By A. S. Kechris

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**Additional resources for Cabal Seminar, 79-81**

**Sample text**

4) Now we consider the following non-homogeneous Cauchy problem: (NP) de = Ax(t) + f (t, (-A)-"y(t)), 0 < t < T, x(0) = x0. But we know that f (t, (-A)-"y(t)) belongs to C([to, tl]; E) fl CQ'((to, tl]; E) where 0 < Q' < min(y,1 - a). 16, we obtain that the function u(t) = U(t - to)xo + f U(t - s) f (s, (-A)-"y(s)) ds tot belongs to C([to, ti]; E) fl C1((to, ti ); E), and is a unique solution of problem (NP). 60) which proves that the function u is a solution of problem (SLP). 60). (5) Finally, we prove the uniqueness of solutions of problem (SLP).

Proof. 3) 1U(X) u(y)I (lu(s) - u(y)I Ix ylIx-yl (lu(s) - u(y)I -_ lx -yl Hence, if 0 < a < b < c < 1, then we have lulb. )(b-a)/(c-a) (hula < (I ul c oo)(c-b)/(c-a) Thus we have only to consider the following two cases: a=0 and c=1. v/nlx - yl. But we remark that 8u 01xi (x)I - lim h-0 u(x + hei) - u(x) I h < sup XOY lu(x) - u(y)l lx - yl (c-b)/(c-a) II. SOBOLEV IMBEDDING THEOREMS 50 so that < sup Iu(x) - u(y)I lull,,,. = max sup I*
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3 lul-n = lull,. )(nA+l)/n(A-0) (lull/µ)(nA+l)/n(A-µ) _ `Y Hence it follows that lull/µ <_ 7 (lull/v)(nµ+l)/(nv+l) ((lull! a)(nµ+l)/n(µ-A) (lull/")(na+l)/n(a-µ) X = (lull/v) (nµ+l)/(nv+l) "ate =,, -+1 A-4 np+l -4 (lull/a) -+1 4-X so that lull/µ _< y (lull/A)(v-µ)/(v-A) (lull/v)(µ-\)/(v-a) .