Sharpness of Zapolsky inequality for quasi-states and Poisson brackets
aa r X i v : . [ m a t h . S G ] J a n Sharpness of Zapolsky inequality forquasi-states and Poisson brackets
Anat AmirNovember 20, 2018
Abstract
Zapolsky inequality gives a lower bound for the L norm of thePoisson bracket of a pair of C functions on the two-dimensional sphereby means of quasi-states. Here we show that this lower bound is sharp. Denote by C ( S ) the Banach algebra of real continuous functions on S taken with the supremum norm.For F ∈ C ( S ), write C ( F ) = { ϕ ◦ F | ϕ ∈ C ( Im ( F )) } . That is, C ( F ) isthe closed sub-algebra generated by F and the constant function 1. Definition 1. A quasi-state on S is a functional ζ : C ( S ) → R satisfying:1. ζ ( F ) ≥ F ≥ ∀ F ∈ C ( S ), ζ is linear on C ( F ).3. ζ (1) = 1.Denote by Q ( S ) the collection of quasi-states on S .1 emark . It was proven in [1] that for a quasi-state ζ and a pair F, G ∈ C ( S ) we have: F ≤ G ⇒ ζ ( F ) ≤ ζ ( G ) . A quasi-state ζ is simple if for every F ∈ C ( S ), ζ is multiplicative on C ( F ). A quasi-state ζ is representable if it is the limit of a net of convexcombinations of simple quasi-states. That is, ζ is an element of the closedconvex hull of the subset of simple quasi-states.Denote by C and O the collections of closed and open subsets of S respec-tively. Write A = C ∪ O . Definition 2. A quasi-measure τ on S is a function τ : A → [0 ,
1] satisfying:1. τ ( S ) = 1.2. For B , B ∈ A with B ⊂ B , τ ( B ) ≤ τ ( B ).3. If { A k } nk =1 ⊂ A is a finite collection of pairwise disjoint subsets whoseunion is in A , then τ ( S nk =1 A k ) = P nk =1 τ ( A k ).4. For U ∈ O , τ ( U ) = sup { τ ( K ) : K ∈ C and K ⊂ U } .Denote by M ( S ) the collection of quasi-measures on S . A quasi-measure is simple if it only takes values of 0 and 1.It was proven in [1] that there exists a bijection between Q ( S ) and M ( S ).For a quasi-state ζ , the corresponding quasi-measure is: τ ( A ) = ( A ∈ C , inf { ζ ( F ) : F ∈ C ( S ) and F ≥ A } A ∈ O , − τ ( S \ A ) , here 1 A is the indicator function on the set A . The corresponding quasi-stateto a quasi-measure τ is defined as follows: ζ ( F ) = Z S F dτ = max S F − Z max S F min S F b F ( x ) dx , b F ( x ) = τ ( { F < x } ). It was proven in [2] that this bijection matchessimple quasi-states with simple quasi-measures. For further details aboutquasi-states and quasi-measures refer to [1] and for details on simple quasi-states and quasi-measures refer to [2].Throughout this paper we will be interested in the extent of non-linearityof a quasi-state. To measure this we will use the following notation: Definition 3.
Let ζ be a quasi-state and take F, G ∈ C ( S ). The extent ofnon-linearity of ζ can be measured by:Π( F, G ) := | ζ ( F + G ) − ζ ( F ) − ζ ( G ) | . Example 1.
One example of a simple quasi-state is Aarnes’ 3-point quasi-state.
Definition 4.
A subset S ⊆ S is called solid if it is connected and itscomplement S c = S \ S is also connected. Denote by C s the set of all closedand solid subsets of S and by O s the set of all open and solid subsets of S .Write A s = C s ∪ O s .Take p , p , p ∈ S to be three distinct points on the sphere. Define τ : C s → { , } by: τ ( C ) = ( , { C ∩ { p , p , p }} ≤ , { C ∩ { p , p , p }} ≥ . As proved in [3], τ can be extended to a quasi-measure on S . It is furthershown in that article that this extension is in-fact a simple quasi-measure.The simple quasi-state corresponding to the extended quasi-measure is calledAarnes’ 3-point quasi-state. We refer the reader to [3] for the full definitionof the extended quasi-measure τ . For our purpose it suffices to note that on A s , τ satisfies: τ ( S ) = ( , { S ∩ { p , p , p }} ≤ , { S ∩ { p , p , p }} ≥ . xample 2. Another example of a simple quasi-state is the median of aMorse function. Let Ω be an area form on S . The median of a Morsefunction F is the unique connected component of a level set of F , m F , forwhich every connected component of S \ m F has area ≤ · R S Ω. Define ζ on the set of Morse functions as ζ ( F ) = F ( m F ). As explained in [5], ζ canbe extended to C ( S ) and is in-fact a quasi-state. For further explanation ofthe concept of the median and the construction of ζ we refer the reader to[5]. It can be easily verified that the quasi-measure corresponding to ζ is theextension of τ : C s → { , } defined as: τ ( C ) = ( , R C Ω < · R S Ω1 , R C Ω ≥ · R S Ωto a quasi-measure on S as in [3]. In-fact, as explained in [3], this extensionis a simple quasi-measure, and hence ζ is a simple quasi-state. Let ω be an area form on S . Given a hamiltonian F : S → R , we definethe hamiltonian vector field IdF : S → T S by the formula: dF ( x )( η ) = ω ( η, IdF ( x )) , ∀ x ∈ S , η ∈ T x S . The hamiltonian flow with hamiltonian function F is the one-parametergroup of diffeomorphisms { g tF } satisfying: ddt (cid:12)(cid:12)(cid:12)(cid:12) t =0 g tF x = IdF ( x ) . If F, G are two hamiltonian functions on S , then their Poisson bracket isdefined as: { F, G } ( x ) = ddt (cid:12)(cid:12)(cid:12)(cid:12) t =0 F ( g tG ( x )) . The Poisson bracket also satisfies the following formula: { F, G } = dF ( IdG ) = − ω ( IdF, IdG ) . Remark . In this paper we are interested in the L -norm of the Poissonbracket. Note that on S we have: dF ∧ dG = − { F, G } · ω , therefore: k{ F, G }k L = Z S |{ F, G }| ω = Z S | dF ∧ dG | . Zapolsky’s inequality ([9], theorem 1.4) relates the extent of non-linearity ofa quasi-state to the L norm of the Poisson bracket. Let ζ be a representablequasi-state on S , then by Zapolsky’s inequality for every F, G ∈ C ( S ) wehave: Π( F, G ) ≤ k { F, G } k L . Note that this result can also be written as:sup
F,G ∈ C ( S ) Π( F, G ) k { F, G } k L ≤ . Our goal in this paper is to show that for some quasi-states Zapolsky’s in-equality is sharp. That is, we will show that there exist quasi-states forwhich: sup
F,G ∈ C ( S ) Π( F, G ) k { F, G } k L = 1 . Theorem 1.
Let ζ be Aarnes’ 3-point quasi-state, then: max F,G ∈ C ∞ ( S ) Π( F, G ) k { F, G } k L = 1 . heorem 2. Let ω be a normalized area form on S , that is R S ω = 1 , and ζ the corresponding median quasi-state. Then we have: sup F,G ∈ C ∞ ( S ) Π( F, G ) k { F, G } k L = 1 . Prior to proving theorem 1 we shall pay attention to the fact that any resultwe can prove for a certain 3-point quasi-state is true for all such quasi-states.
Remark . Let { p , p , p } and { q , q , q } be two sets of three distinct pointson the sphere S , and take ζ and ζ to be the two corresponding Aarnes’3-point quasi-states and Π and Π the corresponding measurements of theirnon-linearity. By a corollary to the isotopy lemma (see [6], 3.6) there existsa diffeomorphism h : S → S satisfying: h ( p i ) = q i , ≤ i ≤ . Since h is a diffeomorphism, both h and h − take solid subsets of the sphereto solid subsets, thus ζ ( F ◦ h ) = ζ ( F ) for every function F ∈ C ( S ). Whichyields: Π ( F, G ) = Π ( F ◦ h, G ◦ h ) . Also, we have: k { F ◦ h, G ◦ h } k L = Z S | d ( F ◦ h ) ∧ d ( G ◦ h ) | = Z S | h ∗ ( dF ∧ dG ) | = Z h ( S ) | dF ∧ dG | = Z S | dF ∧ dG | = k { F, G } k L . Thus: Π ( F ◦ h, G ◦ h ) k { F ◦ h, G ◦ h } k L = Π ( F, G ) k { F, G } k L . Proof of theorem 1
Proof.
Define S = { ( x, y, z ) ∈ R : x + y + z = 1 } ⊂ R . In sphericalcoordinates we have: S = n (cos θ cos φ, cos θ sin φ, sin θ ) : − π ≤ θ ≤ π ≤ φ ≤ π o . Consider the following points on S : p = (1 , , p = (0 , , p = (0 , , . Let ζ and τ be Aarnes’ 3-point quasi-state and quasi-measure correspondingto these points.Denote: D = { ( x, y ) : x + y ≤ } ∆ = { ( u, v ) ∈ R : u, v > u + v < } . We build a continuous function b ψ : D → cl (∆) (see the figure below) satis-fying : • b ψ maps the first quarter homeomorphically to ∆ along the radii. • b ψ maps the second quarter to the segment { } × [0 ,
1] of the y -axis. • b ψ maps the third quarter to the origin (0 , • b ψ maps the fourth quarter to the segment [0 , × { } of the x -axis.7 y ∆ IIIIII IV b ψ ( D ) We now build a smooth function ψ : D → cl (∆) by smoothening b ψ . For theprecise definition of ψ we will need an auxiliary smooth function α : R → [0 , α ( s ) = 0 , ∀ s ≤ α ( s ) = 1 , ∀ s ≥ α ′ ( s ) > , ∀ s ∈ (0 , s α ( s )Then we can define: ρ ( x, y ) = α (cid:0) x + 2 y − (cid:1) · α x + y p x + y ! , ∀ ( x, y ) ∈ D \ { (0 , } , the images below illustrate the behaviour of this function. x y −1 0 1−101 00.51 x y −1 0 1−101 00.51 x y −1 0 1−101 00.51 α (2 x + 2 y − α (cid:18) x + y √ x + y (cid:19) ρ ( x, y )8nd we take ψ : D → cl (∆) to be ψ ( p ) = ( f ( p ) , g ( p )), whereas: f ( x, y ) = x ≤ , < x , ρ ( x, y ) · α (cid:18) x √ x + y (cid:19) α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) g ( x, y ) = y ≤ , < y , ρ ( x, y ) · α (cid:18) y √ x + y (cid:19) α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) . Lemma 1. f, g are smooth functions.Proof.
The proofs of smoothness for f and g are very similar, therefore wewill give the proof only for f . To show that f is smooth, we need to showthat it is smooth on every point of its domain. Take a point ( x , y ) ∈ D ,and consider the following cases: • If x <
0, then f is identically zero in a neighbourhood of x , and hencesmooth. • If x >
0, then x > x , hence p x + y > f is a multiplication of smooth functions divided by smooth α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) . But, since x > α (cid:18) x √ x + y (cid:19) >
0, there-fore the denominator α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) >
0, and f is smooth. • If x = 0 and y <
0, we can find a neighbourhood U of ( x , y ) onwhich y < x + y ≤
0. But then in this neighbourhood we have x + y > x + y √ x + y ≤
0, hence α (cid:18) x + y √ x + y (cid:19) = 0 ⇒ ρ ( x, y ) = 0,9hich yields: f ( x, y ) | U = x > , · α (cid:18) x √ x + y (cid:19) α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) = 0 x ≤ , . Thus f is identically zero in this neighbourhood, and hence smooth. • If x = 0 and y >
0, we can find a neighbourhood of y such that y >
0, thus α (cid:18) y √ x + y (cid:19) >
0. In this neighbourhood the denominator, α (cid:18) x √ x + y (cid:19) + α (cid:18) y √ x + y (cid:19) >
0, thus f will be the multiplication ofsmooth functions for x > x ≤
0. From α ’s smoothnesswe have lim s → α ( m ) ( s ) = 0 for every derivative m ∈ N . If x >
0, everyderivative of f will be a finite sum of products, each of which has amultiplicand of the form α ( m ) (cid:18) x √ x + y (cid:19) for some m ∈ N , therefore:lim ( x,y ) → (0 ,y ) f ( n ) ( x, y ) = 0 , ∀ n ∈ N , and f is smooth. • Finally, if x = 0 and y = 0, we can find a neighbourhood of ( x , y )on which we have x + y < . But then: α (2 x + 2 y −
1) = 0 ⇒ ρ ( x, y ) = 0, and hence f is identically zero in this neighbourhood, thussmooth.We have shown that f is smooth on every point of D , thus f is a smoothfunction. In a similar manner it can be shown that g is also smooth.Denote: A = (cid:26) ( x, y ) ∈ D : x, y > < x + y < (cid:27) . emma 2. The restriction ψ | A is one-to-one and onto ∆ . Also, ψ ( D \ A ) ⊂ ∂ ∆ .Proof. On A we have x, y >
0, thus x + y > p x + y and α (cid:18) x + y √ x + y (cid:19) = 1.Hence: ( f + g ) | A = ρ | A = α (2 x + 2 y − . Similarly: (cid:18) fg (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) A = α (cid:18) x √ x + y (cid:19) α (cid:18) y √ x + y (cid:19) . In spherical coordinates we have: A = n (cos θ cos φ, cos θ sin φ ) : 0 < φ < π < θ < π o . Therefore: ( f + g ) | A = α (2 cos θ −
1) and (cid:18) fg (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) A = α (cos φ ) α (sin φ ) . Note that α is a bijection of (0 ,
1) to (0 ,
1) and (2 cos θ −
1) is a bijectionof (cid:0) , π (cid:1) to (0 , f + g ) | A = α (cos θ −
1) is a bijection of (cid:0) , π (cid:1) to(0 , d (cid:16) fg (cid:17)(cid:12)(cid:12)(cid:12) A dφ = d (cid:16) α (cos φ ) α (sin φ ) (cid:17) dφ = − α ′ (cos φ ) · α (sin φ ) · sinφ + α (cos φ ) · α ′ (sin φ ) · cos φα (sin φ )Recall that α ( s ) , α ′ ( s ) > s ∈ (0 ,
1) and that on A we have 0 < cos φ, sin φ <
1, therefore: d (cid:16) fg (cid:17)(cid:12)(cid:12)(cid:12) A dφ <
0, and (cid:16) fg (cid:17)(cid:12)(cid:12)(cid:12) A is a bijection of (cid:0) , π (cid:1) to (0 , ∞ ). 11e have shown that (cid:16) f + g, fg (cid:17)(cid:12)(cid:12)(cid:12) A is a bijection of A to (0 , × (0 , ∞ ).Since (cid:0) u + v, uv (cid:1) is a bijection of ∆ to (0 , × (0 , ∞ ), ψ | A is a bijection of A to ∆. We still have to show that ψ ( D \ A ) ⊂ ∂ ∆. Note that a point ( x, y ) ∈D \ A satisfies at-least one of these four conditions: • x ≤ f ( x, y ) = 0 and ψ ( x, y ) = (0 , g ( x, y )) ∈ ∂ ∆. • y ≤ g ( x, y ) = 0 and ψ ( x, y ) = ( f ( x, y ) , ∈ ∂ ∆. • x + y ≤ Here ρ ( x, y ) = α (2 x + 2 y − · α ( x + y √ x + y ) = 0 and ψ ( x, y ) = (0 , ∈ ∂ ∆. • x, y > x + y = 1Here ρ ( x, y ) = α (2 x + 2 y − · α (cid:18) x + y √ x + y (cid:19) = 1. hence f ( x, y ) + g ( x, y ) = 1 and ψ ( x, y ) = ( f ( x, y ) , g ( x, y )) ∈ ∂ ∆.Thus we have shown that ψ ( D \ A ) ⊂ ∂ ∆.Let P : S → R be the projection of the sphere to the xy -plane. Define: F, G : S → R by F = f ◦ P and G = g ◦ P . Our goal is to show that:Π ( F, G ) = k {
F, G } k L . We will begin by proving the following lemma:
Lemma 3. Π( F, G ) = 1 . Proof.
Note: (
F, G )( p ) = (1 , F, G )( p ) = (0 , F, G )( p ) = (0 , . p , p ∈ { ( x, y, z ) ∈ S : x ≤ } , and since the half-sphere is asolid subset of the sphere we have τ ( { ( x, y, z ) ∈ S : x ≤ } ) = 1. Also: (cid:8) ( x, y, z ) ∈ S : x ≤ (cid:9) ⊂ F − (0) ⊂ { F < t } , ∀ t > . Therefore: b F ( t ) = τ ( { F < t } ) = 1 , ∀ t > . In the same way we have p , p ∈ { ( x, y, z ) ∈ S : y ≤ } , and as thishalf-sphere is also a solid subset, we get once more τ ( { ( x, y, z ) ∈ S : y ≤ } ) =1. As before: (cid:8) ( x, y, z ) ∈ S : y ≤ (cid:9) ⊂ G − (0) ⊂ { G < t } , ∀ t > . Thus: b G ( t ) = τ ( { G < t } ) = 1 , ∀ t > . Last it should be noted that the arc: (cid:8) ( x, y, ∈ S : x, y ≥ x + y = 1 (cid:9) is also a solid subset of the sphere, and that: p , p ∈ (cid:8) ( x, y, ∈ S : x, y ≥ x + y = 1 (cid:9) . Therefore: τ ( (cid:8) ( x, y, ∈ S : x, y ≥ x + y = 1 (cid:9) ) = 1 . Since: (cid:8) ( x, y, ∈ S : x, y ≥ x + y = 1 (cid:9) ⊂ ( F + G ) − (1) , we have: τ (( F + G ) − (1)) = 1 . { F + G < } is 0. For every0 ≤ t ≤ { F + G < t } is a subset of { F + G < } , thus: b F + G ( t ) = τ ( { F + G < t } ) = 0 , ∀ t ≤ . Hence: ζ ( F ) = 1 − Z b F ( t ) dt = 1 − Z dt = 0 ζ ( G ) = 1 − Z b G ( t ) dt = 1 − Z dt = 0 ζ ( F + G ) = 1 − Z b F + G ( t ) dt = 1 − Z dt = 1 . And we get that:Π(
F, G ) = | ζ ( F + G ) − ζ ( F ) − ζ ( G ) | = | − − | = 1 . We now have to compute k {
F, G } k L . Recall that: k { F, G } k L = Z S | dF ∧ dG | = Z S | ( ψ ◦ P ) ∗ ( dx ∧ dy ) | . From lemma 1, ψ ◦ P is a smooth function, then, as a corollary to the changeof variables formula for a many-to-one function (see [8], theorem F.1) wehave: Z S | ( ψ ◦ P ) ∗ ( dx ∧ dy ) | = Z ψ ◦ P ( S ) n ( x, y ) · dx ∧ dy , with: n ( x, y ) = card (( ψ ◦ P ) − ( x, y )) . Also, by lemma 2, we know that ψ ◦ P covers ∆ exactly twice (since P projectsthe sphere twice onto A ), hence n ( x, y ) = 2 for ( x, y ) ∈ ∆. Thus: Z S | ( ψ ◦ P ) ∗ ( dx ∧ dy ) | = Z cl (∆) n ( x, y ) dx ∧ dy = Z ∆ dx ∧ dy = 1 . p , p , p we have:Π( F, G ) k { F, G } k L = 1 . Remark 3 concludes this proof for any 3-point quasi-state.
In the proof of theorem 2 we will use the fact that diffeomorphisms preservethe relation between the extent of non-linearity of median quasi-states andthe L -norm of the Poisson bracket. Remark . Let h : M → M be a diffeomorphism of surfaces. If ω is an areaform on M then h ∗ ω is an area form on M . Take ζ and ζ to be the medianquasi-states corresponding to h ∗ ω and ω . Recall that m F , the median of afunction F ∈ C ( M ), is the unique connected component of the level set F − ( ζ ( F )) ⊂ M satisfying R B ω ≤ R M ω for each connected component B of M \ m F . Since h, h − are continuous functions, they take connected setsto connected sets, therefore h − ( m F ) is a connected component of the levelset ( F ◦ h ) − ( ζ ( F )) ⊂ M . If A is a connected component of M \ h − ( m F ),then h ( A ) must be a connected component of M \ ( m F ). Therefore: Z A h ∗ ω = Z h ( A ) ω ≤ Z M ω = 12 Z M h ∗ ω . Thus h − ( m F ) must be the median of the function F ◦ h , which yields: ζ ( F ◦ h ) = ζ ( F ) . Therefore if Π and Π are the extents of non-linearity of the quasi-states ζ and ζ , we get: Π ( F ◦ h, G ◦ h ) = Π ( F, G ) . k { F ◦ h, G ◦ h } k L = Z M | d ( F ◦ h ) ∧ d ( G ◦ h ) | = Z M | h ∗ ( dF ∧ dG ) | = Z h ( M ) | dF ∧ dG | = Z M | dF ∧ dG | = k { F ◦ G } k L . Thus: Π ( F ◦ h, G ◦ h ) k { F ◦ h, G ◦ h } k L = Π ( F, G ) k { F, G } k L , and: sup F,G ∈ C ∞ ( M ) Π ( F, G ) k { F, G } k L = sup F,G ∈ C ∞ ( M ) Π ( F, G ) k { F, G } k L . roof of theorem 2 Proof.
Consider the triangle
ABC with vertices: A = (0 , B = (0 , C = (1 , xy -plane. For > ǫ > DK , EJ , IL with: D = (0 , ǫ ) E = (0 , − ǫ ) I = ( ǫ, K = (1 − ǫ, ǫ ) J = (1 − ǫ, L = ( ǫ, − ǫ ) . Let U be the triangle △ ABC after smoothing its corners by curves that donot intersect the segments DK , EJ and IL . Then the segments DK , EJ and IL divide U into seven parts, U , U , . . . , U . xyA B CDE I JKL ǫ − ǫ ǫ − ǫU U U U U U U Note that U ⊂ U ⊂ △ ABC , and hence:(1 − ǫ ) Area ( U ) < Area ( U ) < Area ( △ ABC ) = 12 . (1)17et u : U → [0 , ∞ ) be a function satisfying u − (0) = ∂U with 0 aregular value of u . And take S to be the surface in R defined as S := { z = u ( x, y ) } .Consider the following functions: • P : S → R defined as P ( x, y, z ) = ( x, y ) is the projection of S to theplane. Note that S \ P − ( ∂U ) has two connected components, n ( x, y, ± p u ( x, y )) : ( x, y ) ∈ int ( U ) o , both of which are projected diffeomorphically to int ( U ) by P . • F : S → R defined as F ( x, y, z ) = x . • G : S → R defined as G ( x, y, z ) = y .Then by (1) we get: k { F, G } k L = Z S | dF ∧ dG | = Z S | dx ∧ dy | = 2 · Area ( U ) ∈ ((1 − ǫ ) , . Let σ be an area form on S such that: Z P − ( U ) σ = Z P − ( U ) σ = Z P − ( U ) σ = 210and Z P − ( U ) σ = Z P − ( U ) σ = Z P − ( U ) σ = Z P − ( U ) σ = 110 . Note that σ is a normalized area form on S , and that each of the curves P − ( IL ) , P − ( DK ) and P − ( EJ ) divides S into two disks, one of area:210 + 110 + 210 = 510 = 12and the second of area:110 + 110 + 110 + 210 = 510 = 12 . ζ is the median quasi-state corresponding to σ , we get: ζ ( F ) = F ( IL ) = ǫζ ( G ) = G ( DK ) = ǫζ ( F + G ) = ( F + G )( EJ ) = 1 − ǫ . Therefore: Π(
F, G ) k { F, G } k L ≥ | − ǫ − ǫ − ǫ | −−→ ǫ → , and hence we have: sup F,G ∈ C ∞ ( S ) Π( F, G ) k { F, G } k L = 1 . Note that U is diffeomorphic to a closed disk, hence S is diffeomorphicto the sphere, and there exists a diffeomorphism h : S → S . Recall that σ is a normalized area form on S , hence σ = h ∗ σ is a normalized areaform on S . Let Π be the extent of non-linearity of the median quasi-statecorresponding to σ , then by remark 4 we have:sup F,G ∈ C ∞ ( S ) Π ( F, G ) k { F, G } k L = sup F,G ∈ C ∞ ( S ) Π( F, G ) k { F, G } k L = 1 . If σ is another normalized area form on S , then by Moser’s theorem (see[7], 13.2) there exists a diffeomorphism h : S → S , such that σ = h ∗ σ .If Π is the extent of non-linearity of the median quasi-state correspondingto σ , then by using remark 4 again, we will get:sup F,G ∈ C ∞ ( S ) Π ( F, G ) k { F, G } k L = sup F,G ∈ C ∞ ( S ) Π ( F, G ) k { F, G } k L = 1 . Thus, for every normalized area form ω on S , if Π is the extent of non-linearity of its corresponding median quasi-state, we have:sup F,G ∈ C ∞ ( S ) Π( F, G ) k { F, G } k L = 1 . cknowledgements This work is an M.Sc. thesis prepared under the supervision of Prof. LeonidPolterovich, the department of pure mathematics, Tel-Aviv university.I would like to thank both my supervisor Prof. Leonid Polterovich, and mycolleague Dr. Frol Zapolsky, for their help and support.This research was partially supported by the Israel Science Foundation grant509/07.
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