# Affine Restriction estimates imply Affine Isoperimetric inequalities

One thing I absolutely love about harmonic analysis is that it really has something interesting to say about nearly every other field of Analysis. Today’s example is exactly of this kind: I will show how a Fourier Restriction estimate can say something about Affine Geometry. This was first noted by Carbery and Ziesler (see below for references).

## 1. Affine Isoperimetric Inequality

Recall the Affine Invariant Surface Measure that we have defined in a previous post. Given a hypersurface $\Sigma \subset \mathbb{R}^d$ sufficiently smooth to have a well-defined Gaussian curvature $\kappa_{\Sigma}(\xi)$ (where $\xi$ ranges over $\Sigma$) and with surface measure denoted by $d\sigma_{\Sigma}$, we can define the Affine Invariant Surface measure as the weighted surface measure

$\displaystyle d\Omega_{\Sigma}(\xi) := |\kappa_{\Sigma}(\xi)|^{1/(d+1)} \, d\sigma_{\Sigma}(\xi);$

this measure has the property of being invariant under the action of $SL(\mathbb{R}^d)$ – hence the name. Here invariant means that if $\varphi$ is an equi-affine map (thus volume preserving) then

$\displaystyle \Omega_{\varphi(\Sigma)}(\varphi(E)) = \Omega_{\Sigma}(E)$

for any measurable $E \subseteq \Sigma$.
The Affine Invariant Surface measure can be used to formulate a very interesting result in Affine Differential Geometry – an inequality of isoperimetric type. Let $K \subset \mathbb{R}^d$ be a convex body – say, centred in the origin and symmetric with respect to it, i.e. $K = - K$. We denote by $\partial K$ the boundary of the convex body $K$ and we can assume for the sake of the argument that $\partial K$ is sufficiently smooth – for example, piecewise $C^2$-regular, so that the Gaussian curvature is defined at every point except maybe a $\mathcal{H}^{d-1}$-null set. Then the Affine Isoperimetric Inequality says that (with $\Omega = \Omega_{\partial K}$)

$\displaystyle \boxed{ \Omega(\partial K)^{d+1} \lesssim |K|^{d-1}. } \ \ \ \ \ \ \ (\dagger)$

Notice that the inequality is invariant with respect to the action of $SL(\mathbb{R}^d)$ indeed – thanks to the fact that $d\Omega$ is. Observe also the curious fact that this inequality goes in the opposite direction with respect to the better known Isoperimetric Inequality of Geometric Measure Theory! Indeed, the latter says (let’s say in the usual $\mathbb{R}^d$) that (a power of) the volume of a measurable set is controlled by (a power of) the perimeter of the set; more precisely, for any measurable $E \subset \mathbb{R}^d$

$\displaystyle |E|^{d-1} \lesssim P(E)^d,$

where $P(E)$ denotes the perimeter1 of $E$ – in case $E = K$ a symmetric convex body as above we would have $P(K) = \sigma(\partial K)$. But in the affine context the “affine perimeter” is $\Omega(\partial K)$ and is controlled by the volume instead of viceversa. This makes perfect sense: if $K$ is taken to be a cube $Q$ then $\kappa_{\partial Q} = 0$ and so the “affine perimeter” cannot control anything. Notice also that the power of the perimeter is $d$ for the standard isoperimetric inequality and it is instead $d+1$ for the affine isoperimetric inequality. Informally speaking, this is related to the fact that the affine perimeter is measuring curvature too instead of just area.
So, the inequality should actually be called something like “Affine anti-Isoperimetric inequality” to better reflect this, but I don’t get to choose the names.

The inequality above is formulated for convex bodies since those are the most relevant objects for Affine Geometry. However, below we will see that Harmonic Analysis provides a sweeping generalisation of the inequality to arbitrary hypersurfaces that are not necessarily boundaries of convex bodies. Before showing this generalisation, we need to introduce Affine Fourier restriction estimates, which we do in the next section.

# Some thoughts on the smoothing effect of convolution on measures

A question by Ben Krause, whom I met here at the Hausdorff Institute, made me think back of one of the earliest posts of this blog. The question is essentially how to make sense of the fact that the (perhaps iterated) convolution of a (singular) measure with itself is in general smoother than the measure you started with, in a variety of settings. It’s interesting to me because in this phase of my PhD experience I’m constantly trying to build up a good intuition and learn how to use heuristics effectively.

So, let’s take a measure ${\mu}$ on ${\mathbb{R}^d}$ with compact support (assume inside the unit ball wlog). We ask what we can say about ${\mu \ast \mu}$, or higher iterates ${\mu \ast \mu \ast \mu \ast \ldots}$ and more often1 about ${\mu \ast \tilde{\mu}}$. In particular, we’re interested in the case where ${\mu}$ is singular, i.e. its support has zero Lebesgue measure.

Before starting though, I would like to give a little motivation as to why such convolutions are interesting. Consider the model case where you have an operator defined by ${Tf = \sum_{j\in\mathbb{Z}}{T_j f}:= \sum_{j \in \mathbb{Z}}f\ast \mu_j}$ where ${\mu_j}$ are some singular measures and ${f \in L^2}$. One asks whether the operator ${T}$ is bounded on ${L^2}$, and the natural tool to use is Cotlar-Stein lemma, or almost-orthogonality (from which this blog takes its name). Then we need to verify that

$\displaystyle \sup_{j}{\sum_{k}{\|T_j T_k^\ast\|^{1/2}}} < \infty,$

and same for ${T^\ast_j T_k}$. But what is ${T_j T^\ast_k f}$? It’s simply

$\displaystyle T_j T^\ast_k f = f \ast d\tilde{\mu}_k \ast d\mu_j = f \ast (d\tilde{\mu}_k \ast d\mu_j),$

i.e. another convolution operator. Estimates on the convolution2 ${d\tilde{\mu}_k \ast d\mu_j}$ are likely to help estimate the norm of ${T_j T_k^\ast}$ then. But if this measure is not smooth enough, one can go forward, and since

$\displaystyle \|T T^\ast \| \leq \|T\|^{1/2} \|T^\ast T T^\ast\|^{1/2}$

one sees that estimates on ${d\tilde{\mu}_k \ast d\mu_j \ast d\tilde{\mu}_k }$ are likely to help, and so on, until a sufficient number of iterations gives a sufficiently smooth measure. This isn’t quite the iteration of a measure with itself, but in many cases one has an operator ${Tf = f \ast \mu}$ which then splits into the above sum by a spatial or frequency cutoff at dyadic scales. Then it becomes a matter of rescaling and the case ${d\tilde{\mu}_k \ast d\mu_j}$ can be reduced to that of ${d\tilde{\mu}_0 \ast d\mu_j}$ and further reduced to that of ${d\tilde{\mu}_0 \ast d\mu_0}$ by exploiting an iterate of the above norm inequality, namely that

$\displaystyle \|T_j T_0^\ast\|\leq \|T_j^\ast\|^{1-2^{\ell}}\|T_j (T_0^\ast T_0)^\ell\|^{2^{-\ell}}.$

Another possibility is to write ${d\nu = d\tilde{\mu}_k \ast d\mu_j}$ and consider working with ${d\nu \ast d\nu}$ instead, to obtain results in term of ${|j-k|}$. I will say more in the end, here I just wanted to show that they arise as natural objects.

# Lorentz spaces basics & interpolation

(Updated with endpoint ${q = \infty}$)

I’ve written down an almost self contained exposition of basic properties of Lorentz spaces. I’ve found the sources on the subject to leave something to be desired, and I grew a bit confused at the beginning. Therefore this relatively short note (I might be ruining someone’s assignments out there, but I think the pros of writing down everything in one place balance the cons).

1. Lorentz spaces

In the following take ${1< p,q < \infty}$ otherwise specified, and ${(X, |\cdot|)}$ a ${\sigma}$-finite measure space with no atoms.

The usual definition of Lorentz space is as follows:

Definition 1 The space ${L^{p,q}(X)}$ is the space of measurable functions ${f}$ such that

$\displaystyle \|f\|_{L^{p,q}(X)}:= \left(\int_{0}^{\infty}{t^{q/p}{ f^\ast (t)}^q}\,\frac{dt}{t}\right)^{1/q} < \infty,$

where ${f^\ast}$ is the decreasing rearrangement [I] of ${f}$. If ${q=\infty}$ then define instead

$\displaystyle \|f\|_{L^{p,\infty}(X)} := \sup_{t}{t^{1/p} f^\ast (t)} < \infty.$

# Riemannian geometry I

I had many courses in my undergraduate studies, but for one reason or another never got to see some riemannian geometry. Which is really a shame, because I always wanted to learn some general relativity. So, here I am, studying riemann geometry for the first time. Since there’s lots of things in it, what follows is really just a summary, primarily for my own understanding.

Let ${M}$ be a smooth manifold of dimension ${n}$, ${TM = \sqcup_{p\in M}{T_p M}}$ his tangent bundle, ${T^\ast M}$ the cotangent bundle, and let ${\mathcal{T}(M)}$ denote the space of smooth sections of ${TM}$ (i.e. the smooth vector fields on ${M}$). Analogously, ${\mathcal{T}^{k}_{l}(M)}$ will denote the smooth sections of the tensor bundle ${T^{k}_{l}M}$ – that is, the smooth tensor fields [1]. Tensors fields are multilinear with respect to ${C^{\infty}(M)}$. For example, ${\mathcal{T}M = \mathcal{T}_{1}(M)}$, while ${\mathcal{T}^{1}(M)}$ are the ${1}$-forms. We also impose ${\mathcal{T}^{0}(M)=C^{\infty}(M)}$.

1. Riemannian metrics

A Riemannian metric on ${M}$ is a ${2}$-tensor field (in ${\mathcal{T}^2 (M)}$) – given in local coordinates by ${g_{ij}dx^i \otimes dx^j}$ – which is:

1. symmetric: ${g(X,Y) = g(Y,X)}$ for every ${X, Y \in \mathcal{T}(M)}$;
2. positive definite: ${g(X,X)>0}$ for every ${X\neq 0}$.

It induces an inner product on every tangent space. Continue reading

# Transported measures

The following theorem is a smoothness result for measures trasported onto smaller dimensional spaces. What’s interesting about this technical result is a decomposition needed in the proof which is rather general in principle and useful in that it doesn’t require much knowledge of the geometry of the problem.

Here, suppose ${\Phi:\overline{B_m}\rightarrow \mathbb{R}^n}$ is a ${C^\infty}$ map from the closed unit ball ${\overline{B_m}\subset \mathbb{R}^m}$ in ${\mathbb{R}^n}$, where ${m\geq n}$. Let ${\psi \in C^{(k+1)}\left(\overline{B_m}\right)}$ with compact support in ${B_m}$. Then we can transport measure ${\psi(\tau)d\tau}$ from ${B_m}$ to ${\mathbb{R}^n}$ via ${\Phi}$, i.e.

$\displaystyle d\mu=\Phi_\ast (\psi(\tau)d\tau),$

defined by

$\displaystyle \int_{\mathbb{R}^n}{f(x)}\,d\mu(x)=\int_{B_m}{f(\Phi(\tau))\psi(\tau)}\,d\tau.$

If ${\Phi}$ is a well behaved map we shall prove that the transported measure is indeed quite regular:

Theorem 1 Let ${\Phi}$ be also in ${C^{(k+2)}(\overline{B_m})}$. Let ${J}$ be the determinant of a ${n\times n}$ submatrix of the Jacobian of ${\Phi}$, and suppose that there exists a multiindex ${\alpha}$ such that

$\displaystyle |\partial^{\alpha}_{\tau}J(\tau)|>0$

everywhere on the ball ${\overline{B}_m}$. Then the transported measure ${d\mu=\Phi_\ast (\psi(\tau)d\tau)}$ is absolutely continuous with respect to the Lebesgue measure on ${\mathbb{R}^n}$. Moreover, its density ${\rho}$ has the ${L^1}$-Hölder smoothness property

$\displaystyle \int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}\,dx\leq C \|t\|^{\delta}$

for every ${t\in\mathbb{R}^n}$, where the exponent ${\delta}$ can be choosen in the range ${0<\delta <(2|\alpha|)^{-1}}$. (The constant ${C}$ will depend only on the ${C^{(k+2)}(\overline{B_m})}$-norm of ${\Phi}$, the ${C^1}$-norm of ${\psi}$, a lower bound for ${|\partial^{\alpha}_{\tau}J(\tau)|}$, ${\delta}$ and ${|\alpha|}$).