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Sn, r Now, if r > 0 is given then by (b) we have that there is a 6 > 0 such that s to~ ~ ~ scx) and y ~ X ~ t h IlYll-< I and t ~ x*, t(~) ~ llfll, it Ilx - YI[ > r then f(y) < (I - 6 )l[fll __ __ r Thus, by definition of S " we nsr get that if x s Blk Sn, s, then fn(X) _< I - 6 r 9 Thus l-I/n < --<~ tn(X)d~(x) + (I - 6r Sn, r S ~(Sn, r + (I - 6r ~(BI% Sn, e) r Sn,r Consequently~ we get (*) Now choose n r ~(BI~ Sn;e) < 1/n6r such that if n, m ~ n we have that Sn, r N Sm, r # @. G yields 2/6 e < n, m.

4) easy examples show that x • y and x • z need not imply that + z). =I 25 Theorem 3: Let f 6 X . Then x z y for each y 6 kernel (f) if and only if Ifl (f). (f) we have ~pllx = PllXll Suppose further + yll. As we may assume that x ~ 0 and as under this assumption x + y for y E kernel (f) is somewhat typical of vectors in X(X -- [k(x + y): y 6 scalars, y 6 kernel (f)~,we have I]f[I -< P" But tfI -- I f < ~ ) i " ifa = Ilfll Ilxil, then for any y ~ k e r n e l ~ut then I f < ~ ) l I[xll--< il x + YI[ for each y 6 kernel (f:) we have = if< x + Y) I -< Hftl IIx + ylJ.

We claim that IiulI-< Iiu + ~viI for all ~. Let u = x + y and We prove the claim for ~ > 0 and note that the proof holds with -v replacing v so the 28 desired inequality holds for all ~. Observe that Ilu + vii = IIx + y + = - yll = 211~II = 2, Ilu - vii = I1~ + y - = + yll = 211yll = z, and llull -" II~ + yll = 2 Let 0 < I~--< 1 . Then flu + ~vll = llx + y + ~ - ~yll = II <1 + ~ ) ~ + (1 - ~)yll ffi 2 ffi by the fact that Ilull 0 < 1 - I~ < I + I~ -< 1 a n d for 0 _<. -< 1 we get flu + ~vi] = llull I f .