Lower estimates on the saturation order of approximation of twice continuously differentiable functions by piecewise constants on convex partitions

Розглядається задача про порядок наближення двiчi неперервно дифференцiйовних функцiй багатьох змiнних кусково-сталими сплайнами на опуклих розбиттях. Показано, що порядок насичення наближення кусково-сталими сплайнами в Lp нормi на опуклих розбиттях з N елементiв становить N−2/(d+1), де d – число змiнних. Ключовi слова: двiчi неперервно диференцiйовнi функцiї, кусково-сталi сплайни, порядок насичення, опуклi розбиття. Рассматривается задача про порядок приближения дваджды непрерывно дифференцируемых функций многих переменных кусочно-постоянными на выпуклых разбиениях. Показано, что порядок насыщения приближения кусочно-постоянными в Lp норме на выпуклых разбиениях с N элементами будет N−2/(d+1), где d – число переменных. Ключевые слова: дважды непрерывно дифференцируемые функции, кусочно-постоянные сплайны, порядок насыщения, выпуклые разбиения. We consider the problem of approximation order of twice continuously differentiable functions of many variables by piecewise constants. We show that the saturation order of piecewise constant approximation in Lp norm on convex partitions with N cells is N−2/(d+1) , where d is the number of variables.

We consider the problem of approximation order of twice continuously differentiable functions of many variables by piecewise constants.We show that the saturation order of piecewise constant approximation in L p norm on convex partitions with N cells is N −2/(d+1) , where d is the number of variables.
Let Ω ⊂ R d , d ≥ 1, be a bounded domain.Let L p (Ω), 1 ≤ p ≤ ∞ be the space of measurable functions f : Ω → R such that f p is finite, where For 1 ≤ q ≤ ∞ and r ∈ N, let W r q (Ω) be the standard Sobolev space of measurable functions f : Ω → R endowed with the standard norm For a partition of Ω, we denote by S 0 ( ) the space of piecewise constant functions s : Ω → R that are constant on every cell ω ∈ .For 1 ≤ p ≤ ∞, we define the error of the best L p -approximation of a function f ∈ L p (Ω) by piecewise constant functions on convex partitions from D N : It was established by Birman and Solomyak (see Theorems 3.1 and 3.2 in [1]), that for every f ∈ W 1 q (Ω), as soon as that achieves this order is based on adaptively refined dyadic subdivisions of the cube Ω.Since W 2 q (Ω) is imbedded into W 1 q (Ω) as soon as parameter 1 ≤ q ≤ ∞ satisfies inequality 1  d + 1 q − 1 q > 0, it follows from (1) that, for a function f ∈ W 2 q (Ω), providing that 2 d + 1 p − 1 q > 0. No improvement of the order O N −1/d can be achieved for a smooth non-constant function on "isotropic" partitions (see [2]).
It was shown in [2,3] that on a wider set of all convex partitions D N essentially better order of approximation is possible.More precisely, for every 1 ≤ p ≤ ∞ and q = p, and every function f ∈ W 2 q (Ω), which almost doubles the approximation order in comparison to N −1/d when d is large enough.This improvement in order is achieved on "anisotropic" sequence of partitions { * N } ∞ N =1 that is constructed in the following way: 1.First, the cube Ω is subdivided into m d equal cubes ω i , i = 1, . . ., m d , of side length 1 m , where m = N 1/(d+1) ; 2. Next, each of ω i 's is split into m slices ω ij , j = 1, . . ., m, by equidistant hyperplanes orthogonal to the average gradient ∇f (x) is the gradient of function f at the point x.

LOWER ESTIMATES ON THE SATURATION ORDER OF APPROXIMATION
Moreover, Davydov [3] has shown that N −2/(d+1) is the saturation order of piecewise constant approximation on convex partitions in the following sense: if for a twice continuously differentiable function f : Ω → R, we have that as N → ∞, then the Hesse matrix of function f is neither positive definite nor negative definite on Ω.We remark that similar results in the case d = 2 were obtained by a different method in [4].
Recently in [5] it was established that for 1 ≤ p, q ≤ ∞ which satisfy inequality In this paper we will show that N −2/(d+1) is in fact a saturation order in this case.Let us introduce some additional notation.Let C 2 (Ω) be the space of twice continuously differentiable functions f : , where x ∈ Ω.It's well-known (see [6]), that the Hesse matrix of a convex function is positive definite, and is negative definite for a concave function.For simplicity we will consider only case when the Hesse matrix of function is positive definite.However the presented arguments could be easily adjusted for negative definite matrices.
The main result of this paper is the following This result shows that N −2/(d+1) is the saturation order of piecewise constant approximation on convex partitions in the following sense: if for a twice continuously differentiable function f : Ω → R, we have that E N (f ) p = O(N −2/(d+1) ) as N → ∞, then the Hesse matrix of function f is neither positive, nor negative definite on Ω.
To prove Theorem 1 we need several auxiliary results.The assertions of geometrical Lemmas 1 and 2 below are intuitively clear, and we give their proofs for the completeness of the exposition.Lemma 1.Let T be a simplex in R d , V is a volume of this simplex, a is the length of one of the side of the simplex, and S is a volume of projection of the simplex T on the hyperplane orthogonal to the side a.Then Proof.Let us show how this theorem will look for triangle.And then will prove it in general case.For R 2 , we need to prove that where a is length of the side AB, and S is a length of projection of triangle ABC on the line orthogonal to AB.We can use side AB as Oy axis, and orthogonal line passing through the point A as Ox axis of coordinates.Then we have: where AC(x) and BC(x) are equations of lines AC and BC correspondingly.In geometrical sense this means that the area of the triangle doesn't change if we move vertex C along a line parallel to AB.So we can move C, untill side CA coincides with the projection of triangle ABC on Ox axis.And the area of this triangle could be counted by formula Without lose of generality we can choose side A d A d+1 and hyperplane which is orthogonal to this side and pass through vertex A d .Let B 1 B 2 . . .B d−1 be the projection of vertices A 1 , A 2 , . . ., A d−1 correspondingly.We can use side A d A d+1 as Ox 1 axis.And other axes will be in orthogonal hyperplane.Then we have In geometrical sense this means that the volume of the simplex doesn't change if we move vertices A 1 , A 2 , . . ., A d−1 along a line parallel to A d A d+1 .So we can move them, they coincides with the projection of simplex A 1 A 2 . . .A d+1 on hyperplane.And the volume of this simplex could be counted by formula The following theorem is some kind of generalization of Lemma 1, and it shows how connected volume of arbitrary convex domain Ω with volume of orthogonal projection Ω .Let v is arbitrary direction vector in R d .Denote h(Ω, v) the length of the longest segment inside domain Ω, which is parallel to vector v. Lemma 2. Let Ω ⊂ R d be a bounded convex domain.Let h(Ω, v) be the length of the longest segment in Ω in given direction v, and |Ω | be the volume of projection of the domain Ω on (d − 1)-dimensional hyperplane orthogonal to v. Then Proof.Let us find the longest segment in domain Ω ⊂ R As Ω is convex, that means that polygon According to Lemma 1 we have From the other side length of every segment of Ω parallel to v is less then |A 1 A 2 |.Then |Ω| is less then volume of rectangle with sides A 1 A 2 and B 1 B 2 .So we got following inequalities for Ω ⊂ R 2 .Let the described inequalities holds for Ω ⊂ R d−1 .
For Ω ⊂ R d let us find the longest segment parallel to vector v, and call it A 1 A 2 .Project domain Ω on hyperplane α 1 , which orthogonal to A 1 A 2 and pass through A 1 .We get Ω ⊂ R d−1 -projection of Ω.In hyperplane α 1 we can choose any direction v 1 .Let A 11 A 12 be the longest segment in Ω parallel to vector v 1 .Then project domain Ω on hyperplane α 2 ⊂ R d−2 , where α 2 is orthogonal to v 1 , and α 2 ⊂ α 1 .We get domain Ω ⊂ R d−2 -projection of the Ω on hyperplane α 2 .Inside hyperplane α 2 we choose any direction vector v 2 .Denote the longest segment of Ω parallel to vector v 2 as A 21 A 22 .And so on.When we will get the projection Ω (d−1) this will be just a segment A d−11 A d−12 ⊂ R. Then we denote points on the projections of higher dimensionality points which corresponds to points A k1 , A k2 for k = 1 . . .d − 1.On domain Ω we denote points B k1 , B k2 which corresponds to points A k1 , A k2 for k = 1 . . .d − 1.We will get convex polyhedron with vertices A 1 , A 2 and B k1 , B k2 for k = 1 . . .d − 1, which is inscribed into Ω.We separate this polyhedron on d-dimension simplexes of the form , where i k is 1 or 2 for k = 1, . . ., d − 1.The total number of such simplexes will be 2 d−1 .Projection of this polyhedron on Ω will be polyhedron inscribed into Ω with vertices A 11 , A 12 and points which corresponds to points A k1 , A k2 from Ω (k) on Ω .Let V 1 be the volume of d-dimension polyhedron.According to Lemma 1 we have: where This sum is a volume of polyhedron inscribed in Ω with vertices A 11 , A 12 and points which corresponds to points A k1 , A k2 from Ω (k) on Ω .According to the inductive hypothesis, we obtain where each segment A k1 A k2 is the longest segment in Ω (k) in direction of vector v k .
According to the inductive hypothesis, we have From the other side we have ) is length of the segment, which is the result of intersection of line which pass through point (x 1 , x 2 , . . .x d−1 ) in projection Ω with domain Ω.
Proof.Without loss of generality let f be increasing and convex function.The constant of best approximation in L 1 on interval (a; b) is f a+b 2 (see [7]).Therefore Also for f (u) we have the following inequality Therefore The following theorem is generalization of Lemma 3 for functions defined on R d .
where h(Ω, σ) was defined before Lemma 2. Moreover assume that the Hesse matrix of f is positive definite on Ω, and δ > 0 is such that smallest eigenvalue of ∇ 2 f is at least δ everywhere on Ω.Then where diam(Ω) is the longest segment inside Ω.
Proof.We can change orthonormal basis in R d in such way, that direction of σ and axis Ox 1 coincide.Then according to (4) from Lemma 3 we have where Ω is projection of Ω on the coordinate hyperplane orthogonal to σ, a(x 2 , . . ., x d ) and b(x 2 , . . ., x d ) are the ends of the segment, which is the result of intersection of the line, which pass through point (x 2 , • • • , x d ) ∈ Ω and parallel to σ, with domain Ω.
According to Hölder's inequality we obtain .
According to lower estimate (3) in Lemma 2 we have The first part of theorem is proved.For the second part we can change orthonormal basis in R d in such way, that direction of diam and axis Ox 1 coincide.Then according to (5) from Lemma 3 we have where Ω is projection of Ω on the coordinate hyperplane orthogonal to diam(Ω), a(x 2 , . . ., x d ) and b(x 2 , . . ., x d ) are the ends of the segment, which is the result of intersection of the line, which pass through point (x 2 , • • • , x d ) ∈ Ω and parallel to diam(Ω), with domain Ω.

LOWER ESTIMATES ON THE SATURATION ORDER OF APPROXIMATION
According to Hölder's inequality we obtain .
According to lower estimate (3) in Lemma 2 we have Using the results of Lemma 2 and Lemma 4 we can prove Theorem 1.
Proof of Theorem 1.Let us prove the theorem for case p = 1.First choose δ > 0 and convex domain Q ⊂ Ω such that the smallest eigenvalue of the Hessian of f is at least δ everywhere in Q.Clearly, ∇f (x) = 0, and x ∈ Q.Since the gradient of f is continuous, there is a constant γ and convex domain We assume without loss of generality that Q = Q.
The author expresses his gratitude to Prof. O. Davydov for the statement of the problem and to Prof. D. Skorokhodov for valuable advice and discussions.
the partial differentiation operator, where D r j = ∂ r ∂x r j for r ∈ Z + .A finite collection of subdomains ω ⊂ Ω, referred to as cells in what follows, is called a partition of Ω provided that ω ∩ ω = Ø, for any ω, ω ∈ , ω = ω , and ω∈ |ω| = |Ω|, where |•| is the Lebesgue measure.We call a partition convex if every cell ω ∈ is convex.For N ∈ N, denote by D N the set of all convex partitions of Ω comprising at most N cells.For simplicity reason, we assume in this paper that Ω is a d-dimensional cube.

1 2 .
According to upper estimate in (3) in Lemma 2 we have
2parallel to vector v, and call it A 1 A 2 .Project domain Ω on line orthogonal to A 1 A 2 .Let B 1 and B 2 be the ends of the segment of projection.And B -projection of the A 1 A 2 .Denote on Ω the points C 1 and C 2 that correspond to the points B 1 and B 2 on the projection.If there are several point which correspond to B 1 and B 2 on the projection, then we can choose anyone.