Stability of Affine Approximations on Bounded Domains

Abstract

Let f be an arbitrary function defined on a convex subset G of a linear space. Suppose the restriction of f on every straight line can be approximated by an affine function on that line with a given precision ε>0 (in the uniform metric); what is the precision of approximation of f by affine functionals globally on G? This problem can be considered in the framework of stability of linear and affine maps. We show that the precision of the global affine approximation does not exceed C(logd)ε, where d is the dimension of G, and C is an absolute constant. This upper bound is sharp. For some bounded domains G⊂ℝ^d, it can be improved. In particular, for the Euclidean balls the upper bound does not depend on the dimension, and the same holds for some other domains. As auxiliary results we derive estimates of the multivariate affine approximation on arbitrary domains and characterize the best affine approximations.

Dedicated to Professor Themistocles M. Rassias on the occasion of his 60th birthday.

Appendix

Proof of Proposition 37.1

For every straight line l⊂V, there is an affine functional φ _l :l→ℝ such that ∥f−φ _l ∥≤ε. This functional is unique, up to an addition of a constant. Indeed, if functionals φ _l and $\tilde{\varphi}_{l}$ possess this property, then $\|\varphi_{l} - \tilde{\varphi}_{l}\|_{l} \le 2 \varepsilon$, hence the affine functional $\varphi_{l} - \tilde{\varphi}_{l}$ is identically constant.

Consider now the function φ(x)=φ _l (x)−φ _l (0), where l is a line passing through the points 0 and x. This function is well-defined (does not depend on the choice of φ _l ) and homogeneous. Let us show that φ is linear. It suffices to prove its additivity. Observe first that since |f(x)−φ _l (x)|≤ε and |f(0)−φ _l (0)|≤ε, it follows that

(37.9)

Take arbitrary independent x,y∈V and consider the two-dimensional plane L spanned by them. We need to show that φ(x+y)=φ(x)+φ(y). Let ψ be a linear functional on L such that ψ(x)=φ(x) and ψ(y)=φ(y). Replacing the function f on L by f−ψ, and the function φ by φ−ψ, we assume that φ(x)=φ(y)=0. Since the three points x+y,2x,2y are collinear, from the ε-property it follows that |f(x+y)|≤max{|f(2x)|,|f(2y)|}+2ε. Furthermore, (37.9) implies that |f(2x)|≤2ε and |f(2y)|≤2ε. Thus, |f(x+y)|≤4ε, and, invoking (37.9) again, we conclude that |φ(x+y)|≤6ε. Similarly, |φ(λ x+λ y)|≤6ε for every λ∈ℝ, which, due to the homogeneity, means that φ(x+y)=0. Thus, φ is additive on V, and hence it is linear.

Replacing now f by f−φ on V we assume φ≡0. By (37.9), the function f is uniformly bounded on V. Let us add a constant to f so that sup _x∈V f(x)=−inf _x∈V f(x). If this supremum is greater than ε, then there are points z ₁,z ₂∈V such that f(z ₁)>ε,f(z ₂)<−ε. However, in this case the function φ _l , where l is a line connecting z ₁ and z ₂, cannot be identically constant, otherwise either (f−φ _l )(z ₁) or (f−φ _l )(z ₂) exceeds ε by modulus. Consequently, φ _l grows to +∞ on l, and hence so does f. This contradiction proves that $\sup_{\mathbf{x} \in \mathbb{R}^{d}}f(\mathbf{x}) \le \varepsilon$, and therefore ∥f−φ∥ _V ≤ε. □

Proof of Theorem 37.1

Sufficiency follows immediately from Proposition 37.2 and Corollary 37.1.

(Necessity). We realize the proof for bounded sets K because we only need this case. The proof for general sets is similar. Replacing f by f−φ, it can be assumed that φ≡0. The functional η(g)=∥f−g∥ _K is convex and closed on , therefore it attains its minimum at the point φ=0 iff 0∈∂η(0), where ∂η is the subdifferential (see [20]). Since η(g)=sup _x∈K |f(x)−g(x)|, the set is convex and compact, and the functions |f(x)−g(x)| are uniformly Lipschitz continuous in g on the set , so by the generalized Dubovitskii–Milyutin theorem [2] we have

where [⋅] denotes the closure, and is the delta-function, δ _x (g)=g(x). Since 0∈∂η(0), we see that for every α such that 0<α<∥f∥ _K the convex hull of the set

comes arbitrarily close to the origin. Thus, for every ε>0 there are convex combinations of finitely many points from this set, whose norms are less than ε. Since the space dual to is of dimension d+1, from the Caratheodory theorem, it follows that there are such sets of cardinality ≤d+2. Thus, there exist points a ₁,…,a _m ,b ₁,…,b _m′∈K, where m+m′≤d+2, and nonnegative numbers $\{t_{i}\}_{i=1}^{m}$, $\{s_{i}\}_{j=1}^{m'}$ with ∑ _i t _i +∑ _j s _j =1, such that f(a _i )>α,f(b _j )<−α, and

for any such that ∥g∥≤1. Without loss of generality it can be assumed that ∑ _i t _i ≥1/2. Writing T=∑ _i t _i and $t_{i}' = t_{i} / T$, $s_{j}' = s_{j}/T$, we obtain

for ∥g∥≤1. Substituting g≡1, we see that $|(1 - \sum_{j} s_{j}')| < 2 \varepsilon$. On the other hand,

Thus, for every g such that ∥g∥≤1 we have

Since ∥g∥≤1, it follows that $\|g\|_{K} \le C = \max \{1 , \operatorname{diam} (K)\}$. Therefore,

On the other hand, there is , such that |−g(a)+g(b)|=|a−b|. Thus, |a−b|<2(1+C)ε. In particular, for $\varepsilon = \frac{\rho}{2 ( 1 + C )}$ we obtain $\operatorname{dist} \{ \operatorname{co} (A) , \operatorname{co} (B) \} < \rho$.

It remains to show that either condition (a) is satisfied, or m+m′≤d+1 (by now we have proved that m+m′≤d+2). We take points $\mathbf{x} \in\operatorname{co} (A)$ and $\mathbf{y} \in\operatorname{co} (B)$ such that $|\mathbf{x} - \mathbf{y}| = \operatorname{dist} \{ \operatorname{co} (A) , \operatorname{co} (B) \}$. If x=y, then we have condition (a). If |x−y|>0, then either one of the points x or y lies on the boundary of the corresponding set, or the vector x−y is orthogonal to the affine spans of these sets. In the first case, when, say, x is on the boundary of the polyhedron $\operatorname{co} (A)$, then we take the smallest face of that polyhedron containing x. Replacing the set A by the set of vertices of this face, we reduce the total number of points of A and B at least to d+1, after which property (b) is satisfied. Consider the second case, when the vector x−y is orthogonal to the affine spans of A and B. Denote these spans by $\tilde{A}$ and $\tilde{B}$, respectively. It follows that both $\tilde{A}$ and $\tilde{B}$ are parallel to one hyperplane. Hence, either $\dim (\tilde{A}) + \dim (\tilde{B}) \le d-1$, or there is a straight line parallel to both $\tilde{A}$ and $\tilde{B}$. In the first case, by the Caratheodory theorem, one can choose at most $\dim (\tilde{A}) + 1$ points of the set A and at most $\dim (\tilde{B}) + 1$ points of the set B so that the convex hulls of these sets still contain x and y, respectively. The total number of points is reduced to $\dim (\tilde{A}) + \dim (\tilde{B}) + 2 \le d + 1$, which proves (b). Otherwise, if there is a straight line l parallel to both $\tilde{A}$ and $\tilde{B}$, the two lines passing through x and y parallel to l intersect the polyhedra $\operatorname{co} (A)$ and $\operatorname{co} (B)$ by some segments [x ₁,x ₂] and [y ₁,y ₂], respectively. The distance between those two segments equals to |x−y|, and it is attained at one of the ends of these segments, say, at x ₁. Thus, $\operatorname{dist} \{\mathbf{x}_{1} , [\mathbf{y}_{1}, \mathbf{y}_{2}] \} = |\mathbf{x} - \mathbf{y}|$. However, x ₁ lies on the boundary of $\operatorname{co}(A)$, and we again come to the first case. □

Proof of Lemma 37.6

Let a,b≥0. Since the function p(t) is convex on the segment [0,1], we have $E(p, [a,b]) = \frac{1}{2} \sup_{t \in [a,b]} ( \xi_{a,b}(t) - p(t) )$, where ξ _a,b is the affine function such that ξ _a,b (a)=p(a),ξ _a,b (b)=p(b). It is shown easily that $\sup_{t \in[a,b]} ( \xi_{a,b}(t) - p(t) ) \le \sup_{t \in[0,b-a]} ( \xi_{a,b}(t) - p(t) ) = \frac{b-a}{e}$, hence $E(p, [a,b]) \le \frac{b-a}{2e}$. The case a,b≤0 follows from the symmetry. Finally, in the case a<0<b, assuming |b|≥|a|, we have

 □

Proof of Lemma 37.7

We have

On the other hand, $\frac{1}{4} \sum_{i=1}^{d+1} t_{i}^{2} \ge \frac{1}{4(d+1)}$. Therefore,

Whence, there are i and j such that $t_{i}t_{j}(\mathbf{x}_{i}, \mathbf{x}_{j}) \le - \frac{1}{4d(d+1)^{2}}$. Taking into account that t _i ≤1,t _j ≤1 and |(x _i ,x _j )|≤1, we see that both t _i and t _j are at least $\kappa_{d} = \frac{1}{4d(d+1)^{2}}$. □