Exercise 14, § 3

Let $\mathbf{f}$ be a function which is $n - 1$ times differentiable on an interval $I$ containing 0, and let $\mathbf{f}_n$ be the vector function defined for $x \neq 0$ on $I$ by the relation

a) Show that if $\mathbf{f}$ has an $(n+p)^{th}$ derivative at the point 0 then $\mathbf{f}_n$ has a $p^{th}$ derivative at the point 0 and an $(n+p-1)^{th}$ derivative at all points of a neighbourhood of 0 distinct from 0; moreover, one has $\mathbf{f}_n^{(k)}(0) = \frac{k!}{(n+k)!} \mathbf{f}^{(n+k)}(0)$ for $0 \leq k \leq p$, and $\mathbf{f}_n^{(p+k)}(x)x^k$ tends to 0 with $x$, for $1 \leq k \leq n-1$ (express the derivatives of $\mathbf{f}_n$ with the help of the Taylor expansions of the successive derivatives of $\mathbf{f}$, and use prop. 6 of I, p. 18).

b) Conversely, let $\mathbf{f}_n$ be a vector function admitting an $(n+p-1)^{th}$ derivative on a neighbourhood of 0 in $I$, and such that $\mathbf{f}_n^{(p+k)}(x)x^k$ has a limit for $0 \leq k \leq n-1$. Show that the function $\mathbf{f}_n(x)x^n$ has an $(n+p-1)^{th}$ derivative on $I$; if, further, $\mathbf{f}_n$ admits a $p^{th}$ derivative at the point 0, then $\mathbf{f}_n(x)x^n$ admits an $(n+p)^{th}$ derivative at the point 0.

c) Suppose that $I$ is symmetric with respect to 0 and that $\mathbf{f}$ is even ($\mathbf{f}(-x) = \mathbf{f}(x)$ on $I$). Show, with the help of a), that if $\mathbf{f}$ is $2n$ times differentiable on $I$, then there exists a function $g$ defined and $n$ times differentiable on $I$, such that $\mathbf{f}(x) = g(x^2)$ on $I$.