Bourbaki

Exercise 1, § 2

Functions of a Real Variable, VII, 327

Bourbaki marks this one as harder than the rest.

a) Let gg be a continuous real function for x≥0x \geq 0. Show that if gg satisfies the two identities

∑k=0p−1g(x+kp)=g(x) \sum_{k=0}^{p-1} g\left(\frac{x+k}{p}\right) = g(x)
∑h=0q−1g(x+hq)=g(x) \sum_{h=0}^{q-1} g\left(\frac{x+h}{q}\right) = g(x)
it also satisfies the identity
∑j=0pq−1g(x+jpq)=g(x). \sum_{j=0}^{pq-1} g\left(\frac{x+j}{pq}\right) = g(x).

b) Deduce from this that if a function gg has a continuous derivative for x≥0x \geq 0 and satisfies the identity

∑k=0p−1g(x+kp)=g(x) \sum_{k=0}^{p-1} g\left(\frac{x+k}{p}\right) = g(x)
then it is of the form a(x−12)a(x-\frac{1}{2}), where aa is a constant (remark that gg satisfies the analogous identity where pp is replaced by pnp^n; let nn tend to +∞+\infty and deduce that g′(x)=∫01g′(t) dtg'(x) = \int_0^1 g'(t)\,dt).

c) Conclude from b) that the Γ\Gamma function is the only function having a continuous derivative for x>0x > 0, which satisfies equation (1) of VII, p. 305, and the multiplication formula (12) of VII, p. 318, for a value of pp.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.