A geometric uncertainty principle with an application to Pleijel's estimate
aa r X i v : . [ m a t h . M G ] N ov A GEOMETRIC UNCERTAINTY PRINCIPLE WITHAN APPLICATION TO PLEIJEL’S ESTIMATE
STEFAN STEINERBERGER
Abstract.
One cannot decompose a domain into disks of equal radius: let Ω ⊂ R be anopen, bounded domain and Ω = S Ni =1 Ω i be a partition. Denote the Fraenkel asymmetry by0 ≤ A (Ω i ) ≤ D (Ω i ) := | Ω i | − min ≤ j ≤ N | Ω j || Ω i | with 0 ≤ D (Ω i ) ≤
1. For N sufficiently large depending only on Ω, there is an uncertaintyprinciple N X i =1 | Ω i || Ω | A (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! ≥ . The statement remains true in dimensions n ≥ c n >
0. As an application,we give an (unspecified) improvement of Pleijel’s estimate on the number of nodal domains ofa Laplacian eigenfunction similar to recent work of Bourgain and improve another inequality inthe field of spectral partition problems. Introduction
Motivation.
It is easy to partition R into sets of equal measure that are ’almost’ disks (thehexagonal packing, for example) and it is also possible to decompose R into disks of differentsize (Apollonian packings) – but obviously not both at the same time. We are interested in aquantitative descriptions of this phenomenon. Figure 1.
A partition into sets of equal measure and a partition into disks (onlylarge disks visible).This question turns out to have some relevance in the calculus of variations, in particular inthe study of vibrations of a membrane Ω ⊂ R as well as in spectral partition problems: givenan eigenfunction φ of the Laplacian − ∆ with Dirichlet boundary conditions on Ω, what is themaximal number of connected components of Ω \ { x ∈ Ω : φ ( x ) = 0 } ? Our quantitative study ofthis simple geometric principle in terms of Fraenkel asymmetry and size is very much motivatedby the applicability to nodal domain estimates – it could be of interest to capture the samephenomenon in other geometrically natural quantities. Mathematisches Institut, Universit¨at Bonn, Endenicher Allee 60, 53115 Bonn, Germany, e-mail: [email protected] . Geometric notions.
Let n ≥
2. Consider an open, bounded domain Ω ⊂ R n with a givendecomposition Ω = N [ i =1 Ω i . We require two quantities to measure(1) the deviation of Ω i from a ball(2) the deviation of | Ω i | from min ≤ j ≤ N | Ω j | . In measuring how much a set deviates from a ball, Fraenkel asymmetry has recently become anincreasingly central notion (i.e. [10]): given a domain Ω ⊂ R n , its Fraenkel asymmetry is definedvia A (Ω) := inf B | Ω △ B || Ω | , where the infimum ranges over all disks B ⊂ R n with | B | = | Ω | and △ is the symmetric differenceΩ △ B = (Ω \ B ) ∪ ( B \ Ω) . Fraenkel asymmetry is scale-invariant 0 ≤ A (Ω) ≤ . As for deviation in size, we define the deviation from the smallest element in the partition via D (Ω i ) := | Ω i | − min ≤ j ≤ N | Ω j || Ω i | , which is scale invariant as well and satisfies0 ≤ D (Ω i ) ≤ . Main result.
Our main result states that for partitions of Ω into a large number of sets,an average element of the partition needs to have either its Fraenkel asymmetry A (Ω i ) or itsdeviation from the smallest element D (Ω i ) bounded away from 0 by a universal constant. Thisstatement obviously fails if we only pick one of the two terms: any set can be decomposed into N sets of measure | Ω | /N each or each set can be decomposed into disks of different radii with anarbitrarily small measure of different shape (packings of Apollonian type). Theorem 1.
Suppose Ω ⊂ R n is an open and bounded domain and Ω = N [ i =1 Ω i with measurable sets Ω i satisfying Ω i ∩ Ω j = ∅ for i = j. There exists a universal constant c n > depending only on the dimension and a constant N ∈ N depending only on Ω such that for N ≥ N N X i =1 | Ω i || Ω | A (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! ≥ c n . In particular, c ≥ . Remarks. • Taking Ω to be the union of a finite number of disjoint balls of equal radius shows thatsuch a statement can only hold for N sufficiently large depending on Ω. • There are no assumptions whatsoever on the shape of Ω j – they need not be connected. GEOMETRIC UNCERTAINTY PRINCIPLE 3 • Fraenkel asymmetry turns the problem into a non-local one as the ’missing’ measure Ω △ B can be arbitrarily spread over the plane: this is why we believe that any argument yieldinga substantially improved constant will need to be based on significantly new ideas. Indeed,our proof will essentially only be a ’non-local perturbation’ of a local argument but nottruly non-local itself (hence the small constant). • What can be said about the optimal constant c n ? A natural candidate for an extremizerin R is the hexagonal tiling, which suggests that maybe c ∼ . . . . As packing density of spheres decreases in higher dimensions, we consider it extremelynatural to conjecture that c ≤ c ≤ . . . • The following interesting question is due to Almut Burchard: suppose the hexagonal pack-ing was indeed a minimizer; we can introduce a parameter α > α N X i =1 | Ω i || Ω | A (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! . It seems reasonable to conjecture that the hexagonal packing will then be a minimizer forevery 0 < α ≤
1. However, it is easy to see that there will be some α ≥ α > α . What happens at thetransition? Which configurations minimize the expression then?1.4. Variants and extensions.
There are many possible variations and extensions. We canwrite Fraenkel asymmetry as A (Ω) = inf x ∈ R n | Ω △ ( B + x ) || Ω | , where B is the ball centered at the origin scaled in such a way that | B | = | Ω | . However, thisdefinition can be easily generalized by considering other sets K instead of the ball if one correctsfor the arising lack of rotational symmetry, i.e. A K (Ω) := inf x ∈ R n inf R ∈R | Ω △ ( RK + x ) || Ω | , where K is scaled in such a way that | K | = | Ω | and R is the set of all rotations. The proof of ourmain statement is quite robust: it immediately allows to prove the following variant. Theorem 2.
Let K ⊂ R n be a bounded, convex set with a smooth boundary containing no linesegment. Then there exists a constant c ( K ) > such that for any open, bounded Ω ⊂ R n and anydecomposition Ω = N [ i =1 Ω i with measurable sets Ω i satisfying Ω i ∩ Ω j = ∅ for i = j and N sufficiently large, there is a geometric uncertainty principle N X i =1 | Ω i || Ω | A K (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! ≥ c ( K ) . This is certainly not the most general form of the theorem. Let S be the set of bounded sets in R n such that R n can be partitioned into translations and rotations of S . Suppose K is a boundedset satisfying inf S ∈S A S ( K ) > ε for some ε >
0. Does this already imply a geometric uncertainty principle for A K with a constantdepending only on ε ? STEFAN STEINERBERGER Application to spectral problems
Introduction.
Consider an open, bounded domain Ω ⊂ R . The Laplacian operator withDirichlet conditions gives rise to a sequence of eigenvalues ( λ n ) n ∈ N and associated eigenfunctions( φ n ) n ∈ N , where − ∆ φ n = λ n φ n in Ω φ n = 0 on ∂ Ω . Laplacian eigenfunctions are of great intrinsic interest and have been extensively studied. Onenatural question is to find bounds on the number of connected components ofΩ \ { x ∈ Ω : φ n ( x ) = 0 } . Let us denote this quantity by N ( φ n ). There are no nontrivial lower bounds on N ( φ n ) in general.Denoting the smallest positive zero of the Bessel function by j ∼ . . . . , the known upper boundsare as follows N ( φ n ) ≤ n (Courant, 1924)lim sup n →∞ N ( φ n ) n ≤ (cid:18) j (cid:19) (Pleijel, 1956)lim sup n →∞ N ( φ n ) n ≤ (cid:18) j (cid:19) − · − (Bourgain, 2013) , where (2 /j ) ≤ /
10. Polterovich [15] suggests that the optimal constant might be 2 /π ∼ . ⊂ R giving rise to a large number of nodal domainsneeds to have a completely integrable geodesic flow. Some numerical experiments in this directionhave been carried out by Blum, Gnutzmann & Smilansky [4].2.2. Pleijel’s argument.
Pleijel’s argument [14] is short and simple. Suppose the eigenfunction φ n induces a partition Ω = N [ i =1 Ω i . Then, by the Faber-Krahn inequality, λ n (Ω) ≥ λ (Ω i ) ≥ λ ( B ) , where B is the disk satisfying | B | = | Ω i | . However, λ ( B ) can be explicitely computed and theinequality then implies a lower bound on | B | . Combining this with Weyl’s law λ n ∼ πn/ | Ω | ,yields the result. Of course, this argument is only sharp if we have a decomposition of Ω into disksof equal radius.2.3. Bourgain’s argument.
Bourgain [5] employs a spectral stability estimate due to Hansen &Nadirashvili, which is formulated in terms of the inradius of a domain: for a nonempty, boundeddomain Ω ⊂ R , we have λ (Ω) ≥ " (cid:18) − r i (Ω) r o (Ω) (cid:19) λ (Ω ) , where Ω is the ball with | Ω | = | Ω | , r (Ω) is the radius of Ω and r i the inradius of Ω. The secondingredient is a packing result due to Blind [3]: the packing density of a collection of disks in theplane with radii a , a , . . . satisfying a i ≥ (3 / a j for all i, j is bounded from above by π/ √ . These two results imply the improvement.
GEOMETRIC UNCERTAINTY PRINCIPLE 5
An improved Pleijel estimate.
Exploiting stability estimates for the Faber-Krahn in-equality in terms of Fraenkel asymmetry, we are able to prove the following result.
Corollary.
There exists a constant ε > such that lim sup n →∞ N ( φ n ) n ≤ (cid:18) j (cid:19) − ε . An explicit value for ε would follow from an explicit constant in a Faber-Krahn stabilityresult involving Fraenkel asymmetry (these constants are known to exist but have not yet beendetermined explicitely). Given the general interest in this question, we are confident that sucha result will be eventually obtained. Much like Bourgain, however, we consider the underlyinggeometry more interesting than the actual numerical value – particularly in light of the followingobstruction.2.5. An obstruction.
Take Ω = [0 , of unit measure and cover it using the hexagonal covering(with obvious modifications at the boundary). Numerical computations (e.g. [13]) give that thefirst Laplacian eigenvalue of a hexagon H satisfies λ ( H ) ∼ . | H | . The Weyl law gives λ n (Ω) ∼ πn. We can place N hexagons of size | H | in Ω, where N | H | = 1 . Since we need to have λ n (Ω) ≥ λ ( H ), this implies4 πn ∼ . | H | and thus N = 1 | H | ∼ π . n ∼ . . . . n. As a consequence, any type of argument that leads to an improved Pleijel inequality with a constantsmaller than 0 . . . . will need to employ completely different arguments: the arguments given byPleijel, Bourgain and this paper argue based on the assumption that a partition of Ω into nodaldomains is given. However, such a partition could very well be the hexagonal partition. Argumentsleading to a better constant than 0 . . . . will need to explain why, say, an eigenfunction on adomain will not have eigenfunctions corresponding to a partition into hexagons.2.6. Spectral minimal partitions.
The problem of spectral minimal partitions is as follows:given a smooth, bounded domain Ω ⊂ R n and an integer k ∈ N , find among all partitions of Ωinto k disjoint domains Ω = k [ i =1 Ω i the one minimizing max ≤ i ≤ k λ (Ω i ) . It is conjectured that in two dimensions the minimal partitions should asymptotically behave likehexagonal tilings (with the exception of the boundary, which becomes neglible as k → ∞ ). Werefer to Caffarelli & Lin [7], Helffer & Hoffmann-Ostenhof & Terracini [12] and a survey of Helffer[11]. One basic inequality [12, Proposition 6.1] following immediately from Pleijel’s estimate isthat max ≤ i ≤ k λ (Ω i ) ≥ k πj | Ω | . STEFAN STEINERBERGER
Bourgain remarks that his argument also allows to slightly improve the constant in this inequality.As a second quantity that is sometimes minimized (see e.g. Caffarelli & Lin [7] or B´erard & Helffer[2]), one can consider the average and establish a strenghtened Pleijel-type estimatemax ≤ i ≤ k λ (Ω i ) ≥ k k X i =1 λ (Ω i ) ≥ k πj | Ω | . This inequality, too, can be strenghtened.
Corollary.
There exists a ε > such that for any smooth, bounded domain Ω ⊂ R and all k sufficiently large (depending on Ω ) k k X i =1 λ (Ω i ) ≥ ( πj + ε ) k | Ω | . Proof of Theorem 1 in two dimensions
This section contains a complete proof of the main statement in dimension n = 2: the proofwill track all arising constants. This takes up most of the text and contains all the ideas of thispaper – the argument is robust and the necessary (and rather easy) modifications to obtain themore general results will then be given in subsequent sections.3.1. Two possible strategies.
There seems to be a very natural way to prove the statement,however, we did not manage to fully quantify all the steps and had to find another argument. Werecord our original idea nonetheless in the hope of building additional insight.
Sketch of an idea.
The inequality can be regarded as a probabilistic statement. Pick a randomdomain weighted according to size (i.e. choosing a random point of the domain, the probabilityof picking Ω i is | Ω i | / | Ω | ). Our statement can be read as a lower bound on the expectation of therandom variable A (Ω i ) + D (Ω i ) . This motivates the following argument. Pick a random domain: either it already has large Fraenkelasymmetry (in which case we are done) or it does not and behaves quite disk-like. In the secondcase, we look at its neighbouring domains. If there are few adjacent domains, at least one ofthem touches along a long arc of the boundary meaning that the neighbouring domain has largeFraenkel asymmetry (two disks touch in at most one point). If there are many neighbours, eithermost are significantly smaller (making our randomly chosen domain big in comparison and givingthe statement) or some will need to get squeezed together because there is not enough room (cre-ating a large Fraenkel asymmetry). We believe that such a strategy, properly implemented, couldgive a relatively sharp constant – however, making all these steps quantitative seems complicated.
Sketch of a different idea: our proof.
We chose a different approach of a more global nature:given a decomposition, we immediately switch to a collection of N disks by taking disks realizingthe Fraenkel asymmetry for each partition. Then, we show that • there are few very large elements: the size of neighbourhood of the union of all disks whosesize is bounded away from the smallest element in the partition by a constant factor canbe bounded from above. • ignoring the large sets (of which there are few), the Fraenkel balls of small sets usually donot overlap too much; the exceptional set is small.Removing all large disks and all overlapping disks, we may shrink the remaining disks such thatno two of them overlap: the resulting disk packing cannot have too high a density. GEOMETRIC UNCERTAINTY PRINCIPLE 7
Defining quantities.
The limes inferior in the statement guarantees that boundary effectscoming from ∂ Ω become neglible and we will ignore the boundary throughout the proof (equiva-lently, we could have phrased the statement for periodic partitions of R ).We assume w.l.o.g. that | Ω | = 1. For a point x ∈ R and a set A ⊂ R , we abbreviate k x − A k := inf y ∈ A k x − y k . We introduce two numbers c , c > c = 1250 and c = 7250and the reader can substitute these values throughout the proof if he wishes to. Their role is asfollows: we call Ω i ’big’, if | Ω i | ≥ (1 + c ) min ≤ j ≤ N | Ω j | . (1)The constant c will serve as a measure of overlap: two disks with centers in x, y ∈ R and radii r , r will be considered to have ’large’ overlap if | x − y | ≤ (1 − c )( r + r ) . (2)We define a natural length scale η . Everything in this problem and this proof is scale-invariantand, correspondingly, the actual size of η is completely irrelevant throughout the proof: thevariable cancels in the end. However, we consider it helpful to imagine a fixed length scale η atwhich everything plays out and will phrase all arising quantities in terms of η , which we definevia πη = min ≤ i ≤ N | Ω i | . (3)The proof will be carried out via contradiction, we assume N X i =1 | Ω i || Ω | A (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! ≤ c for some small constant c and show that this will lead to a contradiction if c is small enough. Itmakes sense to be slightly more careful and so we assume that for all d , d ≥ c = d + d N X i =1 | Ω i || Ω | A (Ω i ) ! ≤ d (4)and N X i =1 | Ω i || Ω | D (Ω i ) ! ≤ d . (5)We assign to each of the sets Ω , . . . , Ω n a disk B , B , . . . , B n such that | B i | = | Ω i | and A (Ω i ) = | Ω i △ B i || Ω i | . Note that a disk B i need not be uniquely determined by Ω i (if there is more than one possiblechoice, we pick an arbitrary one and fix it for the rest of the proof). Each of these disks B i has acenter x i and a radius r i ≥ η . STEFAN STEINERBERGER
The union of large sets has small measure.
Here we prove a simple statement: themeasure occupied by ’large’ sets (in the sense of (1)) is small. Note that the statement is indeedfor the measure and not the number of large sets, which could be small.
Lemma.
We have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ | Ω i | > (1+ c ) πη Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ d c + d . (6) Proof.
From (3), (5) and the definition of D (Ω i ), we get that d ≥ N X i =1 | Ω i || Ω | D (Ω i ) = N X i =1 ( | Ω i | − πη ) = 1 − N πη and therefore N ≥ − d πη . Now, let us suppose that 0 ≤ M ≤ N elements of the partition are ’small’ in the sense of satisfying | Ω i | ≤ (1 + c ) πη . We wish to show that M itself has to be big. Trivially, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ | Ω i |≤ (1+ c ) πη Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ M πη . The remaining measure is divided among big sets, hence the number of ’big’ elements is at mostthe remaining measure divided by the smallest possible area a ’big’ set can have N − M ≤ − M πη (1 + c ) πη and thus, in total, 1 − d πη ≤ N = M + ( N − M ) ≤ M + 1 − M πη (1 + c ) πη . Rewriting gives M ≥ c − d − d c c πη , which implies (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ | Ω i |≤ (1+ c ) πη Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c − d − d c c and therefore, since | Ω | = 1, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ | Ω i | > (1+ c ) πη Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ d c + d . (cid:3) A neighbourhood of the union of large sets has small measure.
In the last sectionwe have seen that measure of the set of large disks is small. However, we actually require a slightlystronger statement showing that an entire neighbourhood of that set is still small. For future use,we define the index set I of partition elements with ’big’ measure I = (cid:8) i ∈ { , . . . , N } : | Ω i | ≥ (1 + c ) πη (cid:9) . Lemma. A η − neighbourhood of S i ∈ I B i has small measure: we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)( x ∈ Ω : (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) x − [ i ∈ I B i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ η )(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ d c + 9 d . (7) GEOMETRIC UNCERTAINTY PRINCIPLE 9
Proof.
This argument is very simple: the 2 η − neighbourhood of a disk with radius r has measure( r + 2 η ) π . The worst case is precisely the case, where all B i are well-separated such that their2 η − neighbourhoods do not intersect (otherwise: move the disks apart to create a neighbourhoodwith bigger measure). In this case, the total measure gets amplified by factor( √ c + 2) η π (1 + c ) η π ≤ (cid:3) Most small sets have well-separated balls.
By now, we have a good control on the ’large’disks and their neighbourhood: we can (mentally and later in the proof literally) remove themfrom the stage and consider the remaining small disks. It remains to control their intersections.
Lemma.
The union of ’small’ disks B i , i = I , for which there exists another ’small’ disk suchthat they intersect strongly in the sense of (2) is bounded by (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)[ i/ ∈ I (cid:8) B i : ∃ j / ∈ I i = j : | x i − x j | ≤ (1 − c )( r i + r j ) (cid:9)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ π
37 1 + c c / d (8) Proof.
For simplicity, we introduce the index set J = (cid:8) i / ∈ I : ∃ B i ∃ j / ∈ I i = j : | x i − x j | ≤ (1 − c )( r i + r j ) (cid:9) . We will now derive an upper bound on the measure of the set, which we now can abbreviate as ∪ j ∈ J B j , using nothing but the inequality (5) N X i =1 | Ω i |A (Ω i ) ≤ d . Suppose i ∈ J . Then there exists a j ∈ J such that the balls B i , B j have controlled radius (thisfollows automatically from the fact that both disks are ’small’) η ≤ r i , r j ≤ √ c η and intersect in a quantitatively controlled way | x i − x j | ≤ (1 − c )( r i + r j ) . Then the intersection B i ∩ B j is of interest: if the Fraenkel asymmetry of Ω i is to be small, thenalmost all of its measure should be contained in B i but the very same reasoning also holds for Ω j and B j . In particular, since every point in the intersection can only belong to one of the two sets,we have | Ω i |A (Ω i ) + | Ω j |A (Ω j ) ≥ | B i ∩ B j | . (9)It remains to compute the quantity | B i ∩ B j | . Using scaling invariance, we may assume η = 1.We are then dealing with two disks in the Euclidean plane whose radii r , r are bounded frombelow by 1 and whose centers x , x satisfy d := | x − x | ≤ (1 − c )( r + r ) . Elementary Euclidean geometry yields | B i ∩ B j | = r arccos (cid:18) d + r − r dr (cid:19) + r arccos (cid:18) d + r − r dr (cid:19) − p ( − d + r + r )( d + r − r )( d − r + r )( d + r + r ) . Easy but tedious calculation give that the quantity is decreasing in both radii and as such mini-mized for r = r = 1. This is then a one-dimensional function in c and it is easy to show thatfor c ≤ .
05 the function is2 arccos (1 − c ) − p (2 − c )(1 − c ) c ≥ c . Recalling the normalization η = 1, we get the scale-invariant estimate | B i ∩ B j | ≥ c η . A priori, the intersection patterns of { B i : i ∈ J } can be very complicated. However, there is avery simple monotonicity: we can remove areas, where three or more balls intersect and arrangethe balls in (possibly more than one) chain. This increases the area and decreases the area ofintersection. By the same argument, the area further increases if we assume that any disk in Figure 2.
Increasing area while decreasing average Fraenkel asymmetry { B i : i ∈ J } touches precisely one other disk (i.e. the intersection pattern reduces to that of pairsof disks intersecting each other and no disk). Any such (i.e. intersecting) pair of disks B i , B j satisfies | B i ∪ B j | ≤ c ) πη as well as | B i ∩ B j | ≥ c η , which is connected via (9) to the sum N X i =1 | Ω i |A (Ω i ) ≤ d . Thus (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ j ∈ J B j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:2) c ) πη (cid:3) d c / η = 20 π
37 1 + c c / d . (cid:3) Bounds on the size of the neighbourhood of strongly intersecting small disks.
Byapplying the very same reasoning as in Section 3.4, we could argue that by considering an entire2 η -neighbourhood the measure gets amplified by a factor of at most 9. This is perfectly reasonablebut can actually be improved as we are now dealing with disks intersecting other disks. We arethus studying the following problem: given two disks B , B with radii r , r ≥ η intersecting inprecisely one point, what bounds can be proven on | (cid:8) x ∈ R : k x − ( B ∪ B ) k ≤ η (cid:9) || B | + | B | ≤ ?This problem can be explicitely solved using elementary calculus and reduces to a case-distinctionand two integrations; we leave the details to the interested reader. Carrying out the calculationsgives | (cid:8) x ∈ R : k x − ( B ∪ B ) k ≤ η (cid:9) || B | + | B | ≤
92 + 2 √ π + 9 π arcsin (cid:18) (cid:19) ∼ . · · · ≤ r = r = 1. Arguing as in Section 3.4 and using (8), we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ∈ Ω : (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) x − [ j ∈ J B j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ η (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ π
37 (1 + c ) c / d (10) GEOMETRIC UNCERTAINTY PRINCIPLE 11
Finding a dense disk packing.
We conclude our argument by deriving the existence of adisk packing in the plane with impossible properties. Here, we employ an aforementioned resultof Blind [3] that also played a role in Bourgain’s argument and was mentioned before: the packingdensity of a collection of disks in the plane with radii a , a , . . . satisfying a i ≥ (3 / a j for all i, j is bounded from above by π/ √ . A rough outline of the remainder of the argument is as follows(1) consider the set of Fraenkel disks { B i : 1 ≤ i ≤ N } (2) remove all ’big’ disks(3) remove all remaining ’small’ disks strongly intersecting another small disk(4) shrink all remaining disks by a factor (1 − c ).This leaves us with a set of disjoint disks in the Euclidean plane with roughly the same radiusand we can apply Blind’s result – the argument has one big flaw, of course, removing elementsfrom a set does not increase its packing density (just think of a hexagonal packing of disks: if weremove the little triangle-shaped gaps between the disks, packing density goes up to 1).We counter the problem by not only removing ’big’ disks or ’small’ disks strongly intersectingother small disks but an entire 2 η − neighbourhood of these sets as well. Doing this is equivalentto assuming that while we created holes in the middle of the set, these holes are of such a shapethat within a neighbourhood we can actually achieve packing density 1.From (7) and (10), we get that the setΩ ∗ := Ω \ ( x ∈ Ω : (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) x − [ i ∈ I B i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ η ) ∪ x ∈ Ω : (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) x − [ j ∈ J B j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ η satisfies | Ω ∗ | ≥ − d c + 9 d + 128 π
37 (1 + c ) c / d ! . Ω ∗ consists of disks with radii satisfying η ≤ r i ≤ √ c η and with the additional property that the centers of any two disks are well-seperated | x i − x j | ≥ (1 − c )( r i + r j ) . By shrinking all these disks by a factor of 1 − c while keeping their center in the same place, theybecome disjoint. Thus, from Blind’s result | (1 − c )Ω ∗ | ≤ (1 − c ) " − d c + 9 d + 128 π
37 (1 + c ) c / d ! ≤ π √ . We need to find a set of parameters, for which the inequality fails. Indeed, setting c = 1250 and c = 7250 , we get for any d , d ≥ d + d = 160000 , that (1 − c ) " − d c + 9 d + 128 π
37 (1 + c ) c / d ! ≥ π √
12 + 11000 . This contradiction proves the statement. (cid:3)
Remark.
The weakest point in the argument is certainly the last step, where we remove anentire 2 η − neighhbourhood. Intuition suggests that we should be able that maybe even removing merely a η -neighbourhood should be more than sufficient, however, we have not been able tomake progress on that question, which would certainly be the most natural starting point if onewanted to improve the constant using arguments along these lines.4. Proof of the general case.
Here we give a proof of Theorem 2 in general dimensions (which contains Theorem 1 for n ≥ − c ) " − d c + 9 d + 128 π
37 (1 + c ) c / d ! ≥ π √
12 + 11000 . The crucial point is the following: no matter what actual numerical values are placed in front, bychoosing d ≪ c and d ≪ c , the inequality will always be false for c , c sufficiently close to 1by simple continuity. In the previous proof, it was our goal to keep d , d as large as possible butonce we discard this concern, we can much more wasteful in the actual geometric estimates. Proof.
The argument is again by contradiction. η plays a similar same role as before, we defineit via η = (cid:18) min ≤ j ≤ N | Ω j | (cid:19) /n . The constant c again determines whether a domain is ’big’, which we define to be the case if | Ω i | ≥ (1 + c ) min ≤ j ≤ N | Ω j | . The precise meaning of c is introduced further below. Arguing by contradiction we assume that N X i =1 | Ω i || Ω | A K (Ω i ) ! + N X i =1 | Ω i || Ω | D (Ω i ) ! ≤ c and want to derive a contradiction for c sufficiently small. Following the same argument as before,we again get a bound on the number of large sets (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ | Ω i | > (1+ c ) η n Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ cc + c. Switching again to the Fraenkel bodies K , . . . , K N , we wish to remove a c η neighbourhood ofany ’large’ Fraenkel body K i , where c < ∞ is chosen such that c η is many multiples of thediameter of a ’small’ K i having measure at most (1 + c ) η n . This allows us to bound the size ofa c η neighbourhood of [ | Ω i | > (1+ c ) η n K i by c ( c/c + c ) for some finite constant c . The constant c now measures whether two ’small’Fraenkel bodies have large intersection, writing again I = { i ∈ { , . . . , N } : | Ω i | ≥ (1 + c ) η n } , we consider [ i/ ∈ I { K i : ∃ j i = j / ∈ I : | ( K i ∩ K j ) | ≥ c η n } . The same argument as before implies that for any two elements in the set, we get A K ( K i ) | K i | + A K ( K j ) | K j | ≥ c η n . Since N X i =1 | Ω i || Ω | A K (Ω i ) ! ≤ c, GEOMETRIC UNCERTAINTY PRINCIPLE 13 this implies a bound on the measure of the set (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)[ i/ ∈ I { K i : ∃ j i = j / ∈ I : | ( K i ∩ K j ) | ≥ c η n } (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c c for some constant c < ∞ and a bound of the form c c on the measure of its c η neighbourhood.Finally, since the boundary of the convex body K contains no line segment, we get that for every ε > ε > K , K , . . . of nonoverlapping rotated and scaledtranslates of K in the plane with volumes v , v , . . . satisfyinginf i,j v i v j ≥ − ε has packing density at most 1 − ε . Finally, there exists a constant c such that for any two scaled,translated copies K , K of K with | ( K i ∩ K j ) | ≤ c η n , the rescaled bodies c K , c K (rescaling being done in a way to fix, say, their center of mass)satisfy ( c K ) ∩ ( c K ) = ∅ . Note that the optimal c depends continuously on c and tends to 1 as c tends to 0. Now,following the same argument as before, we can derive the inequality1 − ε ≥ c n (cid:18) − c cc − c c − c c (cid:19) . The dependence is easy: pick some 0 < ε ≪
1. This yields ε >
0. Given ε , pick c ≪ ε . Wepick c so small that c n > − ε . c and c are again externally given but the inequality can nowbe seen to be false if c = 0. By continuity c > (cid:3) Proof of the improved Pleijel estimate.
The Corollary has a very simple proof: as inthe proof of Pleijel’s estimate, we get a lower bound onmin ≤ i ≤ N | Ω i | from the Faber-Krahn inequality. Theorem 1 now implies that either not all elements in thepartition are of that size (in which case some need to be bigger and their requirement for morespaces allows for a smaller number of nodal domains) or that some deviate from the disk in acontrolled way (in which case stability estimates require them to have a larger measure). Proof.
Let Ω = N [ i =1 Ω i be the decomposition introduced by a Laplacian eigenfunction with eigenvalue λ ≫ η = η ( λ ) be chosen in such a way that πη = | B | , where B is the disk such that λ ( B ) = λ .Theorem 1 yields that N X i =1 | Ω i || Ω | ( A (Ω i ) + D (Ω i )) ≥ c for some c ≥ / N X i =1 | Ω i || Ω | D (Ω i ) ≥ c N X i =1 | Ω i || Ω | A (Ω i ) ≥ c . Suppose the first inequality holds. Then c ≤ N X i =1 | Ω i || Ω | D (Ω i ) = 1 | Ω | (cid:0) | Ω | − N πη (cid:1) in which case N ≤ (cid:16) − c (cid:17) | Ω | πη . The fact that Pleijel’s argument is sharp for a partition into equally sized disks (or, equivalently,Weyl’s law) implies lim λ →∞ | Ω | πη = (cid:18) j (cid:19) n and this yields the result. Suppose the second inequality holds. We start with a simple Lemma. Lemma.
We have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≥ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c | Ω | . (11) Proof of the Lemma.
Suppose the statement was false. Then, using A (Ω i ) ≤ c ≤ N X i =1 | Ω i || Ω | A (Ω i ) < | Ω | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≥ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + c | Ω | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≤ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c c c . (cid:3) Now we recall some stability estimates for the Faber-Krahn inequality in terms of Fraenkelasymmetry. Brasco, De Philippis & Velichkov [6] (improving an earlier result of Fusco, Maggi &Pratelli [9]) have shown that λ (Ω) − λ (Ω ) λ (Ω ) & A (Ω) , where Ω is again the disk with | Ω | = | Ω | .Pick any domain Ω i with A (Ω i ) ≥ c/ B to denote the disk such that | B | = | Ω i | . Thestability estimate λ (Ω i ) − λ ( B ) λ ( B ) ≥ C · A (Ω i ) can be rewritten as λ (Ω i ) ≥ (cid:18) C c (cid:19) λ ( B ) (12)Recall that η is chosen such that the disk D of radius η satisfies λ ( D ) = λ n (Ω). However, by(12), we know that λ (Ω i ) is a multiplicative factor larger than the first eigenvalue of the disk ofequal measure. In order for λ (Ω i ) ≤ λ n (Ω) to still be satisfied, we require that | Ω i | πη ≥ C c . (13)We use this as follows: Ω = N [ i =1 Ω i = [ A (Ω i ) ≥ c Ω i ∪ [ A (Ω i ) ≤ c Ω i By Pleijel’s argument, the number of nodal domains in the second set is bounded from above by1 πη (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≤ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) GEOMETRIC UNCERTAINTY PRINCIPLE 15 while (13) implies the number of nodal domains in the first set is bounded by1 πη
11 + C c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≥ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Using (11) and | Ω | = 1, we get1 πη
11 + C c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≥ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 1 πη (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ A (Ω i ) ≤ c Ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ πη
11 + C c c πη (cid:16) − c (cid:17) ≤ (cid:18) − c C
216 + 6 c C (cid:19) πη . By definition of η , the expression ( πη ) − is precisely the upper bound of Pleijel on the numberof nodal domains and thus N ≤ (cid:18) − c C
216 + 6 c C (cid:19) (cid:18) j (cid:19) n for n sufficiently big.4.2. Proof of the Spectral Partition Inequality.
Proof.
Suppose the statement was false. Then there exists some smooth, bounded Ω such that forany ε > k such that there are partitionsΩ = k [ i =1 Ω i with 1 k k X i =1 λ (Ω i ) ≤ ( πj + ε ) k | Ω | . Let us start by re-iterating the proof of the original estimate. The Faber-Krahn inequality implies1 k k X i =1 λ (Ω i ) ≥ k k X i =1 πj | Ω i | . The convexity of x → /x and the fact that k X i =1 | Ω i | = | Ω | immediately imply that 1 k k X i =1 πj | Ω i | ≥ k πj | Ω | with equality if and only if | Ω i | = | Ω | /k for all 1 ≤ i ≤ k . We can quantify the notion of convexitya little bit. Indeed, assuming our desired spectral partition inequality to be false and given any δ >
0, we can find a subsequence of partitions with1 k j { ≤ i ≤ k j : (1 − δ ) | Ω | ≤ k j | Ω i | ≤ (1 + δ ) | Ω |} → . This, however, means that we can find a subsequence of partitions with the propery that k j X i =1 | Ω i || Ω | D (Ω i ) ≤ δ ′ , where δ ′ > k j X i =1 | Ω i || Ω | A (Ω i ) ≥ c − δ ′ , where c > / (cid:3) Acknowledgments.
The author is grateful for various discussions with Bernhard Helffer, whotaught him about spectral partition problems; this interaction took place at a workshop in theBanff International Research Station which is to be thanked for its hospitality. Suggestions froman anonymous referee greatly increased the quality of exposition.