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Algebra I. Lecture Notes by Thomas Keilen

By Thomas Keilen

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Let b > 0 be given, v € No and k G Z". Then Qv/c denotes a closed cube in Rn with sides parallel to the axes, centred at xvJc e R", with |x v *-2- v fc| 0; then rQ is the cube in Rn concentric with Q and with side length r times that of Q. We always tacitly assume in the sequel that d > 0 is chosen in dependence on b so that for all choices v e No and all possible choices of xvJc according to (1), U dQvk = Rn (2) kezn Let K e No.

131, and the Russian translation of [TrijS] (Moskva, 1986), p. 191. As for the "only if"-part we refer again to [SicT]. For our purpose the following observation will be of great service in the sequel. 1, p. 129. The crucial point about (7) is that u and v are unrelated, in contrast to (5). 4 Holder inequalities 45 Hf° C Hi\, S0 G R, 0 < po < Pi < 00, 50 - — = 5i - —. po p Po Pi (8) have Remark In a (-,s)-diagram, see Fig. 3, the assertions (i) correspond to horizontal lines, whereas the embeddings in part (ii) correspond to lines of slope n.

The proof of (8) is complete. Step 3 We prove (hi) and again let P= min(l,p). 4, pp. 31-2, with Lp(/oo). 1/1. 4 Holder inequalities 51 estimated from above by p (16) Now choose e > 0 so small that s > ap + s and apply Holder's inequality to the right-hand side of (16). This gives the right-hand side of (9). The proof is complete. Remark By Step 1 it follows immediately that the scalar case of (7), : c ||/ | hP21| max ||g*+;- | LPl ||, (17) also holds for p = oo. 3/17)); then one can replace Lp(/oo), LPl(/oo), LPl (/«,) in (9) and in Step 3 by Lp(lq), LPl(lq), LP2 (lq), respectively.

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