Uniform Lipschitz Continuity of Bestlp-Approximations by Polyhedral Sets

Uniform Lipschitz Continuity of Bestlp-Approximations by Polyhedral Sets

JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO. 228, 112]118 Ž1998. AY986120 Uniform Lipschitz Continuity of Best l p-Approximations ...

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JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.

228, 112]118 Ž1998.

AY986120

Uniform Lipschitz Continuity of Best l p-Approximations by Polyhedral Sets Martina FinzelU Mathematical Institute, Uni¨ ersity of Erlangen]Nuremberg, D-91054 Erlangen, Germany

and Wu Li † Department of Mathematics and Statistics, Old Dominion Uni¨ ersity, Norfolk, Virginia 23529-0077 Submitted by Joseph D. Ward Received December 30, 1997

In this paper we prove that the metric projection P K , p onto a polyhedral subset K of R n , endowed with the p-norm, is uniformly Lipschitz continuous with respect to p, 1 - p - `. As a consequence the strict best approximation and the natural best approximation are Lipschitz continuous selections for the metric projections P K , ` and P K , 1 , respectively. This extends a recent analogous result in Berens et al. w J. Q 1998 Academic Press Math. Anal. Appl. 213 Ž1997., 183]201x on linear subspaces. Key Words: discrete polyhedral l p-approximation; Lipschitz continuity; strict best approximation; natural best approximation.

1. INTRODUCTION Let R n be endowed with the p-norms 5 ? 5 p defined for a vector y in R n by 5 y 5 p [ ŽÝ nis1 < yi < p .1r p for 1 F p - ` and by 5 y 5 ` [ max 1 F iF n < yi < for p s `. For convenience, we use 5 ? 5 to denote the 2-norm 5 ? 5 2 . Let K denote a closed convex subset of R n. For x g R n, consider the best l p-approximation problem of finding x in K such that 5 x y x p 5 s min  5 x y z 5 p : z g K 4 . * e-mail: [email protected]. † e-mail: [email protected]. 112 0022-247Xr98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.

Ž 1.

UNIFORM LIPSCHITZ CONTINUITY

113

If 1 - p - `, the norms are strictly convex and there exists exactly one element xp in K that satisfies Ž1., which is called the best l p-approximation of x in K and is denoted by P K , p Ž x .. When p s 1 or `, in general, there are infinitely many best approximants. Let P K , p Ž x . be the set of elements xp in K that satisfy Ž1.. It is a compact and convex subset of K. In general, the metric projection P K , ` is not continuous with respect to the Hausdorff metric w7, Theorem 1.1x. If it is, one particular element in P K , `Ž x . is the strict best approximation, denoted by sba K Ž x ., that can be defined as the limit of P K , p Ž x . as p ª `: sba K Ž x . [ lim P K , p Ž x .

for x g R n .

pª`

Ž 2.

For K being a linear subspace, the strict best approximation was introduced by Rice w9x as the best of best approximations. The existence of the limit in Ž2. was first proved by Descloux w2x for a subspace K of R n. Later, Huotari et al. w6x proved Ž2. under a much weaker condition Žcf. also w7, Theorem 1.1x.. In particular, their result implies that Ž2. holds for any polyhedral subset K of R n. Recall that K is a polyhedral subset of R n if K is the intersection of finitely many closed half-spaces, i.e., K s  x g R n : Ax F b4 with an m = n matrix A and b g R m . For x g R n, a special element of P K , 1Ž x . corresponding to sba K Ž x . is the natural best approximation, denoted by nba K Ž x ., which is defined by the limit of P K , p Ž x . as p ª 1q w8x: nba K Ž x . [ limq P K , p Ž x .

for x g R n .

pª1

Ž 3.

Landers and Rogge w8x proved that, for each x, lim p ª 1q P K , p Ž x . exists Ži.e., nba K Ž x . is well defined. when K is a closed convex subset; independently, Fischer w4x proved the existence of this limit for subspaces. For each fixed p g Ž1, `. the metric projection P K , p onto a linear subspace K is a Lipschitz continuous mapping Žsee w5x.. This result has been improved to uniform Lipschitz continuity with respect to p; i.e., in w1x it was proved that if K is a subspace of R n, then there exists a positive constant l such that PK, pŽ x. y PK, pŽ y. F l ? 5 x y y5

for x, y g R n , 1 - p - `. Ž 4 .

As a consequence, we obtain from Ž2., Ž3., and Ž4. that nba K Ž x . y nba K Ž y . F l ? 5 x y y 5

for x, y g R n ,

sba K Ž x . y sba K Ž y . F l ? 5 x y y 5

for x, y g R n ,

Ž 5.

provided that K is a subspace of R n. That is, the strict best approximation and the natural best approximation are Lipschitz continuous when K is a

114

FINZEL AND LI

subspace of R n. For p s ` this was already known Žsee w3x., while for p s 1 it was new. The main purpose of this paper is to show that statements Ž4. and Ž5. hold more generally for any polyhedral subset K of R n. Our proof is based on a result about Lipschitz continuous mappings and a representation of P K , p as best l p-approximation by elements from certain affine subspaces. 2. PIECEWISE LIPSCHITZ CONTINUOUS MAPPINGS In the next lemma we prove that a ‘‘continuous patching’’ of finitely many Lipschitz continuous mappings is also Lipschitz continuous. LEMMA 1. Suppose F1 , F2 , . . . , Fm are Lipschitz continuous mappings from R n to R r , and l is a positi¨ e constant such that Fi Ž x . y Fi Ž y . F l ? 5 x y y 5

for x, y g R n , i s 1, . . . , m.

Ž 6.

If a continuous mapping F: R n ª R r satisfies the condition F Ž x . g  F1 Ž x . , F2 Ž x . , . . . , Fm Ž x . 4

for x g R n ,

then F is also Lipschitz continuous and FŽ x. y FŽ y. F l ? 5 x y y5

for x, y g R n .

Ž 7.

m Proof. Let X i [  x g R n : F Ž x . s Fi Ž x .4 . Then R n s Dis1 X i . For x, y u n g R , define x s u y q Ž1 y u . x and

u 0 [ max  u : 0 F u F 1, F Ž x u . y F Ž x . F l 5 x u y x 5 4 .

Ž 8.

If u 0 s 1, then Ž7. holds. Now assume the contrary, that 0 F u 0 - 1. Since m R n s Dis1 X i , we have

½

xu : u0 - u - u0 q

1 k

m

5 ž l

D Xi is1

/

/B

for k s 1, 2, . . . .

Therefore, there is an index j such that

½

xu : u0 - u - u0 q

1 k

5

l Xj / B

for k s 1, 2, . . . .

As a consequence, there exist 1 ) u 1 ) u 2 ) ??? ) u k ) ??? G u 0 such that x u k g X j and lim x u k s x u 0 .

kª`

UNIFORM LIPSCHITZ CONTINUITY

115

By Ž6. and the continuity of F, we have F Ž x u 0 . y F Ž x u 1 . s lim F Ž x u k . y F Ž x u 1 . kª`

F lim l ? 5 x u k y x u 1 5 F l ? 5 x u 0 y x u 1 5 . kª`

So F Ž x u1 . y F Ž x . F F Ž x u 1 . y F Ž x u 0 . q F Ž x u 0 . y F Ž x . F l ? 5 x u1 y x u 0 5 q l ? 5 x u 0 y x 5 s l ? 5 x u1 y x 5 . Ž 9. By the definition of u 0 , we must have u 0 G u 1 , a contradiction to the choice of u 1. So u 0 s 1 and Ž7. holds. 3. PIECEWISE AFFINE REPRESENTATIONS Now we represent P K , p Ž x . as the best l p-approximation of x by an affine subspace that is generated by a face of K. To do so, we need the following Karush]Kuhn]Tucker characterization of optimal solutions of convex minimization problems with linear constraints. LEMMA 2 w10, Corollary 28.3.1x. Let g Ž z . be a con¨ ex differentiable function on R n and consider the following con¨ ex minimization problem: inf g Ž z . z

subject to

Az F b, Cz s d,

Ž 10 .

where A is an m = n matrix, C is an r = n matrix, b g R m , and d g R r. Then z is a solution of Ž10. if and only if there exist u g R m and ¨ g R r such that =g Ž z . s AT u q C T ¨ , Cz s d,

u G 0,

Az F b, ² u, Az y b : s 0,

Ž 11 .

where =g is the gradient of g, ² u, w : [ Ý m is1 u i wi denotes the inner product in R m , and AT Ž or C T . is the transpose of A Ž or C .. LEMMA 3. Let K be a polyhedral subset of R n. There exist ¨ ectors z , . . . , z s in R n and subspaces G1 , . . . , Gs of R n such that 1

P K , p Ž x . g  z 1 q P G1 , p Ž x y z 1 . , . . . , z s q P Gs , p Ž x y z s . 4 for x g R n , 1 - p - `.

Ž 12 .

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FINZEL AND LI

Proof. Let K s  z g R n : Az F b4 , where A is an m = n matrix and b g R m. Consider the following convex minimization problem: inf 5 x y z 5 pp z

subject to Az F b.

By applying Lemma 2 Žwith C s 0 and d s 0., we have z s P K , p Ž x . if and only if there is a vector ŽLagrange multiplier. u g R m such that =z f p Ž x y z . s AT u,

u G 0,

Az F b,

² u, Az y b : s 0, Ž 13 .

where f p Ž x y z . s 5 x y z 5 pp and =z f p Ž x y z . denotes the gradient of f p Ž x y z . with respect to z. For any index subset J of  1, 2, . . . , m4 , define K J [  z g R n : Az F b, A J z y bJ s 0 4 , where A J Žor bJ . is the matrix Žor the vector. consisting of rows of A Žor components of b . whose corresponding indices are in J. For K J s B let X J s B, and for K J / B define X J [  x g R n : P K , p Ž x . g K J and there exists u g R m such that =z f p Ž x y P K , p Ž x . . s ATJ u J , u J G 0 4 . First note that

D XJ s R n.

Ž 14 .

J

In fact, for x g R n and z s P K , p Ž x ., there exists u g R m such that Ž13. holds. Let J [  i: u i ) 04 . From Ž13. it follows that =z f p Ž x y P K , p Ž x .. s =z f p Ž x y z . s AT u s ATJ u J and A J z y bJ s 0. In particular, we have P K , p Ž x . s z g K J and, therefore x g X J . This proves Ž14.. Next, for x g X J , we deduce a relation between P K , p Ž x . and the metric projection to a linear subspace. When K J s B let z J s 0 and GJ s  04 , while for K J / B, choose z J g K J and define GJ [  z g R n : A J z s 0 4 . Then K J ; z J q GJ , i.e.,

A J Ž z y z J . s 0 for z g K J .

Ž 15 .

Now let x g X J and z s P K , p Ž x . y z J. By the definition of X J and Ž15., A J z s 0, i.e., z g GJ . Moreover, there exists u g R m such that =z f p Ž Ž x y z J . y z . s =z f p Ž x y P K , p Ž x . . s ATJ u J .

Ž 16 .

UNIFORM LIPSCHITZ CONTINUITY

117

Lemma 2 Žwith A s 0, b s 0, C s A J , and d s 0. implies z s P G J , p Ž x y z J ., i.e., P K , p Ž x . s z J q PGJ , p Ž x y z J .

for x g X J .

Ž 17 .

The identity above gives us a representation of P K , p Ž x . in terms of the best l p-approximation of Ž x y z J . by a subspace GJ for x g X J . Let z 1, . . . , z s and G1 , . . . , Gs be a relabeling of z J s and GJ s. Then Ž12. follows from Ž17..

4. UNIFORM LIPSCHITZ CONTINUITY Finally, we use Lemmas 1 and 3 to prove the uniform Lipschitz continuity of P K , p for 1 - p - `. THEOREM 4. Let K be a polyhedral subset of R n. Then there exists a positi¨ e constant h such that PK, pŽ x. y PK, pŽ y. F h ? 5 x y y5

for x, y g R n , 1 - p - `. Ž 18 .

Proof. By Lemma 3, there are vectors z 1 , . . . , z s and subspaces G1 , . . . , Gs such that Ž12. holds. For any index i g  1, . . . , s4 define Fi , p Ž y . [ z i q P G i , p Ž y y z i .

for y g R n .

Ž 19 .

For the subspace Gi of R n, there is a positive constant l i such that Žcf. Ž4.. P G i , p Ž x . y P G i , p Ž y . F li ? 5 x y y 5

for x, y g R n , 1 - p - `.

Ž 20 . This implies z i q PGi, p Ž x y z i . y z i q PGi, p Ž y y z i . s PGi, p Ž x y z i . y PGi, p Ž y y z i . F li ? Ž x y z i . y Ž y y z i . s li ? 5 x y y 5 . Therefore, Fi , p Ž x . y Fi , p Ž y . F l ? 5 x y y 5

for x, y g R n , 1 - p - `, Ž 21 .

where

l s max  l i : 1 F i F s 4 .

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For fixed p ) 1, by Ž19. and Ž12., we have P K , p Ž x . g  Fi , p Ž x . : i s 1, . . . , s 4

for x g R n .

Ž 22 .

Since P K , p is a continuous mapping from R n to K w7, Theorem 1.1x, by Ž21., Ž22., and Lemma 1, we obtain PK, pŽ x. y PK, pŽ y. F l ? 5 x y y5

for x, y g R n .

Ž 23 .

The inequality above holds for any p ) 1, and consequently, Ž18. is satisfied. An immediate consequence of Ž18., Ž2., and Ž3. is the following corollary on Lipschitz continuity of the strict best and the natural best approximations. COROLLARY 5. Let K be a polyhedral subset of R n. Then there is a positi¨ e constant l such that nba K Ž x . y nba K Ž y . F l ? 5 x y y 5

for x, y g R n ,

sba K Ž x . y sba K Ž y . F l ? 5 x y y 5

for x, y g R n .

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