{R}^d), by density of the Schwartz functions. However, it is no longer clear what the relationship is between this new operator that has been operator-theoretically extended and the original operator that had a clear pointwise definition. In particular, a non-trivial question to ask is whether for d\sigma  -a.e. point \xi \in \Sigma we have

\displaystyle \lim_{r \to 0} \frac{1}{|B(0,r)|} \int_{\eta \in B(0,r)} |\widehat{f}(\xi - \eta)| d\eta = \widehat{f}(\xi), \ \ \ \ \ (1)

where B(0,r) is the ball of radius {r} and center {0} . Observe that the Lebesgue differentiation theorem already tells us that for a.e. element of \mathbb{R}^d in the Lebesgue sense the above holds; but the submanifold \Sigma has Lebesgue measure zero, and therefore the differentiation theorem cannot give us any information. In this sense, the question above is about the structure of the set of the Lebesgue points of \widehat{f} and can be reformulated as:

Q: can the complement of the set of Lebesgue points of \widehat{f} contain a copy of the manifold \Sigma ?

The natural approach in regard to these questions, as is well known, is to look at the maximal version of the operators in question. In our case, that would be the operator

\displaystyle R_{\mathrm{max}} f(\xi) := \sup_{r>0}\frac{1}{|B(0,r)|} \int_{\eta \in B(0,r)} |\widehat{f}(\xi - \eta)| d\eta \Big|_{\Sigma},

that is the Hardy-Littlewood maximal function of \widehat{f} restricted to the manifold \Sigma . A standard argument shows that an inequality of the form \|R_{\mathrm{max}} f\|_{L^{q,\infty}(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} implies the d\sigma -a.e. convergence in (1) for functions of L^p(\mathbb{R}^d) (because it is trivial for the dense class of Schwartz functions). Now, working with R_{\mathrm{max}} isn’t actually the best because of that absolute value inside the average that is destroying cancellation, so in practice it is best to work instead with the signed maximal averages

\displaystyle R_{\ast} f(\xi) := \sup_{r>0}\frac{1}{|B(0,r)|} \Big|\int_{\eta \in B(0,r)} \widehat{f}(\xi - \eta) d\eta \Big|;

this operator does not directly address the question of Lebesgue points (although it gives a convergence result nonetheless), but Müller Ricci and Wright have a clever little argument1 to show that one obtains the Lebesgue points result working with R_{\ast} for at least “half” of the range of exponents {p} which you get for the signed averages (actually, Ramos has devised a way to go around this and not lose anything, see links below). In general, one can consider many types of maximal averages combined with the Fourier restriction operator.

Anyway, Müller Ricci and Wright proved that in the 2-dimensional case where \Sigma is a curve in \mathbb{R}^2 with non-vanishing curvature (or the arclength measure is replaced by the affine arclength measure, which automatically dampens the contribution of the points of vanishing curvature) the operator R_{\ast} is L^p(\mathbb{R}^d) \to L^q(\Sigma, d\sigma) bounded for 1\leq p < 4/3 and p' \geq 3q , which is the same range in which {R} is bounded and is optimal (the Fourier restriction conjecture is verified in dimension 2). This is no coincidence of course, since their proof is an adaptation of that of Sjölin for the latter. A few results have followed theirs in the last two years, see for example this by Ramos and this by Kovač and Oliveira e Silva (there should be some work of Fraccaroli & Uraltsev as well but I cannot find it).

2. Kovač’s solution

However, somewhat recently and somewhat surprisingly, Vjeko Kovač has pretty much killed this line of investigation! Indeed, he has shown that there is a general argument with which one can deduce automatically the inequality \|R_{\ast} f\|_{L^q(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} from the corresponding inequality \|R f\|_{L^q(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} , thus reducing the maximal Fourier restriction problem to a corollary of the Fourier restriction problem. Moreover, his proof is very simple and elegant! The succint paper is

Vjekoslav Kovač, “Fourier restriction implies maximal and variational Fourier restriction“, arXiv:1811.05462

In the next section we will discuss Kovač’s argument. Some caveats are in order:

  • The argument only works when {q > p} (more on this later), but this is not really a limitation because this is precisely the “interesting” range of exponents for the Restriction Conjecture (with the endpoint in dimension {d} being2 L^{2d/(d+1)-\epsilon} \to  L^{2d/(d+1)-\epsilon} on the p=q line).
  • The argument is so general and powerful that it works for several types of averages, as long as they satisfy some minimum requirements. For example, it also works for spherical averages when the dimension is at least d \geq 4 !
  • The argument however only works for single-parameter maximal averages (this is not the case for Müller Ricci and Wright, in that their proof can deal with the strong maximal function as well, which is biparameter in 2D).
  • The argument works as well for variation estimates, thus proving a qualitatively stronger a.e. convergence result, in that it is not required to appeal to the existence of a dense subspace of L^p .
  • Lastly, the argument once again works for signed averages, thus to obtain information about the Lebesgue points one has to lose half the range using the argument of Müller, Ricci and Wright (or to try and extend Ramos’ considerations to this case).

In this note, as an exercise for myself, I will illustrate Kovač’s remarkable proof in the slightly simpler case of maximal estimates (instead of the full variation estimates).

3. Kovač’s argument – proofs

The basic idea is to recognise that the lacunary version of the maximal Fourier restriction operator is in some sense a maximal truncation operator – at least for a special kind of averages. The fundamental ingredient in the argument is therefore a (previously unknown, as far as I understand) inequality of Christ-Kiselev type. I have already discussed in the past what we might call “the Christ-Kiselev maximal principle”: it is a general principle that tells us that if we have an L^p \to L^q inequality for a linear operator {T} where crucially {q > p} (that is, in the technical jargon, {T} is L^p-improving), then we can conclude the same inequality for certain maximal truncations of {T} . In that previous post the operator to which the principle was applied was the Fourier transform and the L^p -improving inequality it satisfies is the Hausdorff-Young inequality for {p<2} , while the associated maximal operator was the maximal truncation \sup_{N} \Big|\int_{-\infty}^{N} f(x) e^{-2\pi i \xi x} dx \Big| . The proof was based on a shockingly clever idea: partitioning the real line dyadically according to the L^p mass of the function.
The incarnation of the principle that is needed for Kovač's argument is the following:

Lemma 1 : Let (\Psi_k)_{k\in\mathbb{Z}} be a collection of Schwartz functions with the property that their supports are pairwise disjoint. Then if the inequality \|R f\|_{L^q(\Sigma, d\sigma)} \leq C_{\mathrm{restr}} \|f\|_{L^p(\mathbb{R}^d)} holds, the inequality

\displaystyle \Big\| \sup_{n \in \mathbb{Z}} \Big|\sum_{k = -\infty}^{n} \widehat{f \Psi_k}\Big| \Big\|_{L^q(\Sigma, d\sigma)} \leq C(p,q,C_{\mathrm{restr}}) \Big\|  |f| \Big[\sum_{k\in\mathbb{Z}}|\Psi_k|\Big] \Big\|_{L^p(\mathbb{R}^d)} \ \ \ \ \ (2)

also holds, with constant

\displaystyle C(p,q,C_{\mathrm{restr}}) = \frac{C_{\mathrm{restr}}}{1 - 2^{1/p - 1/q}}.

Notice that the constant C(p,q,C_{\mathrm{restr}}) produced by the proof blows up as p \to q^{-} .
The proof of this Christ-Kiselev type lemma is more straightforward than the other mentioned above and is as follows (apparently the general argument in this form is due to Tao; but Lemma 1 specifically is due to Kovač himself, I think, and was not known before).

Proof: Let {A} denote the constant C(p,q,C_{\mathrm{restr}}) that we have to show is finite. It suffices to assume that the index {k} ranges over the interval [M,N] , as long as the resulting constant A does not depend on either M,N. This will allow us to use induction in the parameter N-M to prove the lemma.
The base case is N-M = 0 and is completely trivial: it’s just the restriction estimate itself! Now assume therefore that it has been proven for all M' < N' such that N' - M' < N - M. For each M \leq n \leq N let \Sigma_n be the set of points \xi \in \Sigma such that {n} is the smallest index for which the supremum \sup_{M \leq n \leq N} \Big|\sum_{k = M}^{n} \widehat{f \Psi_k}(\xi)\Big| is attained. The sets \Sigma_M, \ldots, \Sigma_{N} are measurable, pairwise disjoint, and they partition \Sigma itself. Now look at the term at the right-hand side of (2): there is a smallest index {K} such that

\displaystyle { \int_{\mathbb{R}^d} \Big[ |f| \sum_{k = M}^{K}|\Psi_k| \Big]^p dx > \frac{1}{2}  \int_{\mathbb{R}^d} \Big[  |f| \sum_{k = M}^{N}|\Psi_k| \Big]^p dx.}

Notice that the minimality of {K} implies both that

\displaystyle { \int_{\mathbb{R}^d} \Big[ |f| \sum_{k = M}^{K-1}|\Psi_k| \Big]^p dx \leq \frac{1}{2}  \int_{\mathbb{R}^d} \Big[  |f| \sum_{k = M}^{N}|\Psi_k| \Big]^p dx.}

and that

\displaystyle { \int_{\mathbb{R}^d} \Big[ |f| \sum_{k = K+1}^{N}|\Psi_k| \Big]^p dx \leq \frac{1}{2}  \int_{\mathbb{R}^d} \Big[  |f| \sum_{k = M}^{N}|\Psi_k| \Big]^p dx,}

this latter one being due in particular to the fact that the \Psi_k have disjoint supports and therefore \Big[  |f| \sum_{k = M}^{N}|\Psi_k| \Big]^p = |f|^p  \sum_{k = M}^{N}|\Psi_k|^p .
Now, observe that by our decomposition above we can write the supremum as

\displaystyle \sup_{M\leq n \leq N} \Big| \sum_{k=M}^{n} \widehat{f \Psi_k}\Big| = \Big|\sum_{k,n: M \leq k \leq n \leq N} \widehat{f\Psi_k} \mathbf{1}_{\Sigma_n} \Big|,

and we can split the latter sum into 3 terms:

\displaystyle \sum_{M \leq k \leq K \leq n \leq N} + \sum_{ M \leq k \leq n \leq K-1} + \sum_{K+1 \leq k \leq n \leq N} =: I + II + III.

Terms II, III can be treated similarly by the inductive hypothesis: indeed,

\displaystyle II = \Big|\sum_{ M \leq k \leq n \leq K-1}  \widehat{f\Psi_k} \mathbf{1}_{\Sigma_n} \Big| \leq \sup_{M\leq n \leq K-1} \Big| \sum_{k=M}^{n} \widehat{f \Psi_k}\Big|

and therefore, by inductive hypothesis and the considerations about {K} above,

\displaystyle \|II\|_{L^q(\Sigma,d\sigma)} \leq A \Big\| |f| \sum_{k=M}^{K-1} |\Psi_k|\Big\|_{L^p(\mathbb{R}^d)} \leq A 2^{-1/p} \Big\| |f| \sum_{k=M}^{N} |\Psi_k|\Big\|_{L^p(\mathbb{R}^d)};

and the same bound holds for {III} . Actually, more is true: terms {II} and {III} have disjoint supports in {\Sigma} , and therefore we have the stronger fact that \|II + III\|_{L^q(\Sigma,d\sigma)}  \leq A 2^{1/q - 1/p} \Big\| |f| \sum_{k=M}^{N} |\Psi_k|\Big\|_{L^p(\mathbb{R}^d)} .
The odd term is {I} , but this term actually is simple to treat: the condition k \leq K \leq n effectively decouples the two variables {k,n} , so that \sum_{M \leq k \leq K \leq n \leq N} \widehat{f\Psi_k} \mathbf{1}_{\Sigma_n} = \sum_{M \leq k \leq K} \widehat{f\Psi_k} \cdot \sum_{K \leq n \leq N} \mathbf{1}_{\Sigma_n} . This means that we can apply directly the Restriction estimate to this term, and we have

\displaystyle \|I\|_{L^q(\Sigma,d\sigma)} \leq C_{\mathrm{restr}} \Big\| |f| \sum_{k=M}^{N} |\Psi_k|\Big\|_{L^p(\mathbb{R}^d)}.

The above discussion has shown that we can close the induction if we can find {A} such that

\displaystyle A 2^{1/q - 1/p} + C_{\mathrm{restr}} = A,

and this is indeed possible, but only thanks to the fact that {p} < {q} : take {A} = \frac{C_{\mathrm{restr}}}{1 - 2^{1/q - 1/p} } , as claimed in the statement of the lemma, and we are done. \Box

The Christ-Kiselev type lemma above is particularly well suited for establishing a lacunary version of the maximal Fourier restriction result we want. There are some constraints such as the functions \Psi_k being Schwartz, so we proceed with some caution and content ourselves at first with treating only the special kind of average that follows. Let \psi be a Schwartz function and let \psi_t(x) := \frac{1}{t^d} \psi \Big(\frac{x}{t}\Big) . We will look at averages of the form

\displaystyle \int_{0}^{t} \widehat{f} \ast \psi_s \frac{ds}{s};

the reason for choosing such an expression is easily explained: let \Psi_k be the functions

\displaystyle \Psi_k(x) := \int_{2^k}^{2^{k+1}} \widehat{\psi}(s x) \frac{ds}{s},


\displaystyle \int_{0}^{2^{n+1}} \widehat{f} \ast \psi_s \frac{ds}{s} = \sum_{k \leq n} \widehat{f \Psi_k},

which is precisely the form we want this to have in order to apply the Christ-Kiselev-type lemma (with a little abuse of notation, we are writing \widehat{\psi}  instead of \check{\psi}  ). There is however the support requirement which is very important, and therefore we choose \psi radial (for convenience) and such that its Fourier transform \widehat{\psi} is supported in the annulus \{ \xi \,:\, 1 \leq |\xi| \leq 2\} . It is then immediate to see that \Psi_k(x) is supported in the annulus \{ x \,:\, 2^{-k-1} \leq |x| \leq 2^{-k+1}\} . These annula are not exactly pairwise disjoint, in that \mathrm{Supp}(\Psi_k) intersects \mathrm{Supp}(\Psi_{k-1}) and \mathrm{Supp}(\Psi_{k+1}) (and nothing else), but this is easily fixed by restricting the index {k} to range only over the odd integers and, separately, only on the even integers. An application of Lemma 1 therefore gives us exactly the lacunary maximal Fourier restriction estimate for these special averages:

\displaystyle \Big\| \sup_{n \in \mathbb{Z}} \Big|\int_{0}^{2^n} \widehat{f} \ast \psi_s \frac{ds}{s} \Big| \Big\|_{L^q(\Sigma,d\sigma)} \lesssim_{p,q} C_{\mathrm{restr}} \|\widehat{\psi}\|_{L^\infty}\|f\|_{L^p(\mathbb{R}^d)},

where we have used the fact that, by the hypothesis on the support of {\widehat{\psi}} ,

\displaystyle \sum_{k \in \mathbb{Z}} |\Psi_k(x)| \leq \int_{0}^{\infty} |\widehat{\psi}(sx)|\frac{ds}{s} = \int_{\{|x|^{-1} \leq s \leq 2|x|^{-1}\}} |\widehat{\psi}(sx)|\frac{ds}{s} \leq \|\widehat{\psi}\|_{L^\infty}. \ \ \ \ \ (3)

We are nearly there! A bit like in the proof of Stein’s spherical maximal function theorem which we gave a few days ago, after gaining control of the lacunary supremum it suffices to control the local supremum (when we restrict the parameter to a dyadic interval) to conclude. Indeed, since for any {t \in (0,\infty)} there exists a unique {k \in \mathbb{Z}} and a unique \tau \in [0,1) such that t = 2^k + 2^k\tau , we have that

\displaystyle \sup_{t > 0} \Big|\int_{0}^{t} \widehat{f} \ast \psi_s (\xi) \frac{ds}{s} \Big| \leq \sup_{k \in \mathbb{Z}} \Big|\int_{0}^{2^k} \widehat{f} \ast \psi_s(\xi) \frac{ds}{s} \Big| + \sup_{k \in \mathbb{Z}}\sup_{ 2^{k}\leq t < 2^{k+1} } \Big|\int_{2^k}^{t} \widehat{f} \ast \psi_s(\xi) \frac{ds}{s} \Big|.

The first term on the right-hand side was dealt with just above, so all that remains is to deal with the second term.

As it turns out, this term is very easy to deal with. First of all, consider {k} fixed and observe that

\displaystyle \begin{aligned} \sup_{ 2^{k}\leq t < 2^{k+1} } \Big|\int_{2^k}^{t} \widehat{f} \ast \psi_s \frac{ds}{s} \Big| \lesssim & 2^{-k} \int_{2^k}^{2^{k+1}} |\widehat{f} \ast \psi_s|ds \\ \leq & \Big(2^{-k} \int_{2^k}^{2^{k+1}} |\widehat{f} \ast \psi_s|^q ds \Big)^{1/q} \\ \sim & \Big(\int_{2^k}^{2^{k+1}} |\widehat{f} \ast \psi_s|^q \frac{ds}{s} \Big)^{1/q}, \end{aligned}

where the second inequality is by Hölder. Now we take the supremum in {k} and bound it by the {\ell^q} sum, obtaining

\displaystyle \sup_{k \in \mathbb{Z}}\sup_{ 2^{k}\leq t < 2^{k+1} } \Big|\int_{2^k}^{t} \widehat{f} \ast \psi_s(\xi) \frac{ds}{s} \Big| \lesssim \Big(\int_{0}^{\infty} |\widehat{f} \ast \psi_s|^q \frac{ds}{s}\Big)^{1/q}.

The L^q(\Sigma,d\sigma) norm of the last expression is thus bounded by

\displaystyle \Big(\int_{0}^{\infty} \|\widehat{f} \ast \psi_s\|_{L^q(\Sigma,d\sigma)}^q \frac{ds}{s}\Big)^{1/q},

which by the Restriction inequality is bounded by

\displaystyle \Big(\int_{0}^{\infty} \|f  \widehat{\psi}(s\cdot)\|_{L^p(\mathbb{R}^d)}^q \frac{ds}{s}\Big)^{1/q}.

By Minkowski ({p} < {q} ), we conclude that the above is bounded by

\displaystyle \Big(\int_{\mathbb{R}^d} |f(x)|^p \Big(\int_{0}^{\infty} |\widehat{\psi}(sx)|^q \frac{ds}{s}\Big)^{p/q} dx\Big)^{1/p};

but now observe that as in (3) above we have

\displaystyle \int_{0}^{\infty} |\widehat{\psi}(sx)|^q \frac{ds}{s} \leq \|\widehat{\psi}\|_{L^\infty}^q.

In conclusion, the above discussion has proven

Proposition 2: Let \psi be a Schwartz function such that \mathrm{Supp}(\widehat{\psi}) \subset \{ \xi \,:\, 1 \leq |\xi|\leq 2\} . Then if {q} > {p} and the restriction inequality \|Rf\|_{L^q(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} holds, the maximal restriction inequality

\displaystyle \Big\| \sup_{t> 0} \Big| \int_{0}^{t} \widehat{f}\ast \psi_s \frac{ds}{s} \Big|\Big\|_{L^q(\Sigma,d\sigma)} \lesssim \|\widehat{\psi}\|_{L^\infty} \|f\|_{L^p(\mathbb{R}^d)}

also holds.

Now, this in itself would already be a very nice result on its own, but with a little more effort one can show that the hypotheses on the type of maximal average can actually be widely relaxed. Indeed, the averages we are interested in are of the type g \mapsto g \ast \mu_{t} = \int g(\xi - t y) d\mu(y) for {\mu} a complex Borel measure, in general; the idea is that if {\mu} is sufficiently regular, we can represent averages with respect to {\mu} as averages of averages of the type above.
Suppose then for convenience that the function {\psi} as in the Proposition above is normalised so that \int_{0}^{\infty} \widehat{\psi}(s x) \frac{ds}{s} = 1 (this clearly holds for all x\neq 0 ). We let \Phi^t denote the Schwartz function whose Fourier transform is

\displaystyle \widehat{\Phi^t}(x) := \widehat{\psi}(x)\frac{x}{t} \cdot \nabla \widehat{\mu}\Big(\frac{x}{t}\Big)

(we are therefore assuming that \widehat{\mu} exists and is infinitely smooth). Notice that \widehat{\Phi^t} is supported in the same annulus as {\psi} . This definition is chosen exactly in order for the following to hold:

\displaystyle \begin{aligned} \widehat{\mu}(bx) - \widehat{\mu}(ax) = & \int_{a}^{b} s \frac{d}{ds} \widehat{\mu}(sx) \frac{ds}{s} \\ = & \int_{a}^{b} sx \cdot \nabla\widehat{\mu}(sx) \frac{ds}{s} \\ = & \int_{a}^{b} sx \cdot \nabla\widehat{\mu}(sx) \int_{0}^{\infty} \widehat{\psi}(ts x) \frac{dt}{t} \frac{ds}{s} \\ = & \int_{a}^{b}\int_{0}^{\infty}  sx \cdot \nabla\widehat{\mu}(sx)  \widehat{\psi}(ts x) \frac{dt}{t} \frac{ds}{s} \\ = & \int_{a}^{b}\int_{0}^{\infty}  \widehat{\Phi^t}(ts x)\frac{dt}{t} \frac{ds}{s} \\ = & \int_{0}^{\infty} \int_{ta}^{tb} \widehat{\Phi^t}(s x) \frac{ds}{s}\frac{dt}{t}. \end{aligned}

Observe that in the last line we have used Fubini, and to justify this we will have to assume that the integrand is integrable, which in turn must be deduced from properties of \mu . If the integrand is integrable, we can also take the Fourier transform of f(x) \widehat{\mu}(bx) -  f(x)\widehat{\mu}(ax), and see that it equals

\displaystyle \widehat{f} \ast \mu_b - \widehat{f}\ast \mu_a = \int_{0}^{\infty} \int_{ta}^{tb} \widehat{f} \ast (\Phi^t)_{s} \frac{ds}{s}\frac{dt}{t},

and you can recognise in the inner integral an average of the type we studied above; so it is indeed true that, given some regularity of the measure {\mu} we can see averages with respect to it as averages of our previous averages. It is an easy exercise to see that all the manipulations above are justified if we assume for example that \widehat{\mu} \in C^\infty and that |\nabla\widehat{\mu}(x)|\lesssim (1 + |x|)^{-1-\delta} – the latter having the effect that

\displaystyle \|\widehat{\Phi^t}\|_{L^\infty} \lesssim \min\{t^\delta, t^{-1}\}. \ \ \ \ \ (4)

Taking {a=1} we see by triangle inequality that we can bound \sup_{t>0} |\widehat{f}\ast \mu_t| by the innocuous |\widehat{f} \ast \mu| plus

\displaystyle \int_{0}^{\infty} \sup_{r > 0} \Big| \int_{0}^{r} \widehat{f} \ast (\Phi^t)_s \frac{ds}{s} \Big| \frac{dt}{t}.

Taking the L^q(\Sigma, d\sigma) norm of the latter expression and using Minkowski to exchange the integrals we see by Proposition 2 that we can bound this norm by

\displaystyle \int_{0}^{\infty} \|\widehat{\Phi}^t\|_{L^\infty} \frac{dt}{t} \|f\|_{L^p(\mathbb{R}^d)}.

Finally, the integral in {t} is \lesssim 1 by (4), and this shows Proposition 2 has the corollary

Corollary 3: Let {\mu} be a complex Borel measure such that \widehat{\mu} \in C^\infty and |\nabla\widehat{\mu}(x)|\lesssim (1 + |x|)^{-1-\delta}. Then if {q} > {p} and the restriction inequality \|Rf\|_{L^q(\Sigma,d\sigma)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} holds, the maximal restriction inequality

\displaystyle \Big\| \sup_{t> 0} |\widehat{f} \ast \mu_t| \Big\|_{L^q(\Sigma,d\sigma)} \lesssim_{\mu,p,q} \|f\|_{L^p(\mathbb{R}^d)}

also holds.

4. Concluding remarks

So, what is left to understand in the field of maximal Fourier restriction? It is fair to say that Kovač’s argument provides quite a satisfactory answer to the main question of d\sigma -a.e. convergence of (signed) averages of the Fourier transform. What is left out is the following:

  • the full study of the Lebesgue points – currently one can use the Müller,Ricci,Wright trick above to deduce bounds for the positive averages from bounds for the signed averages, but in doing so half of the range is lost. I think Ramos is working on combining his approach with Kovač’s argument.
  • the study of multiparameter maximal functions – indeed, the Christ-Kiselev maximal principle is, to my limited knowledge, decidedly a one-parameter tool only. It seems that to approach the multiparameter version of the problem one will need an entirely different proof.
  • the study of more singular maximal functions – for example, Fraccaroli and Uraltsev studied the maximal Fourier restriction problem for the curves {\Gamma} in \mathbb{R}^2 where the maximal averages are taken over rectangles with center \xi \in \Gamma whose sides are parallel to the tangent and normal directions to the curve at {\xi}. These are interesting in that they combine features of many different problems at once.
  • single special examples – for instance, as mentioned before, the argument of Kovač does not cover the spherical averages when the dimension is d=2,3 . It might be an interesting little project to sort these out.

1: The idea is to consider f \ast \tilde{f} , which has positive Fourier transform |\widehat{f}|^2 (here \tilde{f}(x) = f(-x) ); after an application of Cauchy-Schwarz, one has R_{\mathrm{max}}f \leq (R_{\ast}(f\ast \tilde{f}))^{1/2} , and that exponent 1/2 is what causes the loss.
2: the \epsilon is because the estimate is known to be false when \epsilon =0 (see this paper of Beckner, Carbery, Semmes and Soria).

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