Smoothness of $f(\sqrt{x})$ How to prove that if a function $f\colon\mathbb R\to\mathbb R$ is of the class $C^{2n}$ and even
then there exists a function $g\colon\mathbb R \to\mathbb R$ of the class $C^n$ such that $f(x)=g(x^2)$.
It is obvious, by using power series expansion in a symmetric neighbourhood of zero, that it holds if $f$ is analytic and even. But how to show that in general?
Richard
 A: Taylor expand $f(x) = P_{2n}(x) + R_{2n}(x)$ at $x = 0$, where $P_{2n}(x)$ is of degree $2n$ and $R_{2n}(x)$ is the remainder. Because $f(x)$ is even, all odd derivatives of $f(x)$ at $x = 0$ are zero and therefore $P_{2n}(x)$ only contains even powers of $x$ and thus can be written as $g_1(x^2)$ for some polynomial $g_1$.
As for $R_{2n}(x)$, since it's the difference $f(x) - P_{2n}(x)$ of two even functions it is even as well. Thus one can write $R_{2n}(x) = g_2(x^2)$ for some $g_2$. The key is to show that $g_2(x)$ is actually $C^n$. On ${\mathbb R} - \{0\}$ it's actually $C^{2n}; g_2(x^2) = f(x) - P_{2n}(x)$ is the difference of two $C^{2n}$ functions and this is preserved under change of variables $x \rightarrow \sqrt{x}$ or $x \rightarrow -\sqrt{x}$.
Thus we must consider $x = 0$ only. Since $R_{2n}(x)$ is the remainder term of the Taylor expansion, it satisfies $|R_{2n}(x)| = o(|x|^{2n})$ as $|x| \rightarrow 0$. So 
$g_2(x) = R_{2n}(\sqrt{|x|})$ satisfies $g_2(x) = o(|x|^n)$ as $|x| \rightarrow 0$, and 
$g_2(x) = 0$. Similarly, using the corresponding properties of derivatives of $R_{2n}(x)$ one can inductively show that for any $k \leq n$, $g_2^{(k)}(x) = o(|x|^{n-k})$ as $|x| \rightarrow 0$. So taking successive difference quotients in the definition of the derivative shows that $g_2^k(0)$ exists and equals zero for all $k \leq n$. 
So we can conclude $g_2(x)$ is a $C^n$ function. Therefore $g(x) = g_1(x) + g_2(x)$ is $C^n$ too.
