By Volkodavov V.F., Radionova I.N., Bushkov S.V.

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**Extra info for A 3D analog of problem M for a third-order hyperbolic equation**

**Example text**

Of course the right side is at most C-yojQ(S)ji 1lGl12, and so JF; :S C-y51Q(S)I. 35) does follow in this case from Carleson's inequality. 35) for a(2). 52k2-t(d+l)2-kd j (w,l)Ee2 d~J(r) dw.

We need to get estimates for the interior integral AZ = AZ(j,p,P,i,k,w) = faip, iq)(L(q- w)dq. 20a). e and JAZj :5 Ch'2"2-t(d+tl. 20). 34) when k :S £, Jp- wj :S 2t+ 1 , and zero otherwise. Set e1 = {(w,f) E e: k;::: for jp- wj;::: 2l+ 1} and e2 = {(w,f) E e: k < f and Jp- wJ < 2l+ 1 }, and split a into a(1) + a(2) accordingly. Thus a(1,2) :5 C J L AZ(I,2 )(p,f,k,w)2-kd J d~-t(r) dw. 35) Ja(1,2Wdp :5 Cf'0 62 JQ(S)J. (p,l)Ea5 We start with a(1). By definitions a(1) :5 CHt(P) J L (w,I:)Eel J 2"( 2 -d)(2" + 21 + Jp- wl)-d- 2 d~-t(r) dw.

It turns out that (LS) is equivalent to (C1)-(C7). We hope that this is as big a surprise for the reader as it was for the authors. The proof will be given in a separate publication. When d = 1 there is a direct construction that shows that (LS) implies (C7). This combines with the arguments given here to provide a much simpler proof of the fact that (C1) or (C2) imply (C7) when d = 1. The argument for showing that (LS) implies (C1)-(C7) when d > 1 is much less direct and it relies in particular on the fact that (C1) implies the other conditions.

### A 3D analog of problem M for a third-order hyperbolic equation by Volkodavov V.F., Radionova I.N., Bushkov S.V.

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