By Kenneth Hoffman

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**Example text**

F (n−1) , (Daν f ))dt = a x |(Daν f )(t)| |F (t, f, f , . . , f (n−1) , (Daν f ))|dt ≤ a n−1 x ≤ |(Daν f )(t)| a qi (t)|f (i) (t)| dt i=0 x n−1 qi (t)|f (i) (t)| |(Daν f )(t)|dt = a i=0 n−1 ≤ x qi i=0 |f (i) (t)| |(Daν f )(t)|dt. 3 Applications 21 That is, ((Daν f )(t))2 2 n−1 x x ≤ a qi ∞ · |f (i) (t)| |(Daν f )(t)|dt. a i=0 That is, n−1 x ((Daν f )(x))2 ≤ A2 + 2 qi |f (i) (t)| |(Daν f )(t)|dt. 31) a for all i = 0, 1, . . , n − 1. 34) all a ≤ x ≤ b. 35) a all a ≤ x ≤ b. Here ρ, Q(x), θ(x) ≥ 0 and Q(a) = 0.

12) 0 Φ(t)q dt. e. e. in (0, x). The function srp+1 (z(s)z (s))1/q is integrable over (0, x) as rp + 1 ≥ 0 and z(s)z (s) is measurable and essentially bounded on (0, x). Applying H¨older’s inequality, we obtain x 0 1/p x srp+1 (z(s)z (s))1/q ds ≤ 0 x(rp+2)/p (z(s))2/q ≤ . 2 Main Results 45 The result then follows when we observe that |Dγ f (s) Dν f (s)| is integrable as Dγ f ∈ AC[0, x] for ν − γ ≥ 1, and Dγ f ∈ C ([0, x]) for ν − γ ∈ (0, 1) , and Dν f ∈ L∞ (0, x). The following result deals with the extreme case of the preceding theorem when p = 1 and q = ∞.

We need the following lemma from [17]. 3. Let f ∈ C([a, b]), μ, ν > 0. Then x0 Jμx0 (Jνx0 f ) = Jμ+ν (f ). 10) 26 3. 4. Let ν ≥ 1, γ ≥ 0, be such that ν − γ ≥ 1, so that γ < ν. Call n := [ν], α := ν − n ; m := [γ], ρ := γ − m. Note that ν − m ≥ 1 and n − m ≥ 1. Let f ∈ Cxν0 ([a, b]) be such that f (i) (x0 ) = 0, i = 0, 1, . . , n − 1. 7) f (x) = (Jνx0 Dxν0 f )(x), for all x ∈ [a, b] : x ≥ x0 . Therefore by Leibnitz’s formula and Γ(p + 1) = pΓ(p), p > 0, we ﬁnd that for all x ≥ x0 . 11) It follows that f ∈ Cxγ0 ([a, b]) and thus x0 (Dxγ0 f )(x) := (DJ1−ρ f (m) )(x) exists for all x ≥ x0 .

### Banach spaces of analytic functions by Kenneth Hoffman

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