By Yoshihiro Mizuta

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We now consider the differentiation of distributions. For a multi-index λ and T ∈ D (G), we define Dλ T by (Dλ T )(ϕ) = (−1)|λ| T (Dλ ϕ) for ϕ ∈ C0∞ (G), which gives a distribution on G. 2. If T is a distribution on G and G is a relatively compact open subset of G, then there exist a function f ∈ L∞ (G ) and a multi-index λ such that T = Dλ f on G . Proof. 1, there exist M > 0 and k > 0 such that |T (ϕ)| M ✺ sup |Dλ ϕ| for every ϕ ∈ C0∞ (G ). , 1). 7) that |T (ϕ)| M ∗ |D(k+1)λ ϕ(y)| dy. 11 Distributions 47 Applying the Hahn-Banach theorem, we extend T to a bounded linear form on L1 (G ).

For ε > 0, consider the compact sets Kj = {x : g(x) Since Kj fj (x) + ε}. Kj = ∅, Kj = ∅ for large j, so that Kj+1 and j 0 g − fj εh. 7 Riemann-Stieltjes integral 27 so that lim λ(fj ) = λ(g), j→∞ as required. 5). What remains is to show that they are Radon measures on G. Let ν denote µ+ or µ− , and γ denote λ+ or λ− , respectively. Let A Rn and ν(A) < ∞. Take a competing sequence {fj } in C0 (G) for ν(A), and consider Vj = {x : fj (x) > 1 − ε} for 0 < ε < 1. Setting V = Vj , we have j ν(V ) = lim ν(Vj ) j→∞ lim (1 − ε)−1 γ(fj ).

Then |µj,j (f ) − µ(fk )| εµj,j (Rn ) + |µj,j (fk ) − µ(fk )|, from which it follows that {µj,j (f )} is a Cauchy sequence. Thus µ(f ) = lim µj,j (f ) j→∞ gives the required measure. 3 (Frostman). Let h be a measure function. 1) µ(B(x, r)) h(r) for all ball B(x, r), then µ(E) Mh (E) for any set E. 1). Proof. Let {B(xj , rj )} be a covering of a set E. Then we have µ(E) µ(B(xj , rj )) j h(rj ), j which gives the first assertion. Conversely, let µj be a measure on F such that µj (Q) = h(2−j ) whenever Q ∈ Gj and Q ∩ F = ∅.