On the theorem of frullani

Theorem 1. Let f: [0, +oo) —► R be a function such that (f(ax)- f(bx))/x is integrable on [0, +oo) for every pair ofpositive real numbers a, b. Then there exists a constant A € R such that

(6)
a


for every a, b > 0.
Proof. First observe that the measurability of f is not a hypothesis of the theorem. Let us take x = e‘, a = log a and /? — log 6. We obtain that (6) is equivalent to
о X b
Г”*"⁰⁰ f , . . п л
/ oo {^{/(e }} ” ^{f(e }}} ^{dt = A(a} " •
So it suffices to prove the theorem that follows.
Theorem 2. Let g: R -» R be a function such that g(x + a) - g(x + /?) is Lebesgue integrable for every pair of real numbers a, p. Then there exists a constant A such that
r-boo

(7)
{g(x + a) — g(x + P)} dx = A(a — P).
J — oo
Proof. Let h_it be defined by hfx) = g(x+a)-g(x). We know that h_a G i’(R) for every real number a.
Moreover, it is obvious that A_ft(x) + h^(x + a) = h_n+p(x).
Hence, A ₍(x) + h_fi(x + a) = h^(x) + h_n(x + P).
All these functions belong to L'(R), so we can apply Fourier’s transform and we ootain
A„(x) + е²”'"^лЛ_д(х) = h_fi(x) + e^2n,fixh₍fx).
Thus we have
hAx) — —h (x).
Let us observe that /г^(х) is continuous and <7()3) = f^{g(x+P)-g(x)} dx = hp(O), and so
G(B) = lim -—(x) = ~G(a).
oi-_e^2n,,,x " a ^v
Now, if we choose A = (7(1), we have G(p) — Aft, which leads to



(a — P)A — G(a —

a-p)-g(x)}dx


J —oo
/4-00
{g(x + a.) — g(x + p)} dx,
-oo
getting Theorem 2.
The next theorem, which contains the example we talked about in the intro­duction, must be compared with C. Burstin’s theorem: A Lebesgue measurable function f: R —> R having arbitrary small periods is a constant a.e. (cf. R. Cig- noli and J. Hounie [6] or J. M. Henle [12]).