Abstract. In this paper, we give Heron's formula in inner product spaces and some properties of this formula. Also, we define a 2-inner product in terms of Heron's formula.The classical Heron's formula says that, for a triangle having sides of the length a, b, c and the area K, the relation
Abstract. In this paper, we give some new characterizations of strict convexity and strict 2-convexity in linear 2-normed spaces. I. IntroductionLet X be a linear space of dimension greater than 1 and ||-,-|| be a real-valued function on X X X satisfying the following conditions:(2N-1) ||a:, 2/|| = 0 if and only if x and y are linearly dependent, (2N-2) ||z,y|| = \\y,x\\, (2N-3) ||aa;,2/|| = |a|||a;,i/|| for any real a, (2N-4) \\x + y,z\\ < ||s,*|| + Il*, -|| is called a 2-norm on X and (X, ||-, -||) is called a linear 2-normed space ([12]). A 2-norm ||-, -|| is a non-negative function and ||a:, y|| = ||x,ax + y|| for any real number. For further details on linear 2-normed spaces, see [12], [18] and [19].Strict convexity and strict 2-convexity in normed linear spaces and linear 2-normed spaces are very important roles in studying some geometric structures of normed linear spaces and linear 2-normed spaces, respectively. For more characterizations of strictly convex and strictly 2-convex linear 2-normed spaces, we may refer to [1]-[11], [13]-[18] and [20].In this paper, we also give some new characterizations of strictly convex and strictly 2-convex linear 2-normed spaces. REMARK.(1) If a linear 2-normed space (A", ||-, -||) is strictly convex, then it is strictly 2-convex. But the converse is not necessarily true ([9]).(2) Every linear 2-normed space of dimension 2 and every 2-inner product space are both strictly convex and strictly 2-convex ([9]).(3) There exist linear 2-normed spaces which are not strictly convex and not strictly 2-convex ([8]). Strict convexityWe give a new characterization of strictly convex linear 2-nornied spaces. Strict 2-convexityIn this section, we give some characterizations of strictly 2-convex linear 2-normed spaces. which is a contradiction and so, z 6 V(x, y), i.e., z = ax f)y for some a,/3.From \\x,y\\ = \\y,z\\ = ||a;,z|| = ±\\x + z,y+z\\ = 1 and \\x,y+ax\\ = \\x,y\\, it follows that a = ¡3 = 1. Therefore, (X, ||-,-||) is strictly 2-convex. This completes the proof.
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