summary:We define the radius of the inscribed and circumscribed circumferences in a triangle located in a real normed space and we obtain new characterizations of inner product spaces

Alsina, C.

Guijarro, P.

Tomás, M. S.

English

Institute of Mathematics AS CR, v. v. i.

Archivum MathematicumClaudi Alsina; Piedad Guijarro Carranza; M. S. TomásCharacterizations of inner product structures involving the radius of the inscribed orcircumscribed circumferenceArchivum Mathematicum, Vol. 32 (1996), No. 3, 233--239Persistent URL: http://dml.cz/dmlcz/107576Terms of use:© Masaryk University, 1996Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czARCHIVUM MATHEMATICUM (BRNO)Tomus 32 (1996), 233 { 239CHARACTERIZATIONS OF INNER PRODUCTSTRUCTURES INVOLVING THE RADIUS OF THEINSCRIBED OR CIRCUMSCRIBED CIRCUMFERENCEC. Alsina, P. Guijarro and M.S. TomásAbstract. We dene the radius of the inscribed and circumscribed circumferencesin a triangle located in a real normed space and we obtain new characterizations ofinner product spaces.1. On the radius of the inscribedcircumference in a triangle in a normed spaceIn an inner product space (i.p.s.) the radius of the inscribed circumference ina triangle of sides x; y and x  y is given by the formula(1) ps(s   kxk)(s  kyk)(s   kx  yk)swhere s is the semiperimeter s = (kxk+ kyk+ kx  yk)2 .Let (E; k k) be a real normed space. If x; y are two independent vectors inEnf0g and (see [4]) if w(x; y) = kykx+ kxkykxk+ kyk is the bisectrix of x and y, in thetriangle of sides x; y and x y, we can consider the bisectrices w(x; y), w( y; x y)and w( x; y   x). It is a straightforward computation to prove that these threelines intersect in a point, i.e. there exist three constants = kxk+ kyk2s ;  = kyk+ kx  yk2s ;  = kxk+ kx  yk2s ;in R such that w(x; y) = y + w( y; x  y) = x+ w( x; y   x)1991 Mathematics Subject Classication : 1980, Rev. 1985: 46C99.Key words and phrases: inner product space, norm derivative 0, bisectrix, perpendicularbisector.Received January 18, 1995.234 C. ALSINA, P. GUIJARRO AND M.S. TOMÁSNow, in order to dene the radius of the inscribed circumference, we need adenition of height in a real normed space. To do that, let us consider (see [5])the functions 0 : E E ! R dened by0(x; y) = limt!0 kx+ tyk2   kxk22t :The mappings 0 play a crucial role in characterizing inner product spaces. Infact, when the norm derives from an inner product (E; ), then 0(x; y) = x  y,i.e., 0 reduce to the given inner product.We quote here some elementary results concerning the functions 0 that wewill use in this paper:(i) 0(x; x) = kxk2 and j0(x; y)j  kxk kyk(ii) 0+(x; y) = 0+(x; y) = 0+(x; y),   0(iii) 0+(x; y) = 0+(x; y) = 0 (x; y),   0(iv) 0+(x; x+ y) = 0+(x; y) + kxk2(v) 0+(x; y) = 0+(y; x) for all x; y in E if and only if E is an inner productspace.In [1] by using the functions 0 we introduce the following denition of heightover the side x  yh(x; y) = y + 0+(y   x; y)kx  yk2 (x  y) for all x; y in E linearly independent;so we can dene \the radius r(x; y) of the inscribed circumference" as the norm ofh (w(x; y); w( y; x  y)), i.e.(2) r(x; y) := kxk kyk2s  xkxk   0  ykyk ; xkxk ykykObserve that r(x; y) is not symmetric in x and y.We want to nd when the expressions (1) and (2) are equal.Theorem 1.1. Let (E; k k) be a real normed space with dim E  2. Then E isan i.p.s. if and only if for all linearly independent vectors x; y in Er(x; y) = ps(s   kxk)(s  kyk)(s   kx  yk)swhere s = kxk+ kyk+ kx  yk2 .Proof. If we assume that r(x; y) dened by (2) can be obtained by means of (1),then we have: r(x; y)2 = (s   kxk)(s  kyk)(s   kx  yk)sCHARACTERIZATIONS OF INNER PRODUCT STRUCTURES : : : 235i.e.kxk2kyk24s2  xkxk   0   ykyk ; xkxk ykyk2 = (s   kxk)(s   kyk)(s   kx  yk)sor equivalentlykykx  0 (y; x) ykyk2 = 4s(s   kxk)(s   kyk)(s   kx  yk):Substituting in this equality y by tz, with t > 0, dividing by t2 and taking limitswhen t tends to zero, we obtainkkzkx   0 (z; x) zkzk2 = limt!0+ 4s(s   kxk)(s  ktzk)(s  kx  tzk)t2= limt!0+ 4t2 kxk+ tkzk+ kx  tzk2  tkzk+ kx  tzk   kxk2  kxk+ kx  tzk   tkzk2  kxk+ tkzk   kx  tzk2= limt!0+ (kx  tzk+ tkzk)2   kxk22t kxk2   (kx  tzk   tkzk)22t == limt!0+ kx  tzk2   kxk22( t) + kzk22 t + kx  tzk kzk kx  tzk2   kxk22( t)   kzk22 t+ kx  tzk kzk=   0 (x; z) + kxk kzk  0 (x; z) + kxk kzk= kxk2kzk2   0 (x; z)2;and therefore kzk2x  0 (z; x)z2 = kzk2  kxk2kzk2   0 (x; z)2 :The substitution x = z + y yieldskzk2  kz + yk2kzk2   0 (z + y; z)2 = kzk2(z + y)   0 (z; z + y)z2= kzk2z + kzk2y   kzk2z   0 (z; y)z2= kzk2y   0 (z; y)z2 == kzk2  kyk2kzk2   0 (y; z)2 ;i.e., kyk2kzk2   0 (y; z)2 = kz + yk2kzk2   0 (z + y; z)2:236 C. ALSINA, P. GUIJARRO AND M.S. TOMÁSThus for all linearly independent vectors u; v in E with kuk = kvk = 1, if wesubstitute in the last equality z = u  v, y = v we obtain:kvk2ku  vk2   0 (v; u  v)2 = kuk2ku  vk2   0 (u; u  v)2;and consequently 0 (v; u  v) = 0 (u; u  v)which in turn implies 0 (v; u)  kvk2 = kuk2   0+(u; v), i.e.,1  0 (v; u) = 1  0+(u; v) :Since 0 (v; u)  kuk kvk = 1 and 0+(u; v)  kuk kvk = 1 we deduce0 (v; u) = 0+(u; v);and interchanging the roles of u and v:0 (u; v) = 0+(v; u);whence 0+(v; u) = 0 (u; v)  0+(u; v) as well as 0+(u; v) = 0 (v; u)  0+(v; u),i.e., 0+(u; v) = 0+(v; u), whenever kuk = kvk = 1, u; v independent vectors and Eis an inner product space. Note. The value r(x; y) as introduced above is not the unique possible denition ofthe radius of the inscribed circumference, since we can consider other heights likeh (w(x; y); w( x; y   x)), h (w( y; x  y); w(x; y)),h (w( y; x  y); w( x; y   x)), h (w( x; y   x); w(x; y)) orh (w( x; y   x); w( y; x  y)) instead of h (w(x; y); w( y; x  y)).2. On the radius of the circumscribedcircumference in a triangle in a normed spaceIf E is an i.p.s. and x; y are two independent vectors in Enf0g, in the triangle ofsides x; y and x y, the center of the circumscribed circumference is the intersectionof the perpendicular bisectors, and the radius R is given by the formulakxk kyk kx  yk4ps(s   kxk)(s   kyk)(s   kx  yk)where s = (kxk+ kyk + kx  yk) =2 is the semiperimeter of the triangle.In a real normed space (E; k k) we can consider the perpendicular bisector ofthe side x  y by takingM (x; y) = x+ y2 + u =  2 RCHARACTERIZATIONS OF INNER PRODUCT STRUCTURES : : : 237where u is a vector in E \orthogonal" to x  y.We can assume that u admits the form u = x + y and having in mind thatin an i.p.s. u  (x  y) = 0 is immediate to prove that  kxk2   x  y =   kyk2   x  y :Thus in the real normed space E, replacing the inner product  by 0 , we canconsider u =  kyk2   0 (x; y) x+  kxk2   0 (y; x) yandM (x; y) = x+ y2 +  (kyk2   0 (x; y))x+ (kxk2   0 (y; x))y =  2 RThe next step is to dene the radius R. First in the triangle of sides x; y andx  y considering the corresponding three perpendicular bisectors is easy to provethat this three lines intersected in a point, i.e., there exist ; ;  in R such thatx2 + w = y2 + v = x+ y2 + uwhere u; v and w are respectively \orthogonal" to x  y, y and x.Then, we dene R(x; y) as the norm of x2 + w and by a straightforward com-putation we obtain that R(x; y) has the following expressionR(x; y) = kyk2  kxk2   0 (x; y)x+ kxk2  kyk2   0 (y; x) y2kxk2kyk2   0 (x; y)2   0 (y; x)2 :This denition is only possible if0 (x; y)2 + 0 (y; x)2 < 2kxk2kyk2i.e., 0 (x; y) < kxk kyk or 0 (y; x) < kxk kyk. For this reason we will assumethat E is strictly convex (in these spaces j0 (x; y)j 6= kxk kyk for all x and y in Elinearly independent (see [7])).Theorem 2.1. Let (E; k k) be a strictly convex real normed space with dimE  3. E is an i.p.s., if and only if, for all x; y independent vectors in Enf0gR(x; y) = kxk kyk kx  yk4ps(s   kxk)(s  kyk)(s   kx  yk)where s = (kxk+ kyk + kx  yk) =2.238 C. ALSINA, P. GUIJARRO AND M.S. TOMÁSProof. If we let y = tz with t > 0 and we take limit when t tends to zero weobtain kxk2 kzk2x  0 (z; x)z2kxk2kzk2   0 (x; z)2   0 (z; x)2 = limt!0+R(x; tz) == 0@ limt!0+ kxk2t2kzk2kx  tzk2[kx  tzk2   (tkzk   kxk)2] h(kxk+ tkzk)2   kx  tzk2i1A1=2 == kxkkzk limt!0+ tkx  tzkkx  tzk2   (tkzk   kxk)21=2  limt!0+ tkx  tzk(kxk+ tkzk)2   kx  tzk2!1=2= kxkkzks kxk 20 (x; z) + 2kxk kzks kxk2kxk kzk+ 20 (x; z) == kzk kxk22pkxk2kzk2   0 (x; z)2and therefore, xkxk   0  zkzk ; xkxk zkzk = 2  0   xkxk ; zkzk2   0   zkzk ; xkxk22r1  0   xkxk ; zkzk2 :If u = xkxk , v = zkzk , then for all unitary and independent vectors u and v inE u  0 (v; u)v = 2  0 (u; v)2   0 (v; u)22p1  0 (u; v)2Now, if 0 (v; u) = 0, then,2q1  0 (u; v)2 = 2  0 (u; v)2and therefore 0 (u; v) = 0 and (see [5], [6]) E is an i.p.s. Note. If we dene the radius R(x; y) as the norm of y2 + v or as the norm ofx+y2 +u the same expression obtained in the initial denition of R(x; y) appears.CHARACTERIZATIONS OF INNER PRODUCT STRUCTURES : : : 239Analogous denitions of R(x; y) can be given by replacing the role of 0  by 0+or by changing the order of the arguments appearing in 0 . For example if weconsider that the radius of the circumscribed circumference is given byR̂(x; y) = kyk2  kxk2   0+(x; y)x+ kxk2  kyk2   0+(y; x) y2  kxk2kyk2   0 (x; y)2which is equal to R(x; y) in an i.p.s. then, a strictly convex real normed space Ewith dim E  2 is an i.p.s. if and only ifR̂(x; y) = kxk kyk kx  yk4ps(s   kxk)(s   kyk)(s   kx  yk) :The proof is immediate using the fact that the symmetry of the second memberof last expression and the symmetry of the numerator of R̂(x; y) imply 0 (x; y)2 =0 (y; x)2 for all x; y in E and E is an i.p.s.References[1] Alsina, C., Guijarro, P. and Tomás, M. S., On heights in real normed spaces and charac-terizations of inner product structures, Jour. Int. Math. & Comp. Sci. Vol. 6, N. 2, 151-159(1993).[2] Alsina, C., Guijarro, P. and Tomás, M. S., A characterization of inner product spaces basedon a property of height's transform, Arch. Math. 61 (1993), 560-566.[3] Alsina, C. and Garcia Roig, J. L., On a functional equation characterizing inner productspaces, Publ. Math. Debrecen 39 (1991), 299-304.[4] Alsina, C., Guijarro, P. and Tomás, M. S., Some remarkable lines of a triangle in realnormed spaces and characterizations of inner product structures, (Accepted for publicationin Aequationes Mathematicae).[5] Amir, D., Characterization of inner product spaces, Basel-Boston (1986).[6] James, R. C., Inner products in normed linear spaces, Bull. Amer. Math. Soc. Vol. 53 (1947),559-566.[7] Tapia, R. A., A characterization of inner product spaces, Proc. Amer. Math. Soc. Vol. 41(1973), 569-574.C. Alsina, P. Guijarro and M.S. TomásSecció de Matematiques, ETSABUniv. Politécnica CatalunyaDiagonal 64908028 Barcelona, SPAIN

Characterizations of inner product structures involving the radius of the inscribed or circumscribed circumference

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Characterizations of inner product structures involving the radius of the inscribed or circumscribed circumference

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Institute of Mathematics AS CR, v. v. i.