Ratio Mathematica Volume 39, 2020, pp. 55-67 New structure of norms on Rn and their relations with the curvature of the plane curves Amir Veisi* Ali Delbaznasab† Abstract Let f1,f2, . . . ,fn be fixed nonzero real-valued functions on R, the real numbers. Let ϕn(Xn) = ( x21f 2 1 + x 2 2f 2 2 + . . . + x 2 nf 2 n )1 2 , where Xn = (x1,x2, . . . ,xn) ∈ Rn. We show that ϕn has properties similar to a norm function on the normed linear space. Although ϕn is not a norm on Rn in general, it induces a norm on Rn. For the nonzero function F : R2 → R, a curvature formula for the implicit curve G(x,y) = F2(x,y) = c 6= 0 at any regular point is given. A similar result is presented when F is a nonzero function from R3 to R. In con- tinued, we concentrate on F(x,y) = ∫ b a ϕ2(x,y)dt. It is shown that the curvature of F(x,y) = c, where c > 0 is a positive multiple of c2. Particularly, we observe that F(x,y) = ∫ π 2 0 √ x2 cos2 t + y2 sin2 tdt is an elliptic integral of the second kind. Keywords: norm; curvature; homogeneous function; elliptic integral. 2010 AMS subject classifications: 53A10. 2010 AMS subject clas- sifications: 53A10. 1 *Faculty of Petroleum and Gas, Yasouj University, Gachsaran, Iran; aveisi@yu.ac.ir †Farhangian University, Kohgiluyeh and Boyer-Ahmad Province, Yasouj, Iran; delbaz- nasab@gmail.com 1Received on October 31st, 2020. Accepted on December 17th, 2020. Published on December 31st, 2020. doi: 10.23755/rm.v39i0.552. ISSN: 1592-7415. eISSN: 2282-8214. ©Amir Veisi et al. This paper is published under the CC-BY licence agreement. 55 Amir Veisi and Ali Delbaznasab 1 Introduction A normed linear space is a real linear space X such that a number ‖x‖, the norm of x, is associated with each x ∈ X, satisfying: ‖x‖≥ 0 and ‖x‖ = 0 if and only if x = 0; ‖λx‖ = |λ|‖x‖ for all λ ∈ R and ‖x + y‖≤‖x‖+‖y‖. For example, let X be a Tychonoff space, C∗(X) the ring of all bounded real- valued continuous functions on X. Then C∗(X) is a normed linear space with the norm ‖f‖ = sup{|f(x)| : x ∈ X} and pointwise addition and scalar multi- plication. This is called the supremum-norm on C∗(X). The associated metric is defined by d(f,g) = ‖f − g‖. A non-empty set C ⊆ Rn is called a convex set if whenever P and Q belong to C, the segment joining P and Q belongs to C. An- alytically the definition can be formulated in this way: if P is represented by the vector x, and Q by the vector y, then C is a convex set if with P and Q it contains also every point with a vector of form λx + (1 −λ)y, where 0 ≤ λ ≤ 1. A point P is an interior point of a set S contained in Rn, if there exists an n-dimensional ball, with center at P , all of whose points lie in S. An open set is a set containing only interior points. A subset C ⊆ Rn is centrally symmetric (or 0-symmetric) if for every point Q ∈ Rn contained in C, −Q ∈ C, where −Q is the reflection of Q through the origin, that is C = −C. Definition 1.1. ([Siegel, 1989, page 5]) A convex body is a bounded, centrally symmetric convex open set in Rn. Example 1.1. The interior of an n-dimensional ball, defined by x21 + x 2 2 + · · · + x2n < a 2 provides an example of a convex body. One of the many important ideas introduced by Minkowski into the study of convex bodies was that of gauge function. Roughly, the gauge function is the equation of a convex body. Minkowski showed that the gauge function could be defined in a purely geometric way and that it must have certain properties analo- gous to those possessed by the distance of a point from the origin. He also showed that conversely given any function possessing these properties, there exists a con- vex body with the given function as its gauge function. Definition 1.2. ([Siegel, 1989, page 6]) Given a convex body B ⊆ Rn containing the origin O, we define a function f : Rn → [0,∞) as follows. f(x) =   1 if x ∈ ∂B, 0 if x = 0, λ if 0 6= x = λy, where λ is the unique positive real number such that the ray through O and the point (whose vector is) x intersects the surface ∂B ( the boundary of B) in a point y. The function f so defined is the gauge function of the convex body B. 56 New structure of norms on Rn and their relations with the curvature of the plane curves Example 1.2. Let f : R → [0,∞) defined by f(x) = max{|x1|, |x2|, . . . , |xn|}, where x = (x1,x2, . . . ,xn) ∈ Rn. Then intB, the interior of the cubic B = {(x1,x2, . . . ,xn) : |xi| ≤ 1} is a convex body and f is a gauge function of it. It is shown in [Siegel, 1989, Theorems 4-7] that a function f : R → [0,∞) is a gauge function if and only if the following conditions hold: f(x) ≥ 0 for x 6= 0, f(0) = 0; f(λx) = λf(x), for 0 ≤ λ ∈ R; and f(x + y) ≤ f(x) + f(y). Moreover, f is continuous and the convex body of f is B = {x : f(x) < 1}. A brief outline of this paper is as follows. In section 2, we introduce a function ϕn on Rn, by the formula ϕn(Xn) = √ x21f 2 1 + x 2 2f 2 2 + · · ·+ x2nf2n, when n fixed nonzero real-valued functions f1,f2, . . . ,fn on R are given. We show that the mappings ϕn have similar properties such as norm functions within difference the ranges of these functions lie in RR while the range of a norm func- tion is in the [0,∞). This definition allows us to define a norm and hence a gauge function on Rn. So it turns Rn into a metric space. In Section 3, we focus on n = 2, ϕ2 and the induced norm on R2. First, we show that if F : R2 → R is a nonzero function, then k, the curvature of the implicit G(x,y) = F2(x,y) = c 6= 0 at every regular point is calculated by this formula: k = |HG|−4F2|HF| 4F ( F2x + F 2 y ) 3 2 , where HF and HG are the Hessian matrices of F and G respectively. It is also shown if F(x,y) = ∫ b a √ x2f2(t) + y2g2(t)dt, then |HF| = 0 and the eigenvalues of HF and HG, where G = F2 are nonnegative. Particularly, when f(t) = cost and g(t) = sint, we prove that ∫ π 2 0 √ x2f2(t) + y2g2(t)dt is an elliptical integral of the second type. 2 A norm on Rn made by the real valued functions on R We begin with the following notation. Notation 2.1. Suppose that f1,f2, . . . ,fn are nonzero real-valued functions on R and define ϕn : Rn → RR with ϕn(Xn) = √ x21f 2 1 + x 2 2f 2 2 + · · ·+ x2nf2n, (∗) 57 Amir Veisi and Ali Delbaznasab where Xn = (x1,x2, . . . ,xn) and RR is the set (in fact, ring) of all real-valued functions on R. The following statement is a key lemma. However, its proof is straightforward and elementary, it will be used in the proof of the triangle inequality in the next results. Lemma 2.1. Let a,b,c and d are nonnegative real numbers. Then √ ac + √ bd ≤ √ (a + b)(c + d). Proposition 2.1. Let Xn,Yn ∈ Rn, n = 1,2 or 3. Then ϕn(Xn+Yn) ≤ ϕn(Xn)+ ϕn(Yn). Proof. The inequality clearly holds when n = 1. Next, we do the proof for n = 2. Take X2 = (x1,y1), Y2 = (x2,y2) ∈ R2 and suppose that f and g are nonzero elements of RR. Then ϕ2(X2 + Y2) = √ (x1 + x2)2f2 + (y1 + y2)2g2 ≤ √ x21f 2 + y21g 2 + √ x22f 2 + y22g 2 = ϕ2(X2) + ϕ2(Y2) if and only if x1x2f 2 + y1y2g 2 ≤ √[ x21f 2 + y21g 2 ][ x22f 2 + y22g 2 ] = ϕ2(X2)ϕ2(Y2). (?) Now, if we let B := x1x2f2 + y1y2g2 and suppose that B ≥ 0, then (?) holds if and only if f2g2(x1y2 −x2y1)2 ≥ 0, which is always true (note, (?) trivially holds if B ≤ 0). Hence, in this case, the proof is complete. Here, we prove the proposition for n = 3. Let X3 = (x1,y1,z1) = (X2,z1) and Y3 = (x2,y2,z2) = (Y2,z2), where X2 = (x1,y1), Y2 = (x2,y2) and let f,g,h be nonzero elements of RR. Then ϕ3(X3 + Y3) = √ (x1 + x2)2f2 + (y1 + y2)2g2 + (z1 + z2)2h2 ≤ √ x21f 2 + y21g 2 + z21h 2 + √ x22f 2 + y22g 2 + z22h 2 = ϕ3(X3) + ϕ3(Y3) 58 New structure of norms on Rn and their relations with the curvature of the plane curves if and only if x1x2f 2 + y1y2g 2 + z1z2h 2 ≤ √ [x21f 2 + y21g 2 + z21h 2][x22f 2 + y22g 2 + z22h 2] = √[ ϕ22(X2) + z 2 1h 2 ][ ϕ22(Y2) + z 2 2h 2 ] Now, if we let a = ϕ22(X2),b = z 2 1h 2,c = ϕ22(Y2) and d = z 2 2h 2, then by (∗) in Notation 2.1, we have x1x2f 2 + y1y2g 2 ≤ √ ac. Moreover, it is clear that z1z2h2 ≤ √ bd. Therefore, x1x2f 2 + y1y2g 2 + z1z2h 2 ≤ √ ac + √ bd. In view of Lemma 2.1, the proof is now complete. Next, we state the general case of Proposition 2.1. Theorem 2.1. Let Xn = (x1,x2, . . . ,xn),Yn = (y1,y2, . . . ,yn) ∈ Rn,λ ∈ R and ϕn be as defined in Notation 2.1. Then the following statements hold. (i) ϕn(Xn) = 0 if and only if Xn = 0, (ii) ϕn(λXn) = |λ|ϕn(Xn), (iii) ϕn(Xn + Yn) ≤ ϕn(Xn) + ϕn(Yn) (triangle inequality). Proof. (i) and (ii) are evident. (iii). The proof is done by induction on n, see Proposition 2.1. If we set Xn−1 = (x1,x2, . . . ,xn−1) and Yn−1 = (y1,y2, . . . ,yn−1) then Xn and Yn can be substituted by (Xn−1,xn) and (Yn−1,yn) respectively. Therefore, ϕn(Xn + Yn) ≤ ϕn(Xn) + ϕn(Yn) if and only if x1y1f 2 1 + · · ·+ xnynf 2 n ≤ ϕn(Xn)ϕn(Yn) = √[ ϕ2n−1(Xn−1) + x 2 nf 2 n ][ ϕ2n−1(Yn−1) + y 2 nf 2 n ] . Now, let a = ϕ2n−1(Xn−1),b = x 2 nf 2 n,c = ϕ 2 n−1(Yn−1) and d = y 2 nf 2 n plus the assumption of induction, we have x1y1f 2 1 + · · ·+ xn−1yn−1f 2 n−1 ≤ √ ac. Moreover, it is obvious that xnynf2n ≤ √ bd. Thus, x1y1f21 + · · · + xnynf2n ≤√ ac + √ bd. Lemma 2.1 now yields the result. 59 Amir Veisi and Ali Delbaznasab Corollary 2.1. If f1,f2, . . . ,fn are nonzero constant functions, then ϕn is a norm (and hence a gauge function) on Rn. By Theorem 2.1, we obtain the following result. Proposition 2.2. Let a,b be real numbers, f1,f2, . . . , and fn the restrictions of some non-zero elements of RR on [a,b] such that each of them is nonzero on this set, and let ϕn be as defined in the previous parts (Notation 2.1). Then the mapping ψn : Rn → [0,∞) defined by ψn(Xn) = ∫ b a ϕn(Xn)dt is a norm on Rn, and hence d(Xn,Yn) = ψ(Xn − Yn) turns Rn into a metric space. Corollary 2.2. The mapping ψn is a gauge function on Rn with the convex body Cn = {Xn ∈ Rn : ψn(Xn) < 1}. 3 F(x,y) = ∫ b a ϕ2(x,y)dt as a norm on R2 and the curvature in the plane Proposition 3.1. ([Goldman, 2005, Proposition 3.1]) For a curve defined by the implicit equation F(x,y) = 0, the curvature of F (denoted by κ) at a regular point (x0,y0) (i.e., the first partial derivatives Fx and Fy at this point are not both equal to 0) is given by the formula κ = |F2y Fxx −2FxFyFxy + F2xFyy|( F2x + F 2 y )3 2 , where Fx denotes the first partial derivative with respect to x, Fy, Fxx denotes the second partial derivative with respect to x, Fyy, and Fxy denotes the mixed second partial derivative (for readability of the above formulas, the argument (x0,y0) has been omitted). We recall that the Hessian matrix of z = F(x,y) and w = F(x,y,z) are defined to be Hz = [ Fxx Fxy Fyx Fyy ] and Hw =  Fxx Fxy FxzFyx Fyy Fyz Fzx Fzy Fzz   at any point at which all the second partial derivatives of F exist. 60 New structure of norms on Rn and their relations with the curvature of the plane curves Theorem 3.1. Let F : R2 → R be a nonzero function and (x0,y0) ∈ R2 a regular point. Suppose that the second partial derivatives of F at (x0,y0) exist and further Fxy = Fyx at this point. Let HF and HG be the Hessian matrices of F and F2 respectively (we assume that G = F2) and let k be the curvature of G(x,y) = F2(x,y) = c 6= 0 at (x0,y0). Then we have k = |HG|−4F2|HF| 4F ( F2x + F 2 y ) 3 2 . Proof. For simplicity, we do the proof without (x0,y0). The partial derivatives of G = F2 are as follows: Gx = 2FFx, Gxx = 2(Fx 2 + FFxx), Gy = 2FFy, Gyy = 2(F 2 y + FFyy), and G 2 xy = 4(FxFy + FFxy) 2. Therefore, |HG| = GxxGyy −G2xy = 4 ( Fx 2 + FFxx )( F2y + FFyy ) −4 ( FxFy + FFxy )2 = 4 [ F2xF 2 y + FF 2 xFyy + FF 2 y Fxx + F 2FxxFyy −F2xF 2 y −2FFxFyFxy −F2F2xy ] = 4 [ F2 ( FxxFyy −F2xy ) + F ( F2xFyy −2FxFyFxy + F 2yFxx )] = 4 [ F2|HF|+ F ( F2xFyy −2FxFyFxy + F 2yFxx )] . In view of Proposition 3.1, we have |HG| = 4 [ F2|HF|+ F ( F2xFyy −2FxFyFxy + F 2yFxx )] = 4 [ F2|HF|+ Fk ( F2x + F 2 y ) 3 2 ] Therefore, k = |HG|−4F2|HF| 4F ( F2x + F 2 y ) 3 2 , and we are done. The next result is a similar consequence for the implicit surface. Theorem 3.2. Let F : R3 → R be a nonzero function and (x0,y0,z0) ∈ R3 a regular point. Suppose that the second partial derivatives of F at (x0,y0,z0) exist 61 Amir Veisi and Ali Delbaznasab and further the mixed partial derivatives at this point are equivalent. If k is the curvature of G(x,y,z) = F2(x,y,z) = c 6= 0 at (x0,y0,z0), then we have k = |HG|−8F3|HF| 8F2 ( F2x + F 2 y + F 2 z ) 3 2 , where HF and HG are the Hessian matrices of F and F2 respectively (we assume that G = F2). Proof. As we did in the previous theorem, the proof is done without (x0,y0,z0). Let K =   Fxx Fxy Fxz Fx Fxy Fyy Fyz Fy Fxz Fyz Fzz Fz Fx Fy Fz 0  . It is known that the curvature k of the implicit surface F(x,y,z) = 0 is k = |K| at every regular point in which the second partial derivatives of F exist. We first calculate the partial derivatives of G and in continued we obtain determinant of HG. Gx = 2FFx, Gxx = 2(Fx 2 + FFxx), G 2 xy = 4(FxFy + FFxy) 2 Gy = 2FFy, Gyy = 2(F 2 y + FFyy), G 2 xz = 4(FxFz + FFxz) 2 Gz = 2FFz, Gzz = 2(F 2 z + FFzz), G 2 yz = 4(FyFz + FFyz) 2. Recall that the Hessian matrices of F and G are HF =  Fxx Fxy FxzFxy Fyy Fyz Fxz Fyz Fzz  , and HG =  Gxx Gxy GxzGxy Gyy Gyz Gxz Gyz Gzz  . Here, we compute the determinant of HG. 1/8|HG| = Fxx ( FyyFzz −F2yz ) −Fxy ( FxyFzz −FxzFyz ) + Fxz ( FxyFyz −FxzFyy ) = FxxFyyFzz −FxxF2yz −FyyF 2 xz −FzzF 2 xy + 2FxyFyzFxz = ( F2x + FFxx )( F2y + FFyy )( F2z + FFzz ) − ( F2x + FFxx )( FyFz + FFyz )2 − ( F2y + FFyy )( FxFz + FFxz )2 −(F2z + FFzz)(FxFy + FFxy)2 + ( FxFz + FFxz )( FyFz + FFyz )( FxFy + FFxy ) = F3 [ FxxFyyFzz −FxxF2yz −FyyF 2 xz −FxxF 2 xy + 2FxyFyzFxz ] + F2 [ FxxFyyF 2 z + FxxFzzF 2 y + FyyFzzF 2 x −2FxyFxzFyFz −2FxyFyzFxFz −2FxzFyzFxFy + F2xyF 2 z + F 2 xzF 2 y + F 2 yzF 2 x ] + F [ 0 ] . 62 New structure of norms on Rn and their relations with the curvature of the plane curves Therefore, we have 1/8|HG| = F3|HF| + F2k(F2x + F2y + F2z ) 3 2 . So the result is obtained, i.e., k = |HG|−8F3|HF| 8F2 ( F2x + F 2 y + F 2 z ) 3 2 . Theorem 3.3. Let f,g be nonzero real-valued functions on R, a,b ∈ R and F : R2 → R defined by F(x,y) = ∫ b a √ x2f2(t) + y2g2(t)d(t). Then (i) The curvature of F(x,y) = c, where c > 0 at any point of the curve is positive multiple of c2. (ii) tr(HF) = Fxx + Fyy ≥ 0. Proof. (i). First, we note that F ≥ 0. The surface F meets the plane z = 0 at the origin only. But the intersection of F with the plane z = c (where c > 0) is the curve F(x,y) = c. Here the partial derivatives of F are calculated (see [Rudin, 1976, Theorem 9.42]). Fx = ∫ b a xf2(t)√ x2f2(t) + y2g2(t) d(t), Fy = ∫ b a yg2(t)√ x2f2(t) + y2g2(t) d(t), Fxx = ∫ b a y2f2(t)g2(t)( x2f2(t) + y2g2(t) )3 2 d(t), Fyy = ∫ b a x2f2(t)g2(t)( x2f2(t) + y2g2(t) )3 2 d(t), and Fxy = − ∫ b a xyf2(t)g2(t)( x2f2(t) + y2g2(t) )3 2 d(t) = Fyx. Let us put ϕ := √ x2f2(t) + y2g2(t). For the simplicity, we set Fx = ∫ xf2 ϕ , Fy = ∫ yg2 ϕ , and so on . . . 63 Amir Veisi and Ali Delbaznasab By formula of the curvature k in Proposition 3.1, we obtain k = 1( F2x + F 2 y )3 2 [( y2 ∫ f2g2 ϕ3 )( y ∫ g2 ϕ )2 + 2 ∫ xyf2g2 ϕ3 ∫ xf2 ϕ ∫ yg2 ϕ + ( x2 ∫ f2g2 ϕ3 )( x ∫ f2 ϕ )2] = ∫ f2g2 ϕ3( F2x + F 2 y )3 2 [ y4 (∫ g2 ϕ )2 + 2x2y2 ∫ f2 ϕ ∫ g2 ϕ + x4 (∫ f2 ϕ )2] = ∫ f2g2 ϕ3( F2x + F 2 y )3 2 [∫ x2f2 ϕ + ∫ y2g2 ϕ ]2 = ∫ f2g2 ϕ3( F2x + F 2 y )3 2 [∫ x2f2 + y2g2 ϕ ]2 = ∫ f2g2 ϕ3( F2x + F 2 y )3 2 [∫ ϕ ]2 = ∫ f2g2 ϕ3( F2x + F 2 y )3 2 F2(x,y). Hence, we observe that the curvature of F(x,y) = c at (x0,y0) is a positive multiple of F2(x0,y0) = c2, and we are done. (ii). Since f2g2(x2 + y2) ϕ3 ≥ 0, it is clear that Fxx + Fyy ≥ 0. So the result holds. Lemma 3.1. Let F : R2 → R be a homogeneous function of degree one. Suppose that the second derivatives of F at (a,b) ∈ R2 exist. Moreover, Fxy = Fyx at this point. Then (i) |HF|(a,b) = 0. (ii) The eigenvalues of HF are 0 and tr(HF) at (a,b). Proof. (i). First, we note that F(λx,λy) = λF(x,y), for all (x,y) ∈ R2 and λ ∈ R. Also, we remind the reader of the following fact, which is known as Euler’s property, xFx + yFy = F(x,y). 64 New structure of norms on Rn and their relations with the curvature of the plane curves Therefore, xFxx + Fx + yFxy = Fx, and xFxy + Fy + yFyy = Fy. Consequently, xFxx = −yFxy and xFxy = −yFyy. Now, consider the Hes- sian matrix HF = [ Fxx Fxy Fxy Fyy ] of F . For the point (0,b), where b 6= 0, we have Fyy(0,b) = 0 = Fxy(0,b). This implies that |HF| = 0. Also, con- sidering the point (a,0), where a 6= 0 gives Fxy(a,0) = 0 = Fxx(a,0), this again yields |HF| = 0. Now, let (a,b) such that a 6= 0 and b 6= 0. Then Fxx(a,b) = −b a Fxy(a,b) and Fyy(a,b) = −a b Fxy(a,b). Hence, |HF| = 0. So we always have |HF| = 0. The proof of (i) is now complete. (ii). Recall that the characteristic equation of HF is λ2 − (tr(HF) = Fxx + Fyy)λ + (|HF| = FxxFyy −F2xy) = 0. So λ2−(Fxx+Fyy)λ = 0. Therefore, λ = 0 or λ = tr(HF), and we are done. Proposition 3.2. Let f,g be nonzero real-valued functions on R and F : R2 → R defined by F(x,y) = ∫ b a √ x2f2(t) + y2g2(t)dt and let G(x,y) = F2(x,y). Then the eigenvalues of HF and HG at any point except the origin are nonnegative. (In fact, the eigenvalues of HF are zero and tr(HF) at that point). Proof. We observe that F is a homogeneous function of degree one. So Lemma 3.1 and Theorem 3.3 (ii) yield the result. For the matrix HG, we look to the Theorem 3.1. Since, F2|HF| = 0, we have |HG| = 4Fk ( F2x + F 2 y )3 2 . We notice that F,k ≥ 0 gives |HG| ≥ 0. On the other hand, tr(HG) = Gxx + Gyy ≥ 0. Therefore, the roots of λ2 − tr(HG)λ + |HG| = 0, which are the eigenvalues of HG, are nonnegative. The proof is finished. In the following result, we present a norm on R2 which is an elliptic integral of the second kind. Corollary 3.1. Let f(t) = cost, g(t) = sint and let F : R2 → R is given by F(x,y) = ∫ π 2 0 √ x2 cos2 t + y2 sin2 tdt. Then the following statements hold. (i) The eigenvalues of HF and HG, where G = F2 at every point except the origin are nonnegative. 65 Amir Veisi and Ali Delbaznasab (ii) F(x,y) is an elliptic integral of the second kind. Proof. (i). It follows from Proposition 3.2. (ii). Notice that F(x,y) = ∫ π 2 0 √ x2(1− sin2 θ) + y2 sin2 θdθ = |x| ∫ π 2 0 √ 1−k2 sin2 θdθ, where k = √ x2−y2 |x| and |x| ≥ |y|. So this gives F(x,y) is an elliptic integral of the second kind and we are done. Corollary 3.2. There are ordered pairs (x,y) with rational coordinates (other than the origin) which satisfy the inequality ∫ π 2 0 √ x2 cos2 θ + y2 sin2 θdθ ≤ r, when 0 < r ∈ Q. Also, if r /∈ Q then (x,y) has irrational coordinates. Proof. It is sufficient to take the pairs (r,0),(0,r),(−r,0) and (0,−r). We end this article with the next results. Proposition 3.3. Let 0 ≤ x,y ∈ R. Then∫ π 2 0 √ x2 cos2 t + y2 sin2 tdt ≤ x + y. Proof. First, note that x2 cos2 t + y2 sin2 t = (xcost + y sint)2 −2xy sintcost, and take 0 ≤ φ ≤ π 2 such that tanφ = y x (if x > 0). Now, (xcost + y sint)2 = x2(cost + y x sint)2 = x2(cost + sinφ cosφ sint)2 = x2(costcosφ + sintsinφ)2 cos2 φ = x2 cos2(t−φ) cos2 φ = (x2 + y2) cos2(t−φ) (note, cos2 φ = x2 x2 + y2 ). Hence, x2 cos2 t + y2 sin2 t ≤ (x2 + y2) cos2(t−φ). Therefore,∫ π 2 0 √ x2 cos2 t + y2 sin2 tdt ≤ ∫ π 2 0 √ (x2 + y2) cos2(t−φ)dt = √ x2 + y2 ∫ π 2 0 |cos(t−φ)|dt = √ x2 + y2 ∫ π 2 −φ −φ cosTdt (T = t−φ) = x + y. 66 New structure of norms on Rn and their relations with the curvature of the plane curves Remark 3.1. We find 4 ∫ π 2 0 √ x2 cos2 t + y2 sin2 tdt ≤ 2(2x+2y). 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