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By Georgi E. Shilov

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It suffices to show that if (f, 99) = 0 for every 99 E K, then f(x) is equal to zero almost everywhere. We proved this in Sec. 3 for n = 1, and we proceed in the general case by induction. As one test function 99(x), we take the product of 991(x 1) and qqn _ 1(x2 , ... , xn), where 99 1. and 99n-1 are test functions of one and n - 1 independent variables, respectively. , xn) 9g1(xl) dxl Rn-1 -00 X 99n - 1(x2 , ... , xn) dx2 ... , x over the is carried. , of Q fixed, we would like to apply the vanishes almost result of Sec.

X;,) dx2 ... (x). 37 ELEMENTARY THEORY OF GENERALIZED FUNCTIONS co) clearly has compact support along with q9(x) and is infinitely differentiable in for each fixed co. Thus, co) is a test function of the argument . With each generalized function f c- K1 we can now associate a functional f. )) = Qf, 0(e; co)). , xR) dx2... &Cn (°''x) _ 0 x'1 =0 and (6(4) ,92) = ((q)() (. 5 ( l) q ax 4 2 R xi =0 The functional f. will also be denoted by f ((cw, x)). 3. Instead of considering test and generalized functions throughout RR, we may confine ourselves to some region G c RR.

If' axl (f 4) _ (f aif) Jo(xi)dx1, where al(x1) is a fixed arbitrary test function such that Jo1(x1) dx1 = 1. -00 Proof. Given al(xl), we can write down the representation a(xe) = m&j) f a(xe) dxl + ao(xj), -00 00 in which f ao(xl) dxl = 0. Therefore, ao(xi) = YI(xl)31 where y(x,) is -00 again a test function (Sec. 1). Hence, (fc9)= (f[ci 00 f a(xe) dxl + Y' -00 00 00 = Cf ails) f a(xi) dxl + (f, Y'#) = f, a' f a(xe) dxl -00 OD 1 inasmuch as (1' axl (y P) I = - (f, \ I 1 I YflJ = \ 0. The lemma is proved.

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