By Katz N.M.

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**Additional info for Algebraic solutions of ODE using p-adic numbers**

**Example text**

We 1 (h ∗ b), where will be sketchy. Setting b(x) = U p+1 (x) we can write Γ(ξ) = − p+1 1 (h · b). ∗ denotes the convolution. Taking the Fourier transform we get Γ = − p+1 Using the Morera Theorem, see [4], it is easy to check that b is analytic in the strip {ζ ∈ C n : |Imζ| < α} for some α > 0 and hence it has at most a countable number of zeroes there. Moreover, since h ∈ Lr (Rn ) for some r ∈ [1, 2] we deduce, by the Hausdorﬀ-Young inequality , that h ∈ Lr (Rn ), where r denotes the conjugate exponent of r (if r = 1, h is continuous), and h ≡ 0, since h ≡ 0.

Let ξε be such that zε = zξε , and set qiε = ∂z/∂ξi |ξε . Without loss of generality we can assume that zε → z ∗ ∈ Zc as ε → 0. 13) has a solution wε (zε ), deﬁned for |ε| < ε0 . In particular, from (i) of that lemma and by continuity, one has that lim (Di wε (zε ) | qjε ) = 0, i, j = 1, . . , d. |ε|→0 Let us consider the matrix B ε = (bεij )ij , where bεij = Di wε (zε ) | qjε . From the above arguments we can choose 0 < ε1 < ε0 , such that | det(B ε )| < 1, ∀ |ε| < ε1 . 16) Fix ε > 0 such that |ε| < min{ε0 , ε1 }.

11-(ii). 26. 1) Du G(ε, z) = o(εβ ), as ε → 0. Then wε (z) = o(εβ ) as ε → 0, uniformly in any compact subset Zc of Z. Proof. 11. Let us set wε = ε−β wε (z) (again, for brevity, in the sequel the dependence on z will be understood). We ﬁrst prove that wε ≤ const. for |ε| small. Precisely let us start by showing Du G(ε, z) = O(εβ ), as ε → 0, =⇒ wε (z) = O(εβ ), as ε → 0. 31) By contradiction, assume lim wε = +∞. 32) and dividing by εβ wε , we ﬁnd I0 (z) wε wε =− Du G(ε, z) 2 − Duu G(ε, z) εβ wε wε wε + o( wε ) + εβ wε Ai,ε qi .

### Algebraic solutions of ODE using p-adic numbers by Katz N.M.

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