Bourbaki

Exercise 3, § 1

Functions of a Real Variable, VII, 326

a) Let gg be a function defined and admitting a continuous kthk^{th} derivative on ]0, +∞[+\infty[, such that g(k)g^{(k)} is decreasing on this interval and that lim⁡x→+∞g(k)(x)=0\lim_{x \to +\infty} g^{(k)}(x) = 0. Show that there exists one and only one solution uu of the equation u(x+1)−u(x)=g(x)u(x + 1) - u(x) = g(x) which admits an increasing kthk^{th} derivative on ]0, +∞[+\infty[ and takes a given value at a given point (use exerc. 1 b)).

b) Let α\alpha be an arbitrary real number. Let SαS_\alpha be the function defined on ]0, +∞[+\infty[ satisfying the relation

Sα(x+1)−Sα(x)=xα S_\alpha(x + 1) - S_\alpha(x) = x^\alpha
such that Sα(1)=0S_\alpha(1) = 0, and, further, that the derivative of SαS_\alpha of order equal to the integer part of (α+1)+(\alpha + 1)^+ is increasing (it is a unique function by aa) ). Show that
Sα′(x)−Sα′(1)=α Sα−1(x), S'_\alpha(x) - S'_\alpha(1) = \alpha\, S_{\alpha - 1}(x),
and
Sα(xp)+Sα(x+1p)+⋯+Sα(x+p−1p)=1pα Sα(x)+Cα S_\alpha \left( \frac{x}{p} \right) + S_\alpha \left( \frac{x + 1}{p} \right) + \cdots + S_\alpha \left( \frac{x + p - 1}{p} \right) = \frac{1}{p^\alpha}\, S_\alpha(x) + C_\alpha
for every integer p≥1p \geq 1, where CαC_\alpha is a constant, which, when α≠−1\alpha \neq -1, is equal to
(1pα−p)Sα+1′(1)α+1 \left( \frac{1}{p^\alpha} - p \right) \frac{S'_{\alpha + 1}(1)}{\alpha + 1}
(cf. VII, p.318, prop. 2 and VI, p. 291, exerc. 3).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.