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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

Asaf Karagila
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Richard
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1 Answers1

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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.

Zarrax
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  • One question, bacause I did not understand: How to prove for example that $g_2^{''}(x)=o(|x|^{n-2})$ as $|x| \rightarrow 0$. – Richard Aug 08 '11 at 14:08
  • Use the chain rule on $R_{2n}(\sqrt{|x|})$ and the fact that the $k$th derivative of $R_{2n}(x)$ is $o(|x|^{2n - k})$ as $x \rightarrow 0$. So the $k$th derivative of $g_2(x)$ is $o(|x|^{n-k})$. – Zarrax Aug 08 '11 at 17:27
  • and the bounds on the $k$th derivative of $R_{2n}(x)$ just come from the fact $R_{2n}(x)$ has its first $2n$ derivatives all zero at $x = 0$. – Zarrax Aug 08 '11 at 17:34
  • $R_{2n}(x)=h(x)x^{2n}$, where $h$ is of the class $C^{2n}(R\setminus 0)$, $h$ is continuous on $R$ and $h(0)=0$. Hence $g_2(x)=R_{2n}(\sqrt {|x|})=k(x)x^n$ for all x, where $k$ is of the class $C^{2n}(R\setminus 0)$, continuous and $k(0)=0$. We can only difference on $R\setminus 0$, and it is not seen whether the derivatives of $g_2$ have a limit in zero. If derivatives $g_2^{(k)}$ exist for $k=0,..,n$ then $g_2$ will be of the class $C^n(R)$. – Richard Aug 08 '11 at 18:25
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    You just need that $R_{2n}(x)$ is $C^{2n}$ on all of ${\mathbb R}$, which is true. So for example suppose you have established $g_2'(0) = 0$ from the definitions. Then for $x \neq 0$, $g_2'(x) = \pm {1 \over 2\sqrt{|x|}} R_{2n}'(\sqrt{|x|})$. You can take difference quotients of $g_2'(x)$, using the bounds on $R_{2n}'(x)$, and you'll get that $g_2''(0) = 0$. You can repeat this idea $n$ times. – Zarrax Aug 08 '11 at 18:42
  • It seems that zarrax's reasoning in comment is wrong. if $g_2'(0) = 0$ then $$\displaystyle g_2''(0)= \lim_{x\to 0}\frac{g_2'(x)-g_2'(0)}x= \lim_{x\to 0} \frac{\frac {\pm R_{2n}'(\sqrt{|x|})}{{2\sqrt{|x|}} }}{x}=\lim_{x\to 0} \frac {\pm R_{2n}'(\sqrt{|x|})}{{2\sqrt{|x|}} x}$$

    and there is no reason for this limit to be zero

     – Pascal Aug 22 '20 at 00:28
  • A good answer here https://mathoverflow.net/questions/72497/if-f-is-smooth-and-even-then-there-exists-a-smooth-function-g-such-that-fx?noredirect=1&lq=1 – Pascal Aug 22 '20 at 00:38