If $f^2$ and $f^3$ are $C^{\infty}(\mathbb R)$ then $f$ is $C^{\infty}(\mathbb R)$

Solution 1:

This result was first proved in:

1. Henri Joris, Une $C^\infty$ application non-immersive qui possède la propriété universelle des immersions, Arch. Math. (Basel), 39, 269-277, 1982.

The proof we provide follows the general framework of the proof which appears in:

2. Robert Myers, An Elementary Proof of Joris's Theorem, Amer. Math. Monthly, 112, no. 9, 829-831, 2005.

We shall use the following notation $$ P_{n,x_0}(x;F)=F(x_0)+F^{(1)}(x_0)(x-x_0)+\frac{F^{(2)}(x_0)}{2!}(x-x_0)^2 +\cdots+\frac{F^{(n)}(x_0)}{n!}(x-x_0)^n. $$ where $F$ is a function $n$ times differentiable in some open interval $(a,b)$ containing $x_0$. We shall be using the following lemmata:

Lemma 1. Let $F: (a,b)\to\mathbb R$ be a function possessing all derivatives up to order $n$, for some $x_0\in(a,b)$. $F^{(1)}(x_0),\ldots,F^{(n)}(x_0)$. Then $$ \lim_{x\to x_0}\frac{F(x)-P_{n,x_0}(x;F)}{(x-x_0)^n}=0. $$

Proof. See Spivak.

Lemma 2. Let $\,F: (a,b)\to \mathbb R\,$ be function which is $k$ times differentiable in $(a,b)$ and $k+1$ times differentiable in $(a,b)\setminus\{x_0\}$, where $x_0\in(a,b)$. If $$ L=\lim_{x\to x_0} \frac{F(x)-P_{k,x_0}(x;F)}{(x-x_0)^{k+1}} \in\mathbb R, $$ then $f$ possesses derivative of $k+1$ order at $x_0$, and $F^{(k+1)}(x_0)=(k+1)! L.$

Proof. Use L' Hôpital's Rule.

Lemma 3. Let $F\in C^{k+n}(a,b),\,$ $x_0\in(a,b),\,$ and $$ G(x)=\left\{\begin{array}{ccl} \dfrac{F(x)-P_{k,x_0}(x;F)}{( x-x_0)^{k+1}} & \text{if} & x\ne x_0, \\ (k+1)! F^{(k+1)}(x_0) & \text{if} & x= x_0, \\ \end{array} \right. $$ then $G\in\ C^n(a,b)$. In particular, if $\,F\in C^\infty(a,b)$, then $\,G\in C^\infty(a,b)$.

Proof. Use induction on the degree of $P_{k,x_0}$ and evoke Lemma 2.

Proof of Joris's Theorem. Let us first note that, if $f(x_0)\ne 0$, for some $x_0\in\mathbb R$, then $f$ maintain sign in a whole open interval $I$, with $x_0\in I$, and as $h=f^3$ is $C^\infty$ in $I$, then $f$ is also $C^\infty$ in $I$, as the composition of $x^{1/3}$, which is $C^\infty$ in $\,\mathbb R\setminus\{0\}\,$ and $f^3$. The hard part is to show that $f$ is $C^\infty$, where $f$ vanishes. We set $$ Z=\{x\in\mathbb R: f(x)=0\} $$ and let $\,x_0\in Z$. We shall study separately the cases where $x_0$ is an isolated point of $Z$ from the case where $x_0$ is an accumulation point of $Z$.

It is important to point out that, in order for $f^{(k+1)}(x_0)$ to be definable, $f$ has to be $k$ times differentiable in an interval containing $x_0$.

Case A. $x_0$ is an isolated point of $Z$.

First sub-case. $\,h^{(n)}(x_0)=0,\,$ for all $n\in\mathbb N$.

We shall show in this sub-case that all the derivatives of $f$ exists at $x=x_0$ and they all vanish. Due to Lemma 1 we have that $$ 0=\lim_{x\to x_0}\frac{h(x)}{(x-x_0)^n}=\lim_{x\to x_0}\frac{f^3(x)}{(x-x_0)^n}, \quad \text{for all $n\in\mathbb N$,} $$ since $P_{n,x_0}(x;h)\equiv 0$. Hence $\,\displaystyle\lim_{x\to x_0}\dfrac{f(x)}{(x-x_0)^n}=0$, for all $n\in\mathbb N$. In particular $$ \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)}{x-x_0}=0, $$ and therefore $f$ is differentiable at $x=x_0$, and $f'(x_0)=0$. We shall show inductively that, for every $n\in\mathbb N$,the function $f$ is $n$ times differentiable at $x=x_0$ and $f^{(0)}(x_0)=f^{(1)}(x_0)=\cdots=f^{(n)}(x_0)=0$. Assume that this is true for some $\,k\in\mathbb N$. Then $P_{k,x_0}(x;f)\equiv 0$ and thus Lemma 2 provides that $f$ is $k+1$ differentiable at $x=x_0$ and $f^{(k+1)}(x_0)=0$.

Second sub-case. There exists an $\,n\in\mathbb N$, such that $\,h^{(n)}(x_0)\ne 0$.

Let $k$ be the least positive integer, such that $\,h^{(k)}(x_0)\ne 0$. Lemma 3 provides that, there exists a function $\,H\in C^\infty(\mathbb R)$, such that $\,h(x)=(x-x_0)^kH(x)$, with $H(x_0)\ne 0$. In particular, there exists an interval $\,(a,b)$, with $x_0\in (a,b)$, where $H$ maintains sign. Meanwhile we have that $$ g^3(x)=h^2(x)=(x-x_0)^{2k}H^2(x), $$ and hence, $$ g(x)\big(H^2(x)\big)^{-1/3}=\big((x-x_0)^{3k}\big)^{1/3}, \quad\text{for all $x\in (a,b)$}. $$ The right hand side of the above is $C^\infty$ in $(a,b)$, which implies that $2k/3$ is an integer, and hence $k=3k_1$, for some positive integer $k_1$. Thus $h(x)=(x-x_0)^{3k_1}H(x)$, and consequently $$ f(x)=(x-x_0)^{k_1}H^{1/3}(x), $$ and hence $f$ is $C^\infty$ in $(a,b)$.

Case B. $x_0$ is an accumulation point of $Z$.

In such case, there exists a sequence $\,\{x_n\}\subset Z$, such that $x_n\to x_0$. We may assume that this sequence in strictly monotone. Without loss of generality we assume that $\{x_n\}$ strictly increasing. Then $$ 0=h(x_1)=h(x_2)=\cdots=h(x_n)=\cdots. $$ Due to Rolle's Theorem, there exist $\xi_i \in(x_i,x_{i+1})$, such that $$ 0=h'(\xi_1)=h'(\xi_2)=\cdots=h'(\xi_n)=\cdots. $$ Thus $\,h'(x_0)=0$, since $h'$ is continuous and $\xi_n\to x_0$. Repeating the same procedure for $h'$ instead of $h$, we obtain that $h''(x_0)=0$, and inductively that $\,h^{(n)}(x_0)=0$, for all $n\in\mathbb N$.

As in the study of the First sub-case, we obtain, by virtue of Lemma 1, that $$ 0=\lim_{x\to x_0}\frac{h(x)}{(x-x_0)^n}=\lim_{x\to x_0}\frac{f^3(x)}{(x-x_0)^n}, \quad \text{for all $n\in\mathbb N$,} $$ and thus \begin{equation*} \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)}{x-x_0}=0. \end{equation*} Therefore, $f$ is differentiable at $x=x_0$, and $f'(x_0)=0$. Next, we shall show inductively that $f$ possesses derivatives of all orders at $x=x_0$, and that they all vanish. Combining all the studied cases so far we have that $f$ is differentiable in the whole of $\mathbb R$. We shall use induction to prove our claim.

Assume that $f$ is $k$ times differentiable in $\mathbb R$ and also $$ f^{(0)}(x_0)=f^{(1)}(x_0)=\cdots=f^{(k)}(x_0)=0, $$ in every $x_0$, which is an accumulation point of $Z$. Then $f$ is $k+1$ times differentiable in the whole of $\mathbb R$, and $f^{(k+1)}(x_0)=0$, for every $x_0$, which is an accumulation point of $Z$. We have that $$ \lim_{x\to x_0}\frac{f(x)}{(x-x_0)^n}=0, $$ for every $x_0$ which is an accumulation point of $Z$, and by virtue of Lemma 2, $f$ possesses derivative of order $k+1$ in every such $x_0$. Combining all the studied cases we obtain that $f$ is $k+1$ times differentiable in the whole of $\mathbb R$. This establishes the induction hypothesis and completes the proof of Joris's Theorem.

Note. This proof can be easily generalised for the case when $\,f^k,f^\ell\in C^\infty(\mathbb R)$ and $(k,\ell)=1$.

Solution 2:

This is not really an answer, I've posted this as a comment on the question above, I just think the information needs to be more prominently displayed for future viewers.

On the answer to the MO question addressing this, Terry Tao commented with the following link to his write up of the solution to the problem based on one of the papers in the answer: http://www.math.ucla.edu/~tao/preprints/Expository/squarecube.dvi. To convert dvi to pdf, see this tex.SE answer.