DEMONSTRATIO MATHEMATICA Vol. XVI No 3 1983 Maciej J. Mqczynski A FUNCTIONAL CHARACTERIZATION OF INNER PRODUCT VECTOR SPACES Let V be a normed vector space with the norm ||*||, and let f(x) = ||x|| for all x e V. Jordan and von Neumann proved in [1] that f(x) = (x,x) for some inner product (.,.) on V if and only if the parallelogram law holds; (1) f(x + y) + tlx - y) = 2f(x) + 2f(y) for all x,y eV. In this note we would like to show that this characterization of inner product spaces holds without the assumption that f(x) arises from a norm, ^n the real case we shall show that it suffices to assume that f(x) is non-negative and real-he- micontinuous at zero. We shall extend this result to complex vector spaces. Let us mention that Jordan and von Neumann's theorem was generalized to abelian groups by S.Gudder in [2], where also a characterization of quadratic forms on real vec- tor spaces has been given. The motivation for the present work is provided by the so-called axiomatic quantum mechanics, where a system of axioms characterizing observatles is aought. A cha- racterization of inner product by means of the parallelogram law in the most general setting, under possibly weakest assump- tions, can be used to solve this problem. On Hilbert space, a characterization of observables by the properties of expect- ed value of the observable has been proposed in [4]. The re- sults of the present paper may be used to give a more general solution. - 79.7 - 2 M.J. M^czyriski Before we state our theorems, we recall some standard ter- minology to be used in the sequel. Let K be either the field of real numbers R or the field of complex numbers C, and let V be a vector space over K. An inner product on V is a func- tion which assigns to each ordered pair of vectors x,y in V a scalar (x,y) in K in such a way that (a) (x + y,z) = (x,z) + (y,z); (b) (cx,y) = c(x,y); (c) (y,x) = (x,y), the bar denoting complex conjugation; (d) (x,x)> 0 if x 4 0; for all x,y,z e V, c e K. Instead of saying that V is a vector space over R we also say that V is a real vector space, and similarly for complex vector space. Clearly, 5f K = R, then condition (c) means that (x,y) is symmetric, i.e. (x,y) = (y,x) for all x,y£ V. If V is a vector space over K and there is no topology on V making V into a topological vector space, then we cannot speak about continuous K-valued functions on V. However, we can define the notion of hemicontinuity for such functions, i.e. continuity with respect to multiplication by scalars only. We shall need only a special case of this notion, namely real-hemicontinuity at zero, as defined in the following de- finition. Definition 1. Let V be a real or complex vec- tor space. We say that a real or complex valued function f on V is real-hemicontinuous at zero, if for every xeV and every sequence of real numbers tending to 0, lim = 0, we have lim fUnx) = f(0). Note that on a topological vector space, every continuous function is real-hemicontinuous at zero. Now we can state our main theorem. Theorem 1. Let V be a vector space over R, and f : V—a real-valued function defined on V, real-hemicon- tinuous at zero, with the following properties: (i) f(x + y) + f(x - y) = 2f(x) + 2f(y) for all x,y e V; (ii) f (x) > 0 if xjiO. - 798 - Inner product vector spaces 3 Then (x,y) = ^ (f(x + y) - f(x) - f(y)) is an inner product on V. We then have f(x) = (x,x) for all xeV, and consequently f(x) is continuous with respect to the topology defined on V by the norm Proof. First observe that putting in (i) x = y = 0 we obtain f(0) = 0. Let (2) h(x,y) = \ (f(x + y) - f(x) - f(y)) for all £ V. It is obvious that h is symmetric. We shall show that h is additive with respect to x, and consequently with respect to y as well (this part of the proof follows closely the method used in Jordan and von Neumann's proof, but we include it for completeness). From (i) we obtain f(x + y) - f(x) - f(y) = -f(x - y) + f(x) + f(y). Taking into account that putting in (i) x = -y we obtain f.(-y) = f(y) for all y eV, we infer from above that h(x,y) = = -h(x,-y) = -h(-x,y). Prom (i) we also obtain f(x + y) - f(x - y) = 2(-f(x - y) + f(x) + f(y)) = = -2(f (x + (-y)) - f(x) - f (-y)) = -4h(x,-y)- = 4h(x,y), so we have (3) f(x + y) - f(x - y) = 4h(x,y) for all x,y eV. Substituting x + z and x - z for x in (i) we obtain f((x + y) + z) + f((x - y) + z) = 2f(x + z) + 2f(y), f((x + y) - z) + f((x - y) - z) = 2f(x - z) + 2f(y). Subtracting side by side and using (3) we obtain (4) h(x + y,z) + h(x - y,z) = 2h(x,z). - 799 - 4 M.J. Myszyiiski Letting y = i in (4) gives h(2x,z) = 2h(x,z). Replacing x + y by x' and x - y by y' in (4) gives h(x',z) + hiy'.z) = h(2x,z) = h(x' + y',z). Hence h is additive with respect to x, and also to y. So we have (5) h(x + y,z) = h(x,z) + h(y,z) for all x,y,z e V. We have shown that h(2x,z) = 2h(x,z). By induction from (5) we infer that h(mx,z) = mh(xfz) for every non-negative integer m. We also have h(-mx,z) = -mh(x,z). Putting mx = y we obtain h(g y,z) = g h(y,z) for m 4 0. Hence h(Ax,y) = = Ah(x,y) for any rational A. Putting x = y in (i) we obtain f(2x) = 4f(x). By a similar o argument as above we infer that f(Ax) = A f(x) for any ratio- nal A. Now for fixed x,y eV, x 4 0, we define the function p : R -*• R by p(A) = f(Ax + y) = 2h(Ax,y) + f(Ax) + f{y), A e R. We have p(A) >0 for all AeR. For all rational A we have OSp(A) = A2f(x) + 2Ah(z,y) + f(y). Since for fixed x,y p(A) is a continuous function of A (a qua- dratic polynomial in A ), we have 0.4 !l2f(x) + 2Ah(x,y) + f(y) for all yeal A. Hence the discriminate must be non-positive, so we have (6) |h(x,y)|2^ f(x)f(y) for all x,yeV. (For x = 0, (6) also holds). 800 Inner product vector spaces 5 Incidentally we have shown that condition (i) implies an abstract form of the Cauchy-Schwarz inequality. Thus for any rational % and any real a we have |h(<xx,y) -ah(x,y)U |h((ot-A)x,y)| + |a-A||h(x,y)| < $(f((a-a)x))1/2(f(y))1/2 + |a-il||h(x,y)| . Since f(x) is real-hemicontinuous at zero, the above expression approaches 0 as % tends to a through rational X . Henoe h(otx,y) ~ah(x,y) = 0 and h(x,y) is real homogeneous. Defining (x,y) = h(x,y) we see that all the axioms of an inner product are satisfied and the theorem has been proved. Let us say that the norm ||»|| on a real vector space V is real unitarp y if there is a real inner product (.,.) on V such that || xf = (x,x) for all x e V. We have the following corolla- ry. Corollary 1. A real-valued mapping N(x) on a real vector space V, taking positive values for non-zero elements of V and real-hemicontinuous at zero, is a real uni- tary norm on V if and only if N2(X + y) + N2(X - y) = 2N2(x) + 2W2(y) for all x,y eV. .Che above results can be generalized to the complex case. Namely, we have the following theorem. Theorem 2. Let V be a vector space over C and f : V —• R a real-valued function on V, real-hemicontinuous at zero, with the following properties: (i) f(x + y) + f(x - y) = 2f(x) + 2f(y) for all x,y e V; (ii) f(x) >0 if x J 0; (iii) f(cox) = f(x) if |u| = 1. - 801 - 6 M.J. M^czydski Theti (x,y) = J (f(x + y) - fix - y)) + j i(f(x + iy) - - f(x - iy)) is an inner product on V. We have f(x) = (x,x) for all x e V, and consequently ||x|| = \f(x) is a norm on V. Proof. If we admit multiplication of vectors in V by real numbers only, V becomes a vector space over R and we can apply Theorem 1. Hence (2) defines a real inner product on V and we have f(x) = h(x,x) for all x e V.