On the converses of the reduction principle in inner product spaces

Costica Mustata
(original), Approximation Theory, best approximation, ISI/JCR, paper

Abstract


Let \(H\) be an inner product space, \(X\) a complete subspace of \(H\), and \(Y\) a closed subspace of \(X\). The main result of this Note is the following converse of the Reduction Principle: if \(x_{0}\in X,\ h\in H\backslash X\) and \(y_{0}\in Y\) is the element of best approximation of both \(x_{0}\) and \(h\), \((x_{0}-h,x_{0}-y_{0})=0\) and \(codim_{X}Y=1\), then \(x_{0}\) is the element of best approximation of \(h\) in \(X\).

Authors

Costică Mustăţa
“Tiberiu Popoviciu” Institute of Numerical Analysis, Romanian Academy, Romania

Keywords

Inner product spaces; the Reduction Principle; best approximation.

Paper coordinates

C. Mustăţa, On the converses of the reduction principle in inner product spaces, Studia Univ. Babeş-Bolyai, Mathematica, 51 (2006) no. 3, 97-104.

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About this paper

Journal

Mathematica

Publisher Name

Studia Universitatis Babes-Bolyai, Mathematica

DOI
Print ISSN

1843-3855

Online ISSN

2065-9490

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[1] Cheney, W., Analysis for Applied Mathematics, Springer-Verlag, New York-BerlinHeidelberg, 2001.
[2] Deutsch, F., Best Approximation in Inner-Product Space, Springer-Verlag, New YorkBerlin-Heidelberg, 2001.
[3] Laurent, P.J., Approximation et Optimisation, Herman, Paris, 1972.

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2006-Mustata-On-the-converses-of-the-reduction-principle-Studia

ON THE CONVERSES OF THE REDUCTION PRINCIPLE IN INNER PRODUCT SPACES

COSTICĂ MUSTĂŢADedicated to Professor Ştefan Cobzaş at his 60 th 60 th  60^("th ")60^{\text {th }}60th  anniversary

Abstract

Let H H HHH be an inner product space, X X XXX a complete subspace of H H HHH, and Y Y YYY a closed subspace of X X XXX. The main result of this Note is the following converse of the Reduction Principle: if x 0 ∈ X , h ∈ H ∖ X x 0 ∈ X , h ∈ H ∖ X x_(0)in X,h in H\\Xx_{0} \in X, h \in H \backslash Xx0∈X,h∈H∖X and y 0 ∈ Y y 0 ∈ Y y_(0)in Yy_{0} \in Yy0∈Y is the element of best approximation of both x 0 x 0 x_(0)x_{0}x0 and h , ( x 0 − h , x 0 − y 0 ) = 0 h , x 0 − h , x 0 − y 0 = 0 h,(x_(0)-h,x_(0)-y_(0))=0h,\left(x_{0}-h, x_{0}-y_{0}\right)=0h,(x0−h,x0−y0)=0 and codim X Y = 1 codim X Y = 1 codim_(X)Y=1\operatorname{codim}_{X} Y=1codimXY=1, then x 0 x 0 x_(0)x_{0}x0 is the element of best approximation of h h hhh in X X XXX.

1. Introduction

Let H H HHH be an inner product space, with real inner product ( ⋅ , ⋅ ) ( ⋅ , ⋅ ) (*,*)(\cdot, \cdot)(⋅,⋅) and the norm ‖ h ‖ = ( h , h ) , h ∈ H ‖ h ‖ = ( h , h ) , h ∈ H ||h||=sqrt((h,h)),h in H\|h\|=\sqrt{(h, h)}, h \in H‖h‖=(h,h),h∈H. For a subset M M MMM of H H HHH and h ∈ H h ∈ H h in Hh \in Hh∈H, the distance of h h hhh to M M MMM is defined by
d ( x , M ) = inf { ‖ h − m ‖ : m ∈ M } . d ( x , M ) = inf { ‖ h − m ‖ : m ∈ M } . d(x,M)=i n f{||h-m||:m in M}.d(x, M)=\inf \{\|h-m\|: m \in M\} .d(x,M)=inf{‖h−m‖:m∈M}.
The set M M MMM is called proximinal if for every h ∈ H h ∈ H h in Hh \in Hh∈H there exists m 0 ∈ M m 0 ∈ M m_(0)in Mm_{0} \in Mm0∈M such that
‖ h − m 0 ‖ = d ( h , M ) . h − m 0 = d ( h , M ) . ||h-m_(0)||=d(h,M).\left\|h-m_{0}\right\|=d(h, M) .‖h−m0‖=d(h,M).
The set
P M ( h ) := { m ∈ M : ‖ h − m ‖ = d ( h , M ) } , h ∈ H P M ( h ) := { m ∈ M : ‖ h − m ‖ = d ( h , M ) } , h ∈ H P_(M)(h):={m in M:||h-m||=d(h,M)},quad h in HP_{M}(h):=\{m \in M:\|h-m\|=d(h, M)\}, \quad h \in HPM(h):={m∈M:‖h−m‖=d(h,M)},h∈H
is called the set of best approximation elements of h h hhh by elements in M M MMM, and the application P M : H → 2 M P M : H → 2 M P_(M):H rarr2^(M)P_{M}: H \rightarrow 2^{M}PM:H→2M is called the metric projection of H H HHH on M M MMM.
If card P M ( h ) = 1 card P M ( h ) = 1 cardP_(M)(h)=1\operatorname{card} P_{M}(h)=1cardPM(h)=1 for every h ∈ H h ∈ H h in Hh \in Hh∈H, then the set M M MMM is called a Chebyshevian set in H H HHH ([2], p.35).
The existence and the uniqueness of best approximation elements are treated in Chapter 3 of [2]: every complete convex set in an inner product space is a Chebyshev set ([2], Th.3.4).
Two elements u , v ∈ H u , v ∈ H u,v in Hu, v \in Hu,v∈H are called orthogonal if ( u , v ) = 0 ( u , v ) = 0 (u,v)=0(u, v)=0(u,v)=0. The cosinus of the angle between the u , v ∈ H ∖ { 0 } u , v ∈ H ∖ { 0 } u,v in H\\{0}u, v \in H \backslash\{0\}u,v∈H∖{0} is defined by the formula
cos ⁡ u , v ^ = ( u , v ) ‖ u ‖ ⋅ ‖ v ‖ cos ⁡ u , v ^ = ( u , v ) ‖ u ‖ ⋅ ‖ v ‖ cos widehat(u,v)=((u,v))/(||u||*||v||)\cos \widehat{u, v}=\frac{(u, v)}{\|u\| \cdot\|v\|}cos⁡u,v^=(u,v)‖u‖⋅‖v‖
Concerning the characterization of best approximation elements, the following result holds ([2], Th.4.9):
Let M M MMM be a subspace of H , h ∈ H H , h ∈ H H,h in HH, h \in HH,h∈H and m 0 ∈ M m 0 ∈ M m_(0)in Mm_{0} \in Mm0∈M. Then m 0 = P M ( h ) m 0 = P M ( h ) m_(0)=P_(M)(h)m_{0}=P_{M}(h)m0=PM(h) iff
( h − m 0 , m ) = 0 h − m 0 , m = 0 (h-m_(0),m)=0\left(h-m_{0}, m\right)=0(h−m0,m)=0
for all m ∈ M m ∈ M m in Mm \in Mm∈M.
The geometric interpretation of this characterization result is that the element h − P M ( h ) h − P M ( h ) h-P_(M)(h)h-P_{M}(h)h−PM(h) is orthogonal to each element of M M MMM. This is the reason why P M ( h ) P M ( h ) P_(M)(h)P_{M}(h)PM(h) is often called the orthogonal projection of h h hhh on M M MMM.
The following result appears in [2], p. 80 under the name "the Reduction Principle":
Let K K KKK be a convex subset of the inner product space H H HHH and let M M MMM be any Chebyshev subspace of H H HHH that contains K K KKK. Then
a) P K ( P M ( h ) ) = P K ( h ) = P M ( P K ( h ) ) , h ∈ H ; P K P M ( h ) = P K ( h ) = P M P K ( h ) , h ∈ H ; P_(K)(P_(M)(h))=P_(K)(h)=P_(M)(P_(K)(h)),h in H;P_{K}\left(P_{M}(h)\right)=P_{K}(h)=P_{M}\left(P_{K}(h)\right), h \in H ;PK(PM(h))=PK(h)=PM(PK(h)),h∈H;
b) d ( h , K ) 2 = d ( h , M ) 2 + d ( P M ( h ) , K ) 2 d ( h , K ) 2 = d ( h , M ) 2 + d P M ( h ) , K 2 d(h,K)^(2)=d(h,M)^(2)+d(P_(M)(h),K)^(2)d(h, K)^{2}=d(h, M)^{2}+d\left(P_{M}(h), K\right)^{2}d(h,K)2=d(h,M)2+d(PM(h),K)2,
for every h ∈ H h ∈ H h in Hh \in Hh∈H.
Obviously, if K K KKK is a closed and convex subset of a complete subspace M M MMM of the inner product space H H HHH, the properties a) and b) are also fulfilled (see Th.4.1 in [2], and Th. 2.2.6 in [3]).
ON THE CONVERSES OF THE REDUCTION PRINCIPLE IN INNER PRODUCT SPACES

2. Results

From now on, we consider the following particular case of the Reduction Principle:
Theorem 1. Let H H HHH be an inner product space, X X XXX a complete subspace of H H HHH, and Y Y YYY a closed subspace of X X XXX. Then
a') P Y ( h ) = P Y ( P X ( h ) ) = P X ( P Y ( h ) ) , h ∈ H ; P Y ( h ) = P Y P X ( h ) = P X P Y ( h ) , h ∈ H ; P_(Y)(h)=P_(Y)(P_(X)(h))=P_(X)(P_(Y)(h)),h in H;P_{Y}(h)=P_{Y}\left(P_{X}(h)\right)=P_{X}\left(P_{Y}(h)\right), h \in H ;PY(h)=PY(PX(h))=PX(PY(h)),h∈H;
b') d ( h , Y ) 2 = d ( h , X ) 2 + d ( P X ( h ) , Y ) 2 d ( h , Y ) 2 = d ( h , X ) 2 + d P X ( h ) , Y 2 d(h,Y)^(2)=d(h,X)^(2)+d(P_(X)(h),Y)^(2)d(h, Y)^{2}=d(h, X)^{2}+d\left(P_{X}(h), Y\right)^{2}d(h,Y)2=d(h,X)2+d(PX(h),Y)2, for every h ∈ H h ∈ H h in Hh \in Hh∈H.
The proof of Theorem 1 is an immediate consequence of the characterization result ([2], Th.4.9) and the Pythagorean Law (see e.g. [1], Th.1, p.70).
A generalization of Theorem 1 is:
Theorem 2. Let H H HHH be an inner product space and M 1 , M 2 , … , M n ( n ≥ 2 ) M 1 , M 2 , … , M n ( n ≥ 2 ) M_(1),M_(2),dots,M_(n)(n >= 2)M_{1}, M_{2}, \ldots, M_{n}(n \geq 2)M1,M2,…,Mn(n≥2) be subspaces of H H HHH with the following properties:
  1. M 1 M 1 M_(1)M_{1}M1 is complete;
  2. M i , i = 2 , 3 , … , n M i , i = 2 , 3 , … , n M_(i),i=2,3,dots,nM_{i}, i=2,3, \ldots, nMi,i=2,3,…,n are closed;
  3. M 1 ⊃ M 2 ⊃ ⋯ ⊃ M n M 1 ⊃ M 2 ⊃ ⋯ ⊃ M n M_(1)supM_(2)sup cdots supM_(n)M_{1} \supset M_{2} \supset \cdots \supset M_{n}M1⊃M2⊃⋯⊃Mn.
    a) For every h ∈ H h ∈ H h in Hh \in Hh∈H the following equalities hold
P M n ( h ) = P M n P M n − 1 … P M 1 ( h ) = P M 1 P M 2 … P M n ( h ) P M n ( h ) = P M n P M n − 1 … P M 1 ( h ) = P M 1 P M 2 … P M n ( h ) P_(M_(n))(h)=P_(M_(n))P_(M_(n-1))dotsP_(M_(1))(h)=P_(M_(1))P_(M_(2))dotsP_(M_(n))(h)P_{M_{n}}(h)=P_{M_{n}} P_{M_{n-1}} \ldots P_{M_{1}}(h)=P_{M_{1}} P_{M_{2}} \ldots P_{M_{n}}(h)PMn(h)=PMnPMn−1…PM1(h)=PM1PM2…PMn(h)
b) Let P M 1 ( h ) = m 1 , P M k P M k − 1 ( h ) = m k , k = 2 , 3 , … , n Let P M 1 ( h ) = m 1 , P M k P M k − 1 ( h ) = m k , k = 2 , 3 , … , n LetP_(M_(1))(h)=m_(1),P_(M_(k))P_(M_(k-1))(h)=m_(k),k=2,3,dots,n\operatorname{Let} P_{M_{1}}(h)=m_{1}, P_{M_{k}} P_{M_{k-1}}(h)=m_{k}, k=2,3, \ldots, nLetPM1(h)=m1,PMkPMk−1(h)=mk,k=2,3,…,n.
The following equality holds:
d ( h , M n ) 2 = ‖ h − m 1 ‖ 2 + ∑ k = 2 h ‖ m k − m k − 1 ‖ 2 d h , M n 2 = h − m 1 2 + ∑ k = 2 h   m k − m k − 1 2 d(h,M_(n))^(2)=||h-m_(1)||^(2)+sum_(k=2)^(h)||m_(k)-m_(k-1)||^(2)d\left(h, M_{n}\right)^{2}=\left\|h-m_{1}\right\|^{2}+\sum_{k=2}^{h}\left\|m_{k}-m_{k-1}\right\|^{2}d(h,Mn)2=‖h−m1‖2+∑k=2h‖mk−mk−1‖2
Proof. For every y ∈ M n y ∈ M n y inM_(n)y \in M_{n}y∈Mn we have
( h − P M n P M n − 1 … P M 1 ( P 1 ) , y ) = ( h − P M 1 ( h ) + P M 1 ( h ) − P M 2 P M 1 ( h ) + … + P M n − 1 P M n − 2 … P M 1 ( h ) − P M n P M n − 1 … P M 1 ( h ) , y ) h − P M n P M n − 1 … P M 1 P 1 , y = h − P M 1 ( h ) + P M 1 ( h ) − P M 2 P M 1 ( h ) + … + P M n − 1 P M n − 2 … P M 1 ( h ) − P M n P M n − 1 … P M 1 ( h ) , y {:[(h-P_(M_(n))P_(M_(n-1))dotsP_(M_(1))(P_(1)),y)],[=(h-P_(M_(1))(h)+P_(M_(1))(h)-P_(M_(2))P_(M_(1))(h)+dots:}],[{:+P_(M_(n-1))P_(M_(n-2))dotsP_(M_(1))(h)-P_(M_(n))P_(M_(n-1))dotsP_(M_(1))(h),y)]:}\begin{gathered} \left(h-P_{M_{n}} P_{M_{n-1}} \ldots P_{M_{1}}\left(P_{1}\right), y\right) \\ =\left(h-P_{M_{1}}(h)+P_{M_{1}}(h)-P_{M_{2}} P_{M_{1}}(h)+\ldots\right. \\ \left.+P_{M_{n-1}} P_{M_{n-2}} \ldots P_{M_{1}}(h)-P_{M_{n}} P_{M_{n-1}} \ldots P_{M_{1}}(h), y\right) \end{gathered}(h−PMnPMn−1…PM1(P1),y)=(h−PM1(h)+PM1(h)−PM2PM1(h)+…+PMn−1PMn−2…PM1(h)−PMnPMn−1…PM1(h),y)
= ( h − P M 1 ( h ) , y ) + ∑ k = 2 n ( P M k − 1 … P M 1 ( h ) − P M k … P M 1 ( h ) , y ) = 0 . = h − P M 1 ( h ) , y + ∑ k = 2 n   P M k − 1 … P M 1 ( h ) − P M k … P M 1 ( h ) , y = 0 . =(h-P_(M_(1))(h),y)+sum_(k=2)^(n)(P_(M_(k-1))dotsP_(M_(1))(h)-P_(M_(k))dotsP_(M_(1))(h),y)=0.=\left(h-P_{M_{1}}(h), y\right)+\sum_{k=2}^{n}\left(P_{M_{k-1}} \ldots P_{M_{1}}(h)-P_{M_{k}} \ldots P_{M_{1}}(h), y\right)=0 .=(h−PM1(h),y)+∑k=2n(PMk−1…PM1(h)−PMk…PM1(h),y)=0.
Using the characterization result ([2], Th.4.9) it follows that the element P M n P M n − 1 … P M 1 ( h ) P M n P M n − 1 … P M 1 ( h ) P_(M_(n))P_(M_(n-1))dotsP_(M_(1))(h)P_{M_{n}} P_{M_{n-1}} \ldots P_{M_{1}}(h)PMnPMn−1…PM1(h) is the orthogonal projection of h h hhh on M n M n M_(n)M_{n}Mn.
On the other hand, ( h − P M n ( h ) , y ) = 0 h − P M n ( h ) , y = 0 (h-P_(M_(n))(h),y)=0\left(h-P_{M_{n}}(h), y\right)=0(h−PMn(h),y)=0 for every y ∈ M n y ∈ M n y inM_(n)y \in M_{n}y∈Mn. Consequently
P M n ( h ) = P M n P M n − 1 … P M 1 ( h ) P M n ( h ) = P M n P M n − 1 … P M 1 ( h ) P_(M_(n))(h)=P_(M_(n))P_(M_(n-1))dotsP_(M_(1))(h)P_{M_{n}}(h)=P_{M_{n}} P_{M_{n-1}} \ldots P_{M_{1}}(h)PMn(h)=PMnPMn−1…PM1(h)
The equality P M n ( h ) = P M 1 P M 2 … P M n ( h ) P M n ( h ) = P M 1 P M 2 … P M n ( h ) P_(M_(n))(h)=P_(M_(1))P_(M_(2))dotsP_(M_(n))(h)P_{M_{n}}(h)=P_{M_{1}} P_{M_{2}} \ldots P_{M_{n}}(h)PMn(h)=PM1PM2…PMn(h) is immediate.
For b) observe that
d ( h , M n ) 2 = ‖ m n − m n − 1 ‖ 2 + ‖ h − m n − 1 ‖ 2 = ‖ m n − m n − 1 ‖ 2 + ‖ m n − 1 − m n − 2 ‖ 2 + ‖ h − m n − 2 ‖ 2 = … = ‖ m n − m n − 1 ‖ 2 + ⋯ + ‖ m 2 − m 1 ‖ 2 + ‖ h − m 1 ‖ 2 d h , M n 2 = m n − m n − 1 2 + h − m n − 1 2 = m n − m n − 1 2 + m n − 1 − m n − 2 2 + h − m n − 2 2 = … = m n − m n − 1 2 + ⋯ + m 2 − m 1 2 + h − m 1 2 {:[d(h,M_(n))^(2)=||m_(n)-m_(n-1)||^(2)+||h-m_(n-1)||^(2)],[=||m_(n)-m_(n-1)||^(2)+||m_(n-1)-m_(n-2)||^(2)+||h-m_(n-2)||^(2)=dots],[=||m_(n)-m_(n-1)||^(2)+cdots+||m_(2)-m_(1)||^(2)+||h-m_(1)||^(2)]:}\begin{gathered} d\left(h, M_{n}\right)^{2}=\left\|m_{n}-m_{n-1}\right\|^{2}+\left\|h-m_{n-1}\right\|^{2} \\ =\left\|m_{n}-m_{n-1}\right\|^{2}+\left\|m_{n-1}-m_{n-2}\right\|^{2}+\left\|h-m_{n-2}\right\|^{2}=\ldots \\ =\left\|m_{n}-m_{n-1}\right\|^{2}+\cdots+\left\|m_{2}-m_{1}\right\|^{2}+\left\|h-m_{1}\right\|^{2} \end{gathered}d(h,Mn)2=‖mn−mn−1‖2+‖h−mn−1‖2=‖mn−mn−1‖2+‖mn−1−mn−2‖2+‖h−mn−2‖2=…=‖mn−mn−1‖2+⋯+‖m2−m1‖2+‖h−m1‖2
Remark. Obviously, Theorem 1 is also valid if H H HHH is a Hilbert space and X , Y X , Y X,YX, YX,Y are closed subspace of H H HHH, with Y ⊂ X Y ⊂ X Y sub XY \subset XY⊂X. Also, Theorem 2 is valid if H H HHH is a Hilbert space and M 1 ⊃ M 2 ⊃ ⋯ ⊃ M n M 1 ⊃ M 2 ⊃ ⋯ ⊃ M n M_(1)supM_(2)sup cdots supM_(n)M_{1} \supset M_{2} \supset \cdots \supset M_{n}M1⊃M2⊃⋯⊃Mn are closed subspaces of H H HHH.
A converse of the Reduction Principle is given in [3], Th.2.2.6:
Let H H HHH be an inner product space, X X XXX a complete subspace of H H HHH and K K KKK a closed and convex subset of X X XXX. If x x xxx is the orthogonal projection of h ∉ X h ∉ X h!in Xh \notin Xh∉X on X , m X , m X,mX, mX,m is the metric projection of h h hhh on K K KKK, then m m mmm is the metric projection of x x xxx on K K KKK.
A first converse of Theorem 1 is:
Theorem 3. Let H H HHH be an inner product space, X X XXX a complete subspace of H H HHH, and Y Y YYY a closed subspace of X X XXX. Let h ∈ H ∖ X h ∈ H ∖ X h in H\\Xh \in H \backslash Xh∈H∖X and let P X ( h ) P X ( h ) P_(X)(h)P_{X}(h)PX(h) and P Y ( h ) P Y ( h ) P_(Y)(h)P_{Y}(h)PY(h) be the orthogonal projections of h h hhh on X X XXX, respectively on Y Y YYY. Then P Y ( h ) P Y ( h ) P_(Y)(h)P_{Y}(h)PY(h) is the orthogonal projection of P X ( h ) P X ( h ) P_(X)(h)P_{X}(h)PX(h) on Y Y YYY.
Proof. Indeed, by hypothesis it follows:
( h − P X ( h ) , x ) = 0 , ∀ x ∈ X , ( h − P Y ( h ) , y ) = 0 , ∀ y ∈ Y , h − P X ( h ) , x = 0 , ∀ x ∈ X , h − P Y ( h ) , y = 0 , ∀ y ∈ Y , {:[(h-P_(X)(h),x)=0","AA x in X","],[(h-P_(Y)(h),y)=0","AA y in Y","]:}\begin{aligned} & \left(h-P_{X}(h), x\right)=0, \forall x \in X, \\ & \left(h-P_{Y}(h), y\right)=0, \forall y \in Y, \end{aligned}(h−PX(h),x)=0,∀x∈X,(h−PY(h),y)=0,∀y∈Y,
ON THE CONVERSES OF THE REDUCTION PRINCIPLE IN INNER PRODUCT SPACES
so that for every y ∈ Y y ∈ Y y in Yy \in Yy∈Y one has:
( P X ( h ) − P Y ( h ) , y ) = ( h − P Y ( h ) − h + P X ( h ) , y ) = ( h − P Y ( h ) , y ) − ( h − P X ( h ) , y ) = 0 P X ( h ) − P Y ( h ) , y = h − P Y ( h ) − h + P X ( h ) , y = h − P Y ( h ) , y − h − P X ( h ) , y = 0 {:[(P_(X)(h)-P_(Y)(h),y)=(h-P_(Y)(h)-h+P_(X)(h),y)],[quad=(h-P_(Y)(h),y)-(h-P_(X)(h),y)=0]:}\begin{gathered} \left(P_{X}(h)-P_{Y}(h), y\right)=\left(h-P_{Y}(h)-h+P_{X}(h), y\right) \\ \quad=\left(h-P_{Y}(h), y\right)-\left(h-P_{X}(h), y\right)=0 \end{gathered}(PX(h)−PY(h),y)=(h−PY(h)−h+PX(h),y)=(h−PY(h),y)−(h−PX(h),y)=0
It follows that P Y ( h ) P Y ( h ) P_(Y)(h)P_{Y}(h)PY(h) is the orthogonal projection of P X ( h ) P X ( h ) P_(X)(h)P_{X}(h)PX(h) on Y Y YYY.
A second converse of Theorem 1 is:
Theorem 4. Let H H HHH be an inner product space, X X XXX a complete subspace of H H HHH, and Y Y YYY a closed subspace of X X XXX with codim X Y = 1 codim X Y = 1 codim_(X)Y=1\operatorname{codim}_{X} Y=1codimXY=1. Let x 0 ∈ X ∖ Y x 0 ∈ X ∖ Y x_(0)in X\\Yx_{0} \in X \backslash Yx0∈X∖Y and P Y ( x 0 ) P Y x 0 P_(Y)(x_(0))P_{Y}\left(x_{0}\right)PY(x0) be the orthogonal projection of x 0 x 0 x_(0)x_{0}x0 on Y Y YYY. If h ∈ H ∖ X , P Y ( h ) = P Y ( x 0 ) h ∈ H ∖ X , P Y ( h ) = P Y x 0 h in H\\X,P_(Y)(h)=P_(Y)(x_(0))h \in H \backslash X, P_{Y}(h)=P_{Y}\left(x_{0}\right)h∈H∖X,PY(h)=PY(x0) and ( h − x 0 , x 0 − P Y ( x 0 ) ) = 0 h − x 0 , x 0 − P Y x 0 = 0 (h-x_(0),x_(0)-:}{:P_(Y)(x_(0)))=0\left(h-x_{0}, x_{0}-\right. \left.P_{Y}\left(x_{0}\right)\right)=0(h−x0,x0−PY(x0))=0, then P Y ( h ) = x 0 P Y ( h ) = x 0 P_(Y)(h)=x_(0)P_{Y}(h)=x_{0}PY(h)=x0.
Proof. If the equality ( h − x 0 , x ) = 0 h − x 0 , x = 0 (h-x_(0),x)=0\left(h-x_{0}, x\right)=0(h−x0,x)=0 is fulfilled for every x ∈ X x ∈ X x in Xx \in Xx∈X, then P X ( h ) = x 0 P X ( h ) = x 0 P_(X)(h)=x_(0)P_{X}(h)=x_{0}PX(h)=x0, i.e. the conclusion of the theorem.
For every y ∈ Y y ∈ Y y in Yy \in Yy∈Y we have
( h − x 0 , y ) = ( h − P Y ( x 0 ) − ( x 0 − P Y ( x 0 ) ) , y ) = ( h − P Y ( x 0 ) , y ) − ( x 0 − P Y ( x 0 ) , y ) = 0 h − x 0 , y = h − P Y x 0 − x 0 − P Y x 0 , y = h − P Y x 0 , y − x 0 − P Y x 0 , y = 0 {:[(h-x_(0),y)=(h-P_(Y)(x_(0))-(x_(0)-P_(Y)(x_(0))),y)],[quad=(h-P_(Y)(x_(0)),y)-(x_(0)-P_(Y)(x_(0)),y)=0]:}\begin{aligned} & \left(h-x_{0}, y\right)=\left(h-P_{Y}\left(x_{0}\right)-\left(x_{0}-P_{Y}\left(x_{0}\right)\right), y\right) \\ & \quad=\left(h-P_{Y}\left(x_{0}\right), y\right)-\left(x_{0}-P_{Y}\left(x_{0}\right), y\right)=0 \end{aligned}(h−x0,y)=(h−PY(x0)−(x0−PY(x0)),y)=(h−PY(x0),y)−(x0−PY(x0),y)=0
It follows that h − x 0 h − x 0 h-x_(0)h-x_{0}h−x0 is orthogonal to Y Y YYY.
Because, by hypothesis, ( h − x 0 , x 0 − P Y ( x 0 ) ) = 0 h − x 0 , x 0 − P Y x 0 = 0 (h-x_(0),x_(0)-P_(Y)(x_(0)))=0\left(h-x_{0}, x_{0}-P_{Y}\left(x_{0}\right)\right)=0(h−x0,x0−PY(x0))=0 it follows that ( h − x 0 , u ) = 0 h − x 0 , u = 0 (h-x_(0),u)=0\left(h-x_{0}, u\right)=0(h−x0,u)=0 for every u ∈ span { x 0 − P Y ( x 0 ) } u ∈ span x 0 − P Y x 0 u in span{x_(0)-P_(Y)(x_(0))}u \in \operatorname{span}\left\{x_{0}-P_{Y}\left(x_{0}\right)\right\}u∈span{x0−PY(x0)}. Because x 0 − P Y ( x 0 ) x 0 − P Y x 0 x_(0)-P_(Y)(x_(0))x_{0}-P_{Y}\left(x_{0}\right)x0−PY(x0) is orthogonal to Y Y YYY and Y Y YYY is a closed subspace of the Hilbert space X X XXX, it follows that X = span { x 0 − P Y ( x 0 ) } ⊕ Y X = span x 0 − P Y x 0 ⊕ Y X=span{x_(0)-P_(Y)(x_(0))}o+YX=\operatorname{span}\left\{x_{0}-P_{Y}\left(x_{0}\right)\right\} \oplus YX=span{x0−PY(x0)}⊕Y, i.e. X X XXX is the direct sum of the subspaces span { x 0 − P Y ( x 0 ) } span x 0 − P Y x 0 span{x_(0)-P_(Y)(x_(0))}\operatorname{span}\left\{x_{0}-P_{Y}\left(x_{0}\right)\right\}span{x0−PY(x0)} and Y Y YYY (see [2], Th.5.9 p. 77 and [1], Th.4, p.65). Consequently ( h − x 0 , x ) = 0 h − x 0 , x = 0 (h-x_(0),x)=0\left(h-x_{0}, x\right)=0(h−x0,x)=0 for every x ∈ X x ∈ X x in Xx \in Xx∈X.
Remark. The condition codim X Y = 1 codim X Y = 1 codim_(X)Y=1\operatorname{codim}_{X} Y=1codimXY=1 in Theorem 4 is essential. Indeed, let { e 1 , e 2 , e 3 } e 1 , e 2 , e 3 {e_(1),e_(2),e_(3)}\left\{e_{1}, e_{2}, e_{3}\right\}{e1,e2,e3} be the orthonormal basis of the Hilbert space R 3 , X = span { e 1 , e 2 } R 3 , X = span e 1 , e 2 R^(3),X=span{e_(1),e_(2)}\mathbb{R}^{3}, X=\operatorname{span}\left\{e_{1}, e_{2}\right\}R3,X=span{e1,e2}, Y = span { 0 } Y = span { 0 } Y=span{0}Y=\operatorname{span}\{0\}Y=span{0} and h = 3 e 1 + e 2 + 5 e 3 h = 3 e 1 + e 2 + 5 e 3 h=3e_(1)+e_(2)+5e_(3)h=3 e_{1}+e_{2}+5 e_{3}h=3e1+e2+5e3. Let x 0 = e 1 + 2 e 2 x 0 = e 1 + 2 e 2 x_(0)=e_(1)+2e_(2)x_{0}=e_{1}+2 e_{2}x0=e1+2e2. Then P Y ( x 0 ) = 0 P Y x 0 = 0 P_(Y)(x_(0))=0P_{Y}\left(x_{0}\right)=0PY(x0)=0 and P Y ( h ) = 3 e 1 + e 2 , P Y ( h ) = 0 P Y ( h ) = 3 e 1 + e 2 , P Y ( h ) = 0 P_(Y)(h)=3e_(1)+e_(2),P_(Y)(h)=0P_{Y}(h)=3 e_{1}+e_{2}, P_{Y}(h)=0PY(h)=3e1+e2,PY(h)=0. The conditions P Y ( x 0 ) = P Y ( h ) P Y x 0 = P Y ( h ) P_(Y)(x_(0))=P_(Y)(h)P_{Y}\left(x_{0}\right)=P_{Y}(h)PY(x0)=PY(h) and ( h − x 0 , x 0 − P Y ( x 0 ) ) = ( 2 e 1 − e 2 , e 1 + 2 e 2 ) = 0 h − x 0 , x 0 − P Y x 0 = 2 e 1 − e 2 , e 1 + 2 e 2 = 0 (h-x_(0),x_(0)-:}{:P_(Y)(x_(0)))=(2e_(1)-e_(2),e_(1)+2e_(2))=0\left(h-x_{0}, x_{0}-\right. \left.P_{Y}\left(x_{0}\right)\right)=\left(2 e_{1}-e_{2}, e_{1}+2 e_{2}\right)=0(h−x0,x0−PY(x0))=(2e1−e2,e1+2e2)=0 are fulfilled, but P X ( h ) = 3 e 1 + e 2 ≠ x 0 = e 1 + 2 e 2 P X ( h ) = 3 e 1 + e 2 ≠ x 0 = e 1 + 2 e 2 P_(X)(h)=3e_(1)+e_(2)!=x_(0)=e_(1)+2e_(2)P_{X}(h)=3 e_{1}+e_{2} \neq x_{0}=e_{1}+2 e_{2}PX(h)=3e1+e2≠x0=e1+2e2. Observe that codim X Y = 2 codim X Y = 2 codim_(X)Y=2\operatorname{codim}_{X} Y=2codimXY=2.
Examples. 1 ∘ 1 ∘ 1^(@)1^{\circ}1∘ Let l 2 = l 2 ( N ) l 2 = l 2 ( N ) l_(2)=l_(2)(N)l_{2}=l_{2}(\mathbb{N})l2=l2(N) be the space of all sequences x = ( x ( i ) ) x = ( x ( i ) ) x=(x(i))x=(x(i))x=(x(i)) of real numbers such that ∑ i = 1 ∞ x 2 ( i ) < ∞ ∑ i = 1 ∞   x 2 ( i ) < ∞ sum_(i=1)^(oo)x^(2)(i) < oo\sum_{i=1}^{\infty} x^{2}(i)<\infty∑i=1∞x2(i)<∞. It is known that l 2 l 2 l_(2)l_{2}l2 is a Hilbert space with respect to the inner product ( x , y ) = ∑ i = 1 ∞ x ( i ) y ( i ) ( x , y ) = ∑ i = 1 ∞   x ( i ) y ( i ) (x,y)=sum_(i=1)^(oo)x(i)y(i)(x, y)=\sum_{i=1}^{\infty} x(i) y(i)(x,y)=∑i=1∞x(i)y(i) and the norm ‖ x ‖ = ( ∑ i = 1 ∞ x 2 ( i ) ) 1 / 2 ‖ x ‖ = ∑ i = 1 ∞   x 2 ( i ) 1 / 2 ||x||=(sum_(i=1)^(oo)x^(2)(i))^(1//2)\|x\|= \left(\sum_{i=1}^{\infty} x^{2}(i)\right)^{1 / 2}‖x‖=(∑i=1∞x2(i))1/2. Let { e 1 , e 2 , … } e 1 , e 2 , … {e_(1),e_(2),dots}\left\{e_{1}, e_{2}, \ldots\right\}{e1,e2,…} be the canonical basis of l 2 l 2 l_(2)l_{2}l2. The closed subspace X = span { e 2 n − 1 ∣ n = 1 , 2 , 3 , … } ― X = span e 2 n − 1 ∣ n = 1 , 2 , 3 , … ¯ X= bar(span{e_(2n-1)∣n=1,2,3,dots})X=\overline{\operatorname{span}\left\{e_{2 n-1} \mid n=1,2,3, \ldots\right\}}X=span{e2n−1∣n=1,2,3,…}― is Chebyshevian in l 2 l 2 l_(2)l_{2}l2 and the orthogonal projection of h = ( h ( 1 ) , h ( 2 ) , … ) ∈ l 2 h = ( h ( 1 ) , h ( 2 ) , … ) ∈ l 2 h=(h(1),h(2),dots)inl_(2)h=(h(1), h(2), \ldots) \in l_{2}h=(h(1),h(2),…)∈l2 is P X ( h ) = ∑ i = 1 ∞ h ( 2 i − 1 ) e 2 i − 1 P X ( h ) = ∑ i = 1 ∞   h ( 2 i − 1 ) e 2 i − 1 P_(X)(h)=sum_(i=1)^(oo)h(2i-1)e_(2i-1)P_{X}(h)=\sum_{i=1}^{\infty} h(2 i-1) e_{2 i-1}PX(h)=∑i=1∞h(2i−1)e2i−1, because h − P X ( h ) = ∑ j = 1 ∞ h ( 2 j ) e 2 j h − P X ( h ) = ∑ j = 1 ∞   h ( 2 j ) e 2 j h-P_(X)(h)=sum_(j=1)^(oo)h(2j)e_(2j)h-P_{X}(h)=\sum_{j=1}^{\infty} h(2 j) e_{2 j}h−PX(h)=∑j=1∞h(2j)e2j is orthogonal on X X XXX.
Let Y = span { e 1 , e 3 + e 5 } Y = span e 1 , e 3 + e 5 Y=span{e_(1),e_(3)+e_(5)}Y=\operatorname{span}\left\{e_{1}, e_{3}+e_{5}\right\}Y=span{e1,e3+e5}. Then Y Y YYY is a Chebyshevian subspace of l 2 l 2 l_(2)l_{2}l2 (and of X ) X ) X)X)X) and
P Y ( h ) = h ( 1 ) e 1 + 1 2 [ h ( 3 ) + h ( 5 ) ] ( e 3 + e 5 ) . P Y ( h ) = h ( 1 ) e 1 + 1 2 [ h ( 3 ) + h ( 5 ) ] e 3 + e 5 . P_(Y)(h)=h(1)e_(1)+(1)/(2)[h(3)+h(5)](e_(3)+e_(5)).P_{Y}(h)=h(1) e_{1}+\frac{1}{2}[h(3)+h(5)]\left(e_{3}+e_{5}\right) .PY(h)=h(1)e1+12[h(3)+h(5)](e3+e5).
By Theorem 1 one obtains
P Y ( h ) = P Y P X ( h ) = P X P Y ( h ) . P Y ( h ) = P Y P X ( h ) = P X P Y ( h ) . P_(Y)(h)=P_(Y)P_(X)(h)=P_(X)P_(Y)(h).P_{Y}(h)=P_{Y} P_{X}(h)=P_{X} P_{Y}(h) .PY(h)=PYPX(h)=PXPY(h).
By Theorem 3, the orthogonal projection of the element
x = ∑ n = 1 ∞ h ( 2 n − 1 ) e 2 n − 1 x = ∑ n = 1 ∞   h ( 2 n − 1 ) e 2 n − 1 x=sum_(n=1)^(oo)h(2n-1)e_(2n-1)x=\sum_{n=1}^{\infty} h(2 n-1) e_{2 n-1}x=∑n=1∞h(2n−1)e2n−1
on Y Y YYY is
y 0 = h ( 1 ) e 1 + 1 2 [ h ( 3 ) + h ( 5 ) ] ( e 3 + e 5 ) . y 0 = h ( 1 ) e 1 + 1 2 [ h ( 3 ) + h ( 5 ) ] e 3 + e 5 . y_(0)=h(1)e_(1)+(1)/(2)[h(3)+h(5)](e_(3)+e_(5)).y_{0}=h(1) e_{1}+\frac{1}{2}[h(3)+h(5)]\left(e_{3}+e_{5}\right) .y0=h(1)e1+12[h(3)+h(5)](e3+e5).
Indeed,
x − y 0 = 1 2 [ h ( 3 ) − h ( 5 ) ] e 3 + 1 2 [ h ( 5 ) − h ( 3 ) ] e 5 + ∑ n = 4 ∞ h ( 2 n − 1 ) e 2 n − 1 x − y 0 = 1 2 [ h ( 3 ) − h ( 5 ) ] e 3 + 1 2 [ h ( 5 ) − h ( 3 ) ] e 5 + ∑ n = 4 ∞   h ( 2 n − 1 ) e 2 n − 1 x-y_(0)=(1)/(2)[h(3)-h(5)]e_(3)+(1)/(2)[h(5)-h(3)]e_(5)+sum_(n=4)^(oo)h(2n-1)e_(2n-1)x-y_{0}=\frac{1}{2}[h(3)-h(5)] e_{3}+\frac{1}{2}[h(5)-h(3)] e_{5}+\sum_{n=4}^{\infty} h(2 n-1) e_{2 n-1}x−y0=12[h(3)−h(5)]e3+12[h(5)−h(3)]e5+∑n=4∞h(2n−1)e2n−1
is orthogonal to Y Y YYY, so y 0 = P Y ( x ) y 0 = P Y ( x ) y_(0)=P_(Y)(x)y_{0}=P_{Y}(x)y0=PY(x).
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘ Let l 2 ( 4 ) = span { e 1 , e 2 , e 3 , e 4 } l 2 ( 4 ) = span e 1 , e 2 , e 3 , e 4 l_(2)(4)=span{e_(1),e_(2),e_(3),e_(4)}l_{2}(4)=\operatorname{span}\left\{e_{1}, e_{2}, e_{3}, e_{4}\right\}l2(4)=span{e1,e2,e3,e4} where e i ( j ) = δ i j , i , j = 1 , 2 , 3 , 4 ( e i ( j ) = δ i j , i , j = 1 , 2 , 3 , 4 e_(i)(j)=delta_(ij),i,j=1,2,3,4(:}e_{i}(j)=\delta_{i j}, i, j=1,2,3,4\left(\right.ei(j)=δij,i,j=1,2,3,4( see 1 ∘ ) 1 ∘ {:1^(@))\left.1^{\circ}\right)1∘), and X = span { e 1 , e 2 , e 3 } , Y = span { e 1 , e 2 } X = span e 1 , e 2 , e 3 , Y = span e 1 , e 2 X=span{e_(1),e_(2),e_(3)},Y=span{e_(1),e_(2)}X=\operatorname{span}\left\{e_{1}, e_{2}, e_{3}\right\}, Y=\operatorname{span}\left\{e_{1}, e_{2}\right\}X=span{e1,e2,e3},Y=span{e1,e2} and Z = span { e 1 } Z = span e 1 Z=span{e_(1)}Z=\operatorname{span}\left\{e_{1}\right\}Z=span{e1}.
If x 0 = 2 e 1 + e 2 + 2 e 3 x 0 = 2 e 1 + e 2 + 2 e 3 x_(0)=2e_(1)+e_(2)+2e_(3)x_{0}=2 e_{1}+e_{2}+2 e_{3}x0=2e1+e2+2e3, then P Y ( x 0 ) = 2 e 1 + e 2 P Y x 0 = 2 e 1 + e 2 P_(Y)(x_(0))=2e_(1)+e_(2)P_{Y}\left(x_{0}\right)=2 e_{1}+e_{2}PY(x0)=2e1+e2. For α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R let h = 2 e 1 + e 2 + α e 3 + β e 4 h = 2 e 1 + e 2 + α e 3 + β e 4 h=2e_(1)+e_(2)+alphae_(3)+betae_(4)h= 2 e_{1}+e_{2}+\alpha e_{3}+\beta e_{4}h=2e1+e2+αe3+βe4. Then P Y ( h ) = 2 e 1 + e 2 P Y ( h ) = 2 e 1 + e 2 P_(Y)(h)=2e_(1)+e_(2)P_{Y}(h)=2 e_{1}+e_{2}PY(h)=2e1+e2 and ( h − x 0 , x 0 − P Y ( x 0 ) ) = 2 ( α − 2 ) = 0 h − x 0 , x 0 − P Y x 0 = 2 ( α − 2 ) = 0 (h-x_(0),x_(0)-P_(Y)(x_(0)))=2(alpha-2)=0\left(h-x_{0}, x_{0}-P_{Y}\left(x_{0}\right)\right)=2(\alpha-2)=0(h−x0,x0−PY(x0))=2(α−2)=0 implies α = 2 α = 2 alpha=2\alpha=2α=2.
ON THE CONVERSES OF THE REDUCTION PRINCIPLE IN INNER PRODUCT SPACES
Every element h = 2 e 1 + e 2 + 2 e 3 + β e 4 , β ∈ R h = 2 e 1 + e 2 + 2 e 3 + β e 4 , β ∈ R h=2e_(1)+e_(2)+2e_(3)+betae_(4),beta inRh=2 e_{1}+e_{2}+2 e_{3}+\beta e_{4}, \beta \in \mathbb{R}h=2e1+e2+2e3+βe4,β∈R has as orthogonal projection on X X XXX
P X ( h ) = 2 e 1 + e 2 + 2 e 3 = x 0 P X ( h ) = 2 e 1 + e 2 + 2 e 3 = x 0 P_(X)(h)=2e_(1)+e_(2)+2e_(3)=x_(0)P_{X}(h)=2 e_{1}+e_{2}+2 e_{3}=x_{0}PX(h)=2e1+e2+2e3=x0
Observe that codim X Y = 1 codim X Y = 1 codim_(X)Y=1\operatorname{codim}_{X} Y=1codimXY=1.
Consider now the orthogonal projections on Z ( codim X Z = 2 ) Z codim X Z = 2 Z(codim_(X)Z=2)Z\left(\operatorname{codim}_{X} Z=2\right)Z(codimXZ=2). Then P Z ( x 0 ) = 2 e 1 , P Z ( h ) = 2 e 1 P Z x 0 = 2 e 1 , P Z ( h ) = 2 e 1 P_(Z)(x_(0))=2e_(1),P_(Z)(h)=2e_(1)P_{Z}\left(x_{0}\right)=2 e_{1}, P_{Z}(h)=2 e_{1}PZ(x0)=2e1,PZ(h)=2e1 and ( h − x 0 , x 0 − P Z ( x 0 ) ) = α + β − 3 = 0 h − x 0 , x 0 − P Z x 0 = α + β − 3 = 0 (h-x_(0),x_(0)-P_(Z)(x_(0)))=alpha+beta-3=0\left(h-x_{0}, x_{0}-P_{Z}\left(x_{0}\right)\right)=\alpha+\beta-3=0(h−x0,x0−PZ(x0))=α+β−3=0 implies α + β = 3 α + β = 3 alpha+beta=3\alpha+\beta=3α+β=3.
Choosing the element h = 2 e 1 + 2 e 2 + e 3 + 2 e 4 h = 2 e 1 + 2 e 2 + e 3 + 2 e 4 h=2e_(1)+2e_(2)+e_(3)+2e_(4)h=2 e_{1}+2 e_{2}+e_{3}+2 e_{4}h=2e1+2e2+e3+2e4 one obtains
P X ( h ) = 2 e 1 + 2 e 2 + e 3 ≠ 2 e 1 + e 2 + 2 e 3 = x 0 P X ( h ) = 2 e 1 + 2 e 2 + e 3 ≠ 2 e 1 + e 2 + 2 e 3 = x 0 P_(X)(h)=2e_(1)+2e_(2)+e_(3)!=2e_(1)+e_(2)+2e_(3)=x_(0)P_{X}(h)=2 e_{1}+2 e_{2}+e_{3} \neq 2 e_{1}+e_{2}+2 e_{3}=x_{0}PX(h)=2e1+2e2+e3≠2e1+e2+2e3=x0
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘ Let L 2 [ − 1 , 1 ] L 2 [ − 1 , 1 ] L_(2)[-1,1]L_{2}[-1,1]L2[−1,1] be the Hilbert space of all (Lebesgue) measurable realvalued functions on [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1,1][−1,1] with the property that ∫ − 1 1 h 2 ( t ) d t < ∞ ∫ − 1 1   h 2 ( t ) d t < ∞ int_(-1)^(1)h^(2)(t)dt < oo\int_{-1}^{1} h^{2}(t) d t<\infty∫−11h2(t)dt<∞. The inner product on L 2 [ − 1 , 1 ] L 2 [ − 1 , 1 ] L_(2)[-1,1]L_{2}[-1,1]L2[−1,1] is ( x , y ) = ∫ − 1 1 x ( t ) y ( t ) d t ( x , y ) = ∫ − 1 1   x ( t ) y ( t ) d t (x,y)=int_(-1)^(1)x(t)y(t)dt(x, y)=\int_{-1}^{1} x(t) y(t) d t(x,y)=∫−11x(t)y(t)dt and the associated norm is ‖ h ‖ = ( ∫ − 1 1 h 2 ( t ) d t ) 1 / 2 ‖ h ‖ = ∫ − 1 1   h 2 ( t ) d t 1 / 2 ||h||=(int_(-1)^(1)h^(2)(t)dt)^(1//2)\|h\|=\left(\int_{-1}^{1} h^{2}(t) d t\right)^{1 / 2}‖h‖=(∫−11h2(t)dt)1/2. Consider also the Legendre polynomials (see [2])
p 0 ( t ) = 1 2 , p 1 ( t ) = 6 2 t , p 2 ( t ) = 10 4 ( 3 t 2 − 1 ) , p 3 ( t ) = 14 4 ( 5 t 3 − 3 t ) p 0 ( t ) = 1 2 , p 1 ( t ) = 6 2 t , p 2 ( t ) = 10 4 3 t 2 − 1 , p 3 ( t ) = 14 4 5 t 3 − 3 t p_(0)(t)=(1)/(sqrt2),p_(1)(t)=(sqrt6)/(2)t,p_(2)(t)=(sqrt10)/(4)(3t^(2)-1),p_(3)(t)=(sqrt14)/(4)(5t^(3)-3t)p_{0}(t)=\frac{1}{\sqrt{2}}, p_{1}(t)=\frac{\sqrt{6}}{2} t, p_{2}(t)=\frac{\sqrt{10}}{4}\left(3 t^{2}-1\right), p_{3}(t)=\frac{\sqrt{14}}{4}\left(5 t^{3}-3 t\right)p0(t)=12,p1(t)=62t,p2(t)=104(3t2−1),p3(t)=144(5t3−3t)
and in general
p n ( t ) = ( − 1 ) n 2 n + 1 2 n ⋅ 2 ⋅ n ! ⋅ d n d t n [ ( 1 − t 2 ) n ] p n ( t ) = ( − 1 ) n 2 n + 1 2 n ⋅ 2 ⋅ n ! ⋅ d n d t n 1 − t 2 n p_(n)(t)=((-1)^(n)sqrt(2n+1))/(2^(n)*sqrt2*n!)*(d^(n))/(dt^(n))[(1-t^(2))^(n)]p_{n}(t)=\frac{(-1)^{n} \sqrt{2 n+1}}{2^{n} \cdot \sqrt{2} \cdot n!} \cdot \frac{d^{n}}{d t^{n}}\left[\left(1-t^{2}\right)^{n}\right]pn(t)=(−1)n2n+12n⋅2⋅n!⋅dndtn[(1−t2)n]
for n ≥ 0 n ≥ 0 n >= 0n \geq 0n≥0.
The set { p 0 , p 1 , … , p n } , n ≥ 0 p 0 , p 1 , … , p n , n ≥ 0 {p_(0),p_(1),dots,p_(n)},n >= 0\left\{p_{0}, p_{1}, \ldots, p_{n}\right\}, n \geq 0{p0,p1,…,pn},n≥0 is orthonormal in L 2 [ − 1 , 1 ] L 2 [ − 1 , 1 ] L_(2)[-1,1]L_{2}[-1,1]L2[−1,1]. Consider the following subspaces of L 2 [ − 1 , 1 ] L 2 [ − 1 , 1 ] L_(2)[-1,1]L_{2}[-1,1]L2[−1,1] :
X = span { p 0 , p 1 , p 2 , p 3 } , Y = span { p 0 , p 1 , p 2 } and Z = span { p 0 , p 1 } X = span p 0 , p 1 , p 2 , p 3 , Y = span p 0 , p 1 , p 2  and  Z = span p 0 , p 1 {:[X=span{p_(0),p_(1),p_(2),p_(3)}","quad Y=span{p_(0),p_(1),p_(2)}quad" and "],[Z=span{p_(0),p_(1)}]:}\begin{gathered} X=\operatorname{span}\left\{p_{0}, p_{1}, p_{2}, p_{3}\right\}, \quad Y=\operatorname{span}\left\{p_{0}, p_{1}, p_{2}\right\} \quad \text { and } \\ Z=\operatorname{span}\left\{p_{0}, p_{1}\right\} \end{gathered}X=span{p0,p1,p2,p3},Y=span{p0,p1,p2} and Z=span{p0,p1}
For every h ∈ L 2 [ − 1 , 1 ] h ∈ L 2 [ − 1 , 1 ] h inL_(2)[-1,1]h \in L_{2}[-1,1]h∈L2[−1,1] one obtains ([2], Th.4.14)
P X ( h ) = ( h , p 0 ) p 0 + ( h , p 1 ) p 1 + ( h , p 2 ) p 2 + ( h , p 3 ) p 3 P Y ( h ) = ( h , p 0 ) p 0 + ( h , p 1 ) p 1 + ( h , p 2 ) p 2 and P X ( h ) = h , p 0 p 0 + h , p 1 p 1 + h , p 2 p 2 + h , p 3 p 3 P Y ( h ) = h , p 0 p 0 + h , p 1 p 1 + h , p 2 p 2  and  {:[P_(X)(h)=(h,p_(0))p_(0)+(h,p_(1))p_(1)+(h,p_(2))p_(2)+(h,p_(3))p_(3)],[P_(Y)(h)=(h,p_(0))p_(0)+(h,p_(1))p_(1)+(h,p_(2))p_(2)quad" and "]:}\begin{gathered} P_{X}(h)=\left(h, p_{0}\right) p_{0}+\left(h, p_{1}\right) p_{1}+\left(h, p_{2}\right) p_{2}+\left(h, p_{3}\right) p_{3} \\ P_{Y}(h)=\left(h, p_{0}\right) p_{0}+\left(h, p_{1}\right) p_{1}+\left(h, p_{2}\right) p_{2} \quad \text { and } \end{gathered}PX(h)=(h,p0)p0+(h,p1)p1+(h,p2)p2+(h,p3)p3PY(h)=(h,p0)p0+(h,p1)p1+(h,p2)p2 and 
P Z ( h ) = ( h , p 0 ) p 0 + ( h , p 1 ) p 1 . P Z ( h ) = h , p 0 p 0 + h , p 1 p 1 . P_(Z)(h)=(h,p_(0))p_(0)+(h,p_(1))p_(1).P_{Z}(h)=\left(h, p_{0}\right) p_{0}+\left(h, p_{1}\right) p_{1} .PZ(h)=(h,p0)p0+(h,p1)p1.
Obviously, Z ⊂ Y ⊂ X ⊂ L 2 [ − 1 , 1 ] Z ⊂ Y ⊂ X ⊂ L 2 [ − 1 , 1 ] Z sub Y sub X subL_(2)[-1,1]Z \subset Y \subset X \subset L_{2}[-1,1]Z⊂Y⊂X⊂L2[−1,1] and P Z ( h ) = P Z P Y P X ( h ) P Z ( h ) = P Z P Y P X ( h ) P_(Z)(h)=P_(Z)P_(Y)P_(X)(h)P_{Z}(h)=P_{Z} P_{Y} P_{X}(h)PZ(h)=PZPYPX(h).
Let x 0 = p 0 + 2 p 1 + 2 p 2 + p 3 x 0 = p 0 + 2 p 1 + 2 p 2 + p 3 x_(0)=p_(0)+2p_(1)+2p_(2)+p_(3)x_{0}=p_{0}+2 p_{1}+2 p_{2}+p_{3}x0=p0+2p1+2p2+p3. If h ∈ L 2 [ − 1 , 1 ] ∖ X h ∈ L 2 [ − 1 , 1 ] ∖ X h inL_(2)[-1,1]\\Xh \in L_{2}[-1,1] \backslash Xh∈L2[−1,1]∖X then P Y ( h ) = P Y ( x 0 ) P Y ( h ) = P Y x 0 P_(Y)(h)=P_(Y)(x_(0))P_{Y}(h)=P_{Y}\left(x_{0}\right)PY(h)=PY(x0) iff ( h , p 0 ) = 1 , ( h , p 1 ) = 2 h , p 0 = 1 , h , p 1 = 2 (h,p_(0))=1,(h,p_(1))=2\left(h, p_{0}\right)=1,\left(h, p_{1}\right)=2(h,p0)=1,(h,p1)=2 and ( h , p 2 ) = 2 h , p 2 = 2 (h,p_(2))=2\left(h, p_{2}\right)=2(h,p2)=2. The condition ( x 0 − P Y ( x 0 ) , h − x 0 ) = 0 x 0 − P Y x 0 , h − x 0 = 0 (x_(0)-P_(Y)(x_(0)),h-x_(0))=0\left(x_{0}-P_{Y}\left(x_{0}\right), h-x_{0}\right)=0(x0−PY(x0),h−x0)=0 implies ( p 3 , h − x 0 ) = 0 p 3 , h − x 0 = 0 (p_(3),h-x_(0))=0\left(p_{3}, h-x_{0}\right)=0(p3,h−x0)=0 and, consequently, ( p 3 , h ) = ( p 3 , x 0 ) = 1 p 3 , h = p 3 , x 0 = 1 (p_(3),h)=(p_(3),x_(0))=1\left(p_{3}, h\right)=\left(p_{3}, x_{0}\right)=1(p3,h)=(p3,x0)=1. It follows P X ( h ) = x 0 P X ( h ) = x 0 P_(X)(h)=x_(0)P_{X}(h)=x_{0}PX(h)=x0. Observe that codim X Y = 1 codim X Y = 1 codim_(X)Y=1\operatorname{codim}_{X} Y=1codimXY=1.
Now P Z ( x 0 ) = p 0 + 2 p 1 P Z x 0 = p 0 + 2 p 1 P_(Z)(x_(0))=p_(0)+2p_(1)P_{Z}\left(x_{0}\right)=p_{0}+2 p_{1}PZ(x0)=p0+2p1 and P Z ( h ) = P Z ( x 0 ) P Z ( h ) = P Z x 0 P_(Z)(h)=P_(Z)(x_(0))P_{Z}(h)=P_{Z}\left(x_{0}\right)PZ(h)=PZ(x0) implies ( h , p 0 ) = 1 , ( h , p 1 ) = 2 h , p 0 = 1 , h , p 1 = 2 (h,p_(0))=1,(h,p_(1))=2\left(h, p_{0}\right)=1,\left(h, p_{1}\right)=2(h,p0)=1,(h,p1)=2. The condition ( x 0 − P Z ( x 0 ) , h − x 0 ) = 0 x 0 − P Z x 0 , h − x 0 = 0 (x_(0)-P_(Z)(x_(0)),h-x_(0))=0\left(x_{0}-P_{Z}\left(x_{0}\right), h-x_{0}\right)=0(x0−PZ(x0),h−x0)=0 implies
( 2 p 2 + p 3 , h − x 0 ) = 2 ( p 2 , h ) + ( p 3 , h ) − 5 = 0 2 p 2 + p 3 , h − x 0 = 2 p 2 , h + p 3 , h − 5 = 0 (2p_(2)+p_(3),h-x_(0))=2(p_(2),h)+(p_(3),h)-5=0\left(2 p_{2}+p_{3}, h-x_{0}\right)=2\left(p_{2}, h\right)+\left(p_{3}, h\right)-5=0(2p2+p3,h−x0)=2(p2,h)+(p3,h)−5=0
Let h 1 = p 0 + 2 p 1 + p 2 + 3 p 3 + p 4 h 1 = p 0 + 2 p 1 + p 2 + 3 p 3 + p 4 h_(1)=p_(0)+2p_(1)+p_(2)+3p_(3)+p_(4)h_{1}=p_{0}+2 p_{1}+p_{2}+3 p_{3}+p_{4}h1=p0+2p1+p2+3p3+p4 and h 2 = p 0 + 2 p 1 + 1 2 p 2 + 4 p 3 + p 4 h 2 = p 0 + 2 p 1 + 1 2 p 2 + 4 p 3 + p 4 h_(2)=p_(0)+2p_(1)+(1)/(2)p_(2)+4p_(3)+p_(4)h_{2}=p_{0}+2 p_{1}+\frac{1}{2} p_{2}+4 p_{3}+p_{4}h2=p0+2p1+12p2+4p3+p4. Then P Z ( h i ) = P Z ( x 0 ) , i = 1 , 2 P Z h i = P Z x 0 , i = 1 , 2 P_(Z)(h_(i))=P_(Z)(x_(0)),i=1,2P_{Z}\left(h_{i}\right)=P_{Z}\left(x_{0}\right), i=1,2PZ(hi)=PZ(x0),i=1,2 and ( x 0 − P Z ( x 0 ) , h i − x 0 ) = 0 , i = 1 , 2 x 0 − P Z x 0 , h i − x 0 = 0 , i = 1 , 2 (x_(0)-P_(Z)(x_(0)),h_(i)-x_(0))=0,i=1,2\left(x_{0}-P_{Z}\left(x_{0}\right), h_{i}-x_{0}\right)=0, i=1,2(x0−PZ(x0),hi−x0)=0,i=1,2, but P X ( h 1 ) ≠ P X ( h 2 ) ≠ x 0 P X h 1 ≠ P X h 2 ≠ x 0 P_(X)(h_(1))!=P_(X)(h_(2))!=x_(0)P_{X}\left(h_{1}\right) \neq P_{X}\left(h_{2}\right) \neq x_{0}PX(h1)≠PX(h2)≠x0. Observe that codim X Z = 2 codim X Z = 2 codim_(X)Z=2\operatorname{codim}_{X} Z=2codimXZ=2.

References

[1] Cheney, W., Analysis for Applied Mathematics, Springer-Verlag, New York-BerlinHeidelberg, 2001.
[2] Deutsch, F., Best Approximation in Inner-Product Space, Springer-Verlag, New York-Berlin-Heidelberg, 2001.
[3] Laurent, P.J., Approximation et Optimisation, Herman, Paris, 1972.
"T. Popoviciu" Institute of Numerical Analysis, O.P.1, C.P. 68 , Cluj-Napoca, RomANIA
E-mail address: cmustata@ictp.acad.ro
  1. Received by the editors: 21.03.2006.
    2000 Mathematics Subject Classification. 41A65, 46C05.
    Key words and phrases. Inner product spaces, the Reduction Principle, best approximation.
2006

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