Filomat 9:7 (05), 587 599 DOI 0.98/FIL507587M Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat Some Equivalent Characterizations of Inner Product Spaces and Their Consequences Dan Ştefan Marinescu a, Mihai Monea b, Mihai Opincariu c, Marian Stroe d a Colegiul Naţional Iancu De Hunedoara Hunedoara, Romania b Colegiul Naţional Decebal Deva, Romania c Colegiul Naţional Avram Iancu Brad, Romania d Colegiul Economic Emanoil Gojdu Hunedoara, Romania Abstract. In this paper, first we prove the equivalence of the Frechet and the Jordan-Neumann relations in normed spaces. Next, we will present some characterizations of the inner product spaces and we will show their equivalence. Finally, we will prove some consequences of our main results.. Introduction The problem of finding necessary and sufficient conditions for a normed space to be an inner product space has been previously investigated by many mathematicians. The first results were obtained by M. Frechet [6] in 935. Proposition.. (Frechet) A complex normed space (X, for all x, y, z X. ) is an inner product space if and only if x y z x y z x y y z z y, In the same year, Jordan and von Neumann [9] have given another characterization of the inner product spaces. This result is also known as the parallelogram law. Proposition.. (Jordan, von Neumann) A complex normed space (X, ) is an inner product space if and only if for all x, y X. x y x y x y, 00 Mathematics Subject Classification. Primary 46C5. Keywords. normed spaces, inner product spaces Received: 06 January 04; Accepted: 0 July 04 Communicated by Hari M. Srivastava Email addresses: marinescuds@gmail.com (Dan Ştefan Marinescu), mihaimonea@yahoo.com (Mihai Monea), opincariumihai@yahoo.com (Mihai Opincariu), maricu_stroe@yahoo.com (Marian Stroe)
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 588 These two characterizations are equivalent, as we will see in the next paragraph. More similar results are known today. A large collection of the results can be found in []. In this paper we want to present other characterizations of inner product spaces. The results will be detailed in the third section and they are more general than most of those mentioned in the bibliography. Finally, we will present a consistent lot of consequences.. The Equivalence of the First Two Characterizations The characterizations described in the Propositions. and. are the best known at this moment. But these results are equivalent and this equivalence exists in any normed spaces, not necessarily inner product spaces. Proposition.. Let (X, ) be a real or complex normed space. The next two statements are equivalent: (FR) x y z x y z x y z y z y, for all x, y, z X, (JN) x y x y x y, for all x, y X. Proof. (FR) (JN) If we choose z y in (FR), we observe that we obtain (JN). (JN) (FR) We apply (JN) three times and we obtain and But x y z x y z x y z y z By adding these equalities, we obtain (FR). x y z y z y z. ( x y ) (x z) x y x z. y z ( x y ) (x z) 3. Main Results In this section we present six characterizations of an inner product space. These results are presented in Theorem 3.. Firstly, we consider (X, ) a real or complex normed space. Theorem 3.. The following seven statements are equivalent: a) X is an inner product space; b) For all n N, n, for all a, a,..., a n R and for any x, x, x,..., x n X x x k x x k c) For all n N, n, for all a, a,..., a n R and for any x, x,..., x n X x k (a a... a n ) x k k<l n k<l n a l x l x k ; a l x l x k ;
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 589 d) For all n N, n, for all a, a,..., a n R, with n and for any x, x,..., x n X x k x k k<l n a l x l x k ; e) For all n N, n, for all a, a,..., a n R with n 0 and for any x, x,..., x n X x k x k f) For all n N, n and for any x, x,..., x n X n x k x k n g) (Hlawka) For all n N, n and for any x, x,..., x n X x k (n ) x k k<l n k<l n k<l n x l a j x k ; a j x l x k ; x l x k. Proof. We will prove the next implications : a) b) c) d) e) f ) ) a) a) b) We use the identity, x x y, x z x z, x y y x z y z, which is true for all x, y, z X. Then, we obtain (a a... a n ) x x k (x x k ) a k x x k a l x x k, x x l a l x x l, x x k k<l n k<l n a k x x k a l ( x x k, x x l x x l, x x k ) k<l n a k x x ( k a l x xk x x l x k x l ) k<l n a k a l a l x x k a l x k x l k<l n l<k n k<l n
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 590 x x k a l x l x k. k<l n b) c) We choose x 0. c) d) It is obvious from the statement c). d) e) Firstly, we make the substitution. The relation from d) becomes ( ) x k a k x k a k x k x k ( ) k<l n Now, we replace x k with x k and the conclusion follows immediately. e) f ) We apply the relations from d) in the case a a... a n n. f ) ) Firstly, we apply f) for any x k and x l, with n. We obtain Then, the relation from f) becomes x k n n k<l n x k x l x k x l x l x k. x k x k n k<l n k<l n x k (n ) and we obtain the conclusion. ) a) If we choose n 3, we obtain a l x l x k a l x l x k. ( xk x l x k x l ) ( xk x l ) x k k<l n k<l n x k x l x k x l x x x 3 x x x 3 x x x x 3 x x 3 for all x, x, x 3 X. Proposition. concludes our proof. Remarks.. The characterizations from b), c) or d) are more general than other characterizations of the same type because the coefficients a, a,.., a n do not have to be nonnegative numbers. For example, the characterizations from b), but in more restrictive conditions, represents the main result from [4] (see Lemma.3.). The characterizations from e) represents a transformation for a normed space of the Theorem 3 from [5]. 3. The characterizations from f) or g) are known but are very useful for our chain of implications.
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 59 4. Consequences In this final section we want to show the usefulness of the results from the previous section. We will present some results from our references, but with solutions that use the Theorem 3.. We will start with Corollary. from [] and we will give it in the next improved form: Corollary 4.. A real or complex normed space (X, ) is an inner product space if and only if for any x, y X and a, b R. ax by bx ay (a b )( x y ), Proof. If the normed space X is an inner product space we have, from c), Theorem 3., the next identity for any x, y X and a, b R. This is equivalent to ax by (a b)(a x b y ) ab x y, () ax by a x b y ab ( x y x y ). () If we replace in (), y with y and reverse a with b we obtain bx ay (b a)(b x a y ) ab x y, bx ay b x a y ab ( x y x y ). (3) If we choose a b in () we obtain the parallelogram law. Now, the conclusion follows if we add the relations () and (3). The second part of the proof is obvious, if we choose a b in the hypothesis. The next results are represented by Corollary.. from [3]. We recall its content here, but we will present a different solution. Corollary 4.. (Moslehian, Rassias) A real or complex normed space (X, ) is an inner product space if and only if x x i x x i, {,} i for any n N, n and for all x, x,..., x n X. {,} i Proof. Firstly, it is easy to observe that x {,} x i i n x i. ( ) Now we apply c) from Theorem 3.. and we obtain x x i x x i i i i x x i a j x i x j. i i<j n
We do the summation for {, } and we obtain D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 59 x x i n i x i. We use (*) and we obtain the conclusion for the first part. The second part is obtained if we choose n. The results from Proposition 4.3 and 4.4. can be considered as an extension for the distance formula in the inner product space. Proposition 4.3. Let X be an inner product space and a, a,..., a n R, b, b,..., b n R with n and n b i. Let be x, x,..., x n X. If u, v X so that u n x i and v n b i x i then u v ( bj a j ) (bi ) xj x i Proof. Using various properties we obtain the following: u v b i u b i x i b i u x i xj b j b i x i b i a j x j x i b i j a j x j a j x i j xj b j b i x i xj b j b i x i xj xj b i a j x i a j x i xj b j b i x i j xj xj xj b i a j x i a j x i b j b i x i j ( ) bj a j (bi ) xj x i, q.e.d.
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 593 Proposition 4.4. Let X be an inner product space and a, a,..., a n R with n 0. Let x, x,..., x n X and y, y,..., y n X. If u, v X such that u n x i and v n y i then u v xi ( xi yi ) a j y j a j x j y j. i,j Proof. Firstly, we will prove the result in the case n. Then u v u y i u yi a j yj a j y j j yi a j x i y j a j y j j yi a j x i y j a j y j j xi a j a i y j xi a j x j j yi a j y j xi ( xi yi ) a j y j a j x j y j i,j Next, we consider the general case, so n 0. We assume now b, b,..., b n R so that b i all i {,,.., n}. It is easy to observe that n b i. Now, we consider u, v X so that u n b i x i and v n b i y i. Then u u and v v. By using first part of the proof, we obtain u v xi ( xi yi ) b j b i y j b i b j x j y j. i,j for Then a j i,j u v u v x i y j a j ( xi x j yi y j )
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 594 xi ( xi yi ) a j y j a j x j y j. i,j As a consequence of Proposition 4.4., we obtain a very short proof for the next corollary. This result represents another characterization of the inner product spaces and it is known as a version of the generalized parallelogram law (see [], pag. 3). Corolarry 4.5. (Moslehian) Let (X, ) be a real or complex normed space. Then the norm is provided from an inner product space if and only if ( ) x i x j y i y j x i y j xi y i, i,j i,j for any x, x,..., x n X and y, y,..., y n X. Proof. Let u n x i and v n x i. Then ( ) xi y i x i y j i,j x i y j i,j x i x j i,j i,j i,j u v ( xi x j yi y j ) ( xi yi ) x j y j i,j ( ) y i y j x i y j xi y i. For the reverse implication, we choose n and y y 0 and we obtain Now, our proof is ready due to Proposition.. i,j x x x x 4 x 4 x x x. Another version of the generalized parallelogram law could be considered the Corollary.7. from []. In this paper we will give another proof for this result. Corollary 4.6.(Moslehian) The real or complex normed space (X, ) is an inner product space if and only if for any n N, n we have ri rj x i x j ri rj r j r i y i y j r j r i i,j ri rj x i y j ( ) r j r i xi y i, for all x, x,..., x n X, y, y,..., y n X and r, r,.., r n (0, ).
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 595 Proof. We apply Proposition 4.4. and for i, n, we replace r i, x i with r i x i and y i with r i y i. We obtain which is equivalent to r i r j ri x i r j x j i,j r i r j ri y i r j y j ri x i r j y j r i r j (r i x i r i y i ) r i, ( ) ri rj ri r r i r j j x i x j r j r i i,j ( ) ri rj ri r r i r j j x i y j r j r i ( ) ri rj ri r r i r j j y i y j r j r i (x i y i ). We perform the simplification in the left term an we obtain the conclusion. Second part of the proof follows the same idea from the proof of the previous proposition. We choose n, y y 0 and r r 0. Remark. In fact, the Corollary 4.6. represents a more general result than the previous. If we choose r r... r n we obtain x i x j y i y j which is equivalent to the identity from Corollary 4.5. ( ) x i y j xi y i, A very interesting applications of the main result is Corrolary 4.8. We will give a version for an inner product space for an inequality of G. Dospinescu (see pag. 3 from [3]). Firstly, we will prove the next useful lemma: Lemma 4.7. Let (X, ) a real or complex normed space and n N, n. For any a, a,..., a n (0, ), x, x,..., x n X and p [, ) we put s(k) i,j k p x i p k (a a... ) p x i. Then for all k {,, 3,.., n }. s(k ) s(k), Proof. We evaluate the diference s(k ) s(k) and we obtain s(k ) s(k) x k p k p p k (a a... ) p x i (a a... ) p x i.
Because the function is convex, we obtain and the conclusion follows. D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 596 x p k p k k (a a... ) p x i x i k a i k k x i k a a p kx k k i a i k k a k i k a x i k i a a p k k i a x k i k k a i k k a a p ix i k i a a n k i a x n p i k k a p ix i k a a n x n p i p k (a a... ) p x i x k p p Corollary 4.8. Let X an inner product space and n, k N so that k < n. Then x i x j n x i x j, k i<j k for any x, x,..., x n X. Moreover, the number n k is the best possible. Proof. First, let α R so that x i x j α i<j k x i x j, for any x, x,..., x n X. If we choose x 0 and x x 3... x n 0 we obtain that α n k. Now, we want to prove the validity of the inequality for α n k. We apply the point c) of Theorem 3.. and we obtain x i x j n x i x i. Then, the inequality which we want to prove is equivalent to n x i x i n k k k k x i x i x i n x i k x i k k x i. From previous Lemma, the last inequality is true because is equivalent to s(n) s(k) in the case p and a a... a n.
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 597 The result from the next proposition represents a version for an inner product space of a Hayashi s inequality. (see [7] or [0], pag. 97) Proposition 4.9. Let X be an inner product space. Then z z z z z z z z z z 3 z z 3, cyclic for all z, z, z, z 3 X. The equality holds if z {z, z, z 3 } or z z 3 z z (z z ) z z 3 z z (z z ) z z z z 3 (z z 3) 0. Proof. The case of equality for z {z, z, z 3 } is evident. From this consideration, we choose z, z, z, z 3 X, but z {z, z, z 3 } and we apply b) from Theorem 3.. for n 3. We obtain 3 3 3 (z z k ) z z k k<j 3 a j z k z l, for any a, a, a 3 R. Then 3 3 z z k a j z k z l. k<l 3 With a z z 3 zz, a z z 3 zz and a 3 z z zz 3, we have zj z l z z k zj z l z j z l z z k z j z k z k z l z z k z z l zj z l z z k z z z z 3 z z 3 z z z z z z 3 zj z l z z k z j z l z z k z j z l z z k z z z z 3 z z 3, z z z z z z 3 which is equivalent to conclusion. From the proof we also obtain the equality in the case which is equivalent to 3 (z z k ) 0, z z 3 z z (z z ) z z 3 z z (z z ) z z z z 3 (z z 3) 0.
D. Ş Marinescu et al. / Filomat 9:7 (05), 587 599 598 Finally, we will use the Theorem 3.. to give a short proof for the next result. It represents the version for the inner product space of Zarantonello s inequality and it can be found in [7]. Proposition 4.0. (Zarantonello) Let X be a Hilbert space and f : X X a function such that ( ) f (x) f y x y for all x, y X. Let be n N, n. Then, for all a, a,..., a n 0 with n and for any x, x,..., x n X we have: f a j x j j f (x i ) i<j a j ( xi x j f (x i ) f ( x j ) ). Proof. We have f a j x j j j f a j x j j f (x i ) f (x i ) f a j x j f (x i ) f a j x j f (x i ) j a j f (xi ) f (x j ). Now it is sufficient to prove that f a j x j f (x i ) j xj a j x i. j But f a j x j f (x i ) j a j x j j a j x i j a j x j x i j j ( ) a j xj x i. Using the statement d) of Theorem 3.. we obtain ( ) a j xj x i j xj a j x i. j
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