If $f$ is smooth and even then there exists a smooth function $g$ such that $f(x)=g(x^2)$

Question

Assume that a function $f: R \rightarrow R$ is smooth and even. Does there exist a smooth function $g:R \rightarrow R$ such that $f(x)=g(x^2)$ ?

Answer 1

Incidentally, this follows easily from a simple integral formula for the derivatives of $g$ (see this question that I recently posted). Indeed, the Taylor formal power series in $0$ of an even $C^{\infty}$ function $f$ is of course in $\mathbb{R}[[x^2]]\, .$

(more details). Such a function $g$ necessarily has to be defined as $g(x):=f(\sqrt x)$ for $x > 0$. This makes it $C^\infty$ on $\mathbb{R_+}$ as a composition of smooth functions. However, it is not immediately obvious from the composition formula that the derivatives for $x > 0$, $\, g^{(k)}(x),$ do have a limit for $x\to 0$, which is of course a necessary condition for $g$ to be extendable to a $C^\infty$ function on $\mathbb{R}$. But this is clear from the representation

$$\frac{g^{(k)}}{k!} (x^2)=(2x)^{-2k+1}k {2k \choose k}\, \int_0^x (x^2-t^2)^{k-1}\frac{f^{(2k)}}{(2k)!}(t) dt\, $$

that exhibits the $k$-th Taylor coefficient of $g$ as an integral mean of the $2k$-th Taylor coefficients of $f$ on the interval on $[0,x]$, so that $$\lim_{x > 0\atop x \to 0} \frac{g^{(k)}}{k!} (x) = \frac{f^{(2k)}}{(2k)!}(0)\, .$$ In general, for a function $g\in C^\infty(\mathbb{R_+})$, all derivatives being continuously extendable at $0$ is also sufficient condition for $g$ to be smoothly extendable on $\mathbb{R}\, $ (as an easy instance of the Whitney extension theorem; or by Borel's theorem, extending $g$ on the left half-line by any $h\in C^\infty(\mathbb{R})$ with prescribed derivatives at $0$, or by more elementary arguments ad hoc).

Answer 2

Yes, this is a theorem of Hassler Whitney:

Whitney, Hassler Differentiable even functions. Duke Math. J. 10, (1943). 159–160.

Answer 3

See the answer (here) for more information: Smooth functions which are invariant under a compact Lie group representation factor smoothly over the basic invariant polynomials.

Answer 4

Basic proof,

By Taylor formula $f(x) = \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{2k} + x^{2n }\epsilon (x)$ with $\epsilon(x)=\int_0^1 \frac{(1-t)^{2n-1}}{(2n-1)!} f^{(2n)}(tx)dt$ , $\epsilon\in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ because $f \in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ then all dérivatives of $\epsilon $ are bounded in neighborhood of 0.

$$\forall x\geq0,\quad g(x)= \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{k} + x^{n }\epsilon (\sqrt x)=g_1(x)+g_2(x)$$ with $g_2(x)=x^{n }\epsilon (\sqrt x)$. To show $g^{(n)}(0)$ exists , it suffices to show $g_2^{(n)}(0)$ exists $$g_2'(x)=x^{n-1}(n\epsilon(\sqrt x)+\sqrt x \epsilon'(\sqrt x))=x^{n-1 }\epsilon_1 (\sqrt x)$$ with $\epsilon_1(x)= n\epsilon (x)+x\epsilon'(x)\to 0$ because $\epsilon '$ is bounded in neighborhood of 0. By induction $$g_2^{(k)}(x)=x^{n-k}\epsilon_k(\sqrt x),\quad \forall k\leq n $$ thus $g_2^{(n)}(x)=\epsilon_n(\sqrt x)$ and $g_2^{(n)}(0)$ exists