Games for eigenvalues of the Hessian and concave/convex envelopes. - PowerPoint PPT Presentation

Games for eigenvalues of the Hessian and concave/convex envelopes. Julio D. Rossi (joint work with Pablo Blanc) U. Buenos Aires (Argentina) jrossi@dm.uba.ar www.dm.uba.ar/ ∼ jrossi U. Carlos III, Madrid, Spain, 2018 beamer-tu-logo

Eigenvalues of D 2 u For a function u : Ω ⊂ R n �→ R we denote � ∂ 2 u � D 2 u = ∂ x i ∂ x j i , j and λ 1 ( D 2 u ) ≤ λ 2 ( D 2 u ) ≤ .... ≤ λ j ( D 2 u ) ≤ ....λ n ( D 2 u ) the ordered eigenvalues of the Hessian D 2 u . Notice that ∆ u = λ 1 ( D 2 u ) + ... + λ n ( D 2 u ) . beamer-tu-logo

Main goals For the problem � λ j ( D 2 u ) = 0 , in Ω , u = F , on ∂ Ω . • Relate solutions to convex/concave envelopes of the boundary datum F . • Find a necessary and sufficient condition on the domain Ω in such a way that this problem has a viscosity solution that is continuous in Ω for every F ∈ C ( ∂ Ω) . • Show a connection with probability (game theory). beamer-tu-logo

Convex envelopes A function u : Ω ⊂ R n �→ R is convex if u ( λ x + ( 1 − λ ) y ) ≤ λ u ( x ) + ( 1 − λ ) u ( y ) . Given F : ∂ Ω �→ R the convex envelope of F in Ω is u ∗ ( x ) = u ( x ) . sup u convex , u | ∂ Ω ≤ F That is, u ∗ is the largest convex function that is below F on ∂ Ω beamer-tu-logo

Concave envelopes u is concave if u ( λ x + ( 1 − λ ) y ) ≥ λ u ( x ) + ( 1 − λ ) u ( y ) . Given F : ∂ Ω �→ R the concave envelope of F in Ω is u ∗ ( x ) = u concave , u | ∂ Ω ≥ F u ( x ) . inf Notice that in an interval ( a , b ) ⊂ R , it holds that u ∗ ( x ) = u ∗ ( x ) = ( u ( b ) − u ( a )) ( x − a ) + u ( a ) . b − a beamer-tu-logo

Convex envelopes If u ∈ C 2 is convex then D 2 u ( x ) must be positive semidefinite, � D 2 u ( x ) v , v � ≥ 0 . In terms of the eigenvalues of D 2 u this can be written as λ 1 ( D 2 u ( x )) = inf | v | = 1 � D 2 u ( x ) v , v � ≥ 0 . Moreover, the convex envelope of F in Ω is the largest viscosity solution to � λ 1 ( D 2 u ) = 0 , in Ω , u ≤ F , on ∂ Ω . A. Oberman – L. Silvestre (2011). beamer-tu-logo

Concave / convex envelopes Let H j be the set of functions v such that v ≤ F on ∂ Ω , and have the following property: for every S affine of dimension j and every j − dimensional domain D ⊂ S ∩ Ω it holds that v ≤ z in D where z is the concave envelope of v | ∂ D in D . beamer-tu-logo

Concave / convex envelopes Theorem The function u ( x ) = sup v ( x ) v ∈ H j is the largest viscosity solution to λ j ( D 2 u ) = 0 in Ω , with u ≤ F on ∂ Ω . The equation for the concave envelope of F | ∂ Ω in Ω is just λ n = 0; while the equation for the convex envelope is λ 1 = 0. beamer-tu-logo

Condition (H) A comparison principle (hence uniqueness) for the equation λ j ( D 2 u ) = 0 was proved in F .R. Harvey, H.B. Jr. Lawson, (2009). For the existence, it is assumed Condition (H): the domain is smooth (at least C 2 ) and such that κ 1 ≤ κ 2 ≤ ... ≤ κ n − 1 , the main curvatures of ∂ Ω , verify κ j ( x ) > 0 and κ n − j + 1 ( x ) > 0 , ∀ x ∈ ∂ Ω This condition is used to construct barriers. beamer-tu-logo

Condition (G) Our geometric condition on the domain reads as follows: Given y ∈ ∂ Ω we assume that for every r > 0 there exists δ > 0 such that for every x ∈ B δ ( y ) ∩ Ω and S ⊂ R n a subspace of dimension j , there exists v ∈ S of norm 1 such that ( G j ) { x + tv } t ∈ R ∩ B r ( y ) ∩ ∂ Ω � = ∅ . We say that Ω satisfies condition (G) if it satisfies both ( G j ) and ( G N − j + 1 ) . beamer-tu-logo

Condition (G) Figure: Condition in R 3 . We have ∂ Ω in blue, B δ ( y ) in green and S in beamer-tu-logo red.

Theorem The problem � λ j ( D 2 u ) = 0 , in Ω , u = F , on ∂ Ω . has a continuous solution (up to the boundary) for every continuous data F if and only if Ω satisfies condition (G) . beamer-tu-logo

The Laplacian and a random walk Let us consider a final payoff function F : R n \ Ω �→ R . In a random walk with steps of size ǫ from x the position of the particle can move to x ± ǫ e j , each movement being chosen at random with the same probability, 1 / 2 n . We assumed that Ω is homogeneous and that every time the movement is independent of its past history. beamer-tu-logo

The Laplacian, ∆ Let u ǫ ( x ) = E x ( F ( x N )) be the expected final payoff when we move with steps of size ǫ . Applying conditional expectations we get � 1 n 2 nu ǫ ( x + ǫ e j ) + 1 � � u ǫ ( x ) = 2 nu ǫ ( x − ǫ e j ) . j = 1 That is, n � � � 0 = u ǫ ( x + ǫ e j ) − 2 u ǫ ( x ) + u ǫ ( x − ǫ e j ) . j = 1 beamer-tu-logo

The Laplacian, ∆ Now, one shows that u ǫ converge as ǫ → 0 to a continuous function u uniformly in Ω . Then, we get that u is a viscosity solution to the Laplace equation � − ∆ u = 0 in Ω , u = F on ∂ Ω . beamer-tu-logo

Tug-of-War games Rules Two-person, zero-sum game: two players are in contest and the total earnings of one are the losses of the other. Player I, plays trying to minimize his expected outcome. Player II is trying to maximize. Ω ⊂ R n , bounded domain and F : R n \ Ω → R a final payoff function. Starting point x 0 ∈ Ω . At each turn, Player I chooses a subspace S of dimension j and then Player II chooses v ∈ S with | v | = 1. The new position of the game is x ± ǫ v with probability (1/2–1/2). Game ends when x N �∈ Ω , Player I earns F ( x N ) (Player II beamer-tu-logo earns − F ( x N ) )

Remark The sequence of positions { x 0 , x 1 , · · · , x N } has some probability, which depends on The starting point x 0 . The strategies of players, S I and S II . Expected result Taking into account the probability defined by the initial value and the strategies: E x 0 S I , S II ( F ( x N )) ”Smart” players Player I tries to choose at each step a strategy which minimizes the result. Player II tries to choose at each step a strategy which beamer-tu-logo maximizes the result.

Extremal cases E x u I ( x ) = sup inf S I , S II ( F ( x N )) S II S I E x u II ( x ) = inf sup S I , S II ( F ( x N )) S II S I The game has a value ⇔ u I = u II . Theorem This game has a value u ǫ ( x ) . beamer-tu-logo

Dynamic Programming Principle Main Property (Dynamic Programming Principle) � 1 � 2 u ǫ ( x + ǫ v ) + 1 u ǫ ( x ) = 2 u ǫ ( x − ǫ v ) inf sup dim ( S )= j v ∈ S , | v | = 1 beamer-tu-logo

Idea If λ 1 ≤ ... ≤ λ N are the eigenvalues of X , with corresponding orthonormal eigenvectors v 1 , ..., v N and v = � N i = 1 a i v i , then N � ( a i ) 2 λ i . � Xv , v � = i = 1 From this expression it can be easily deduced that the j − st eigenvalue verifies min v ∈ S , | v | = 1 � Xv , v � = λ j . max dim ( S )= j beamer-tu-logo

Condition (F) Given y ∈ ∂ Ω we assume that there exists r > 0 such that for every δ > 0 there exists T ⊂ R n a subspace of dimension j , v ∈ R n of norm 1, λ > 0 and θ > 0 such that { x ∈ Ω ∩ B r ( y ) ∩ T λ : � v , x − y � < θ } ⊂ B δ ( y ) ( F j ) where T λ = { x ∈ R n : d ( x , T ) < λ } . For our game we assume that Ω satisfies both ( F j ) and ( F N − j + 1 ). Theorem : ( H ) ⇒ ( F ) ⇒ ( G ) . beamer-tu-logo

Theorem Assume (F) . Then u ǫ → u , as ǫ → 0 , uniformly in Ω . The limit u is the unique viscosity solution to � λ j ( D 2 u ) = 0 , in Ω , u = F , on ∂ Ω . beamer-tu-logo

References I. Birindelli, G. Galise and I. Ishii, to appear. L. Caffarelli, L. Nirenberg and J. Spruck, (1985). M.G. Crandall, H. Ishii and P .L. Lions. (1992). F .R. Harvey, H.B. Jr. Lawson, (2009) J. J. Manfredi, M. Parviainen and J. D. Rossi, (2010). A. M. Oberman and L. Silvestre, (2011). beamer-tu-logo

THANKS !!!. GRACIAS !!!. beamer-tu-logo