By Hille E.

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**Example text**

3. 48), we obtain ∞ m−1 x − a)m ψ(x) δ(m−1) (x − a) = ψk cj (x − a)k+m δ (j) (x − a) = 0. 8. 48). For a particular solution we may choose f1 (x) = H(z) . 48) are only determined up to a generalized delta-hyperfunction of order m − 1. This will have implications if we are looking for interpretations of certain familiar ordinary functions in terms of hyperfunctions. Take for example the ordinary function x → 1/xm . Suppose now that we have deﬁned a suitable interpretation f1 (x) of x → 1/xm . Reasonably it should satisfy the equation xm f (x) = 1.

7. , f (x) = fe (x) + fo (x), where fe (x) is even and fo (x) is odd. Proof. We simply set fo (x) := f (x) + f (−x) f (x) − f (−x) , fe (x) := . 8. Prove that f (−(−x)) = f (x). 9. Prove: (i) If f (x) is an even (odd) hyperfunction, then f (x) is an odd (even) hyperfunction. (ii) Let φ(x) be a real analytic function. If φ(x) and f (x) have the same (opposite) parity, then the hyperfunction φ(x)f (x) is even (odd). 10. Show that if f (x) = [F+ (z), F− (z)] is an even hyperfunction, we can write F+ (z) − F− (−z) F− (z) − F+ (−z) , , f (x) = 2 2 and if f (x) is an odd hyperfunction, we can write f (x) = F+ (z) + F+ (−z) F− (z) + F+ (−z) , .

Ii) Let φ(x) be a real analytic function. If φ(x) and f (x) have the same (opposite) parity, then the hyperfunction φ(x)f (x) is even (odd). 10. Show that if f (x) = [F+ (z), F− (z)] is an even hyperfunction, we can write F+ (z) − F− (−z) F− (z) − F+ (−z) , , f (x) = 2 2 and if f (x) is an odd hyperfunction, we can write f (x) = F+ (z) + F+ (−z) F− (z) + F+ (−z) , . 2 2 Our hyperfunctions are supposed to be complex-valued at the points where they have values. Let us now tackle the problem how the complex-conjugate hyperfunction f (x) has to be deﬁned if f (x) = [F+ (z), F− (z)] is given.

### An Integral Equality and its Applications by Hille E.

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