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

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4 Answers 4

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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).

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    $\begingroup$ (on the integral formula: by density, it suffices to prove it for polynomials, and by linearity, it suffices to consider $f(x)=x^n$. Substituting $s:=t^2/x^2$ in the integral, the identity reduces to the classical relation between the Beta and the Gamma functions). $\endgroup$ Aug 10, 2011 at 7:56
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Yes, this is a theorem of Hassler Whitney:

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

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  • $\begingroup$ Wow -- I wouldn't have been surprised if the theorem were 100 years older than this! It would seem the necessary tools were available earlier... Perhaps it was necessary for people to start thinking systematically about differential topology for such a question to attain relevance? $\endgroup$
    – Tim Campion
    Aug 24, 2020 at 20:58
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    $\begingroup$ @TimCampion Whitney contributed much to structures of germs of smooth functions. This fact is saying that a Z/2-invariant germ (R, 0) -> (R, 0) lifts along x^2 -- this is a consequence of Malgrange preparation theorem, which among many other things, proves that germ of a generic smooth mapping (R^2, 0) -> (R^2, 0) is equivalent to either identity, (x^2, y) (fold) or (x^3 + xy, y) (cusp), a theorem also originally by Whitney. These questions gained relevance only after people started thinking systematically about singularities rather than smoothness. $\endgroup$ Jul 26, 2021 at 13:01
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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.

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  • $\begingroup$ As a representation theorist, I love this answer. $\endgroup$
    – LSpice
    Aug 24, 2020 at 19:56
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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

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