When is the Composite with Cube Root Smooth

Let $f:\mathbf R\to \mathbf R$ be a smooth map and $g:\mathbf R\to \mathbf R$ be defined as $g(x)=f(x^{1/3})$ for all $x\in \mathbf R$.

Problem. Then $g$ is smooth if and only if $f^{(n)}(0)$ is $0$ whenever $n$ is not an integral multiple of $3$.

One direction is easy. Assume $g$ is smooth. Then we have $f(x)=g(x^3)$ for all $x$. Differentiating and using the chain rule gives that the required derivatives of $f$ vanish.

I am struggling with the converse. Assume $f^{(n)}(0)=0$ whenever $n$ is not an integral multiple of $3$. We need to show that $g$ is smooth. Since $x^{1/3}$ is smooth at all points except $x=0$, we see that $g$ too is so. So the only problem is at $0$. I am only able to show that the first derivative of $g$ at $0$ exists. Here is what I have done.

Let $x>0$ be arbitrary. By Taylor we know that $$f(x)= f(0)+f'(0)x+ \frac{f''(0)}{2}x^2+ \frac{f'''(\lambda_x x)}{6}x^3$$ for some $0<\lambda_x<1$.

This gives by hypothesis that $$f(x) - f(0) = \frac{f'''(\lambda_x x)}{6}x^3$$ Thus $$g(x)-g(0)=\frac{f'''(\lambda_{x^{1/3}} x^{1/3})}{6} x$$ and therefore $$\frac{g(x)-g(0)}{x}=\frac{f'''(\lambda_{x^{1/3}} x^{1/3})}{6}$$ Since $\lim_{x\to 0}f'''(\lambda_{x^{1/3}} x^{1/3}))$ exists, we see that $g'(0)$ exists.

But I am not able to extend this argument to show that $g''(0)$ etc. also exist.

Solution 1:

$\newcommand{\Reals}{\mathbf{R}}$The "missing link" in my earlier sketch was the principle that if $R$ is a smooth function with $R(x) = O(|x|^{N})$ near $0$, and if we write $R(x) = x^{N} r(x)$, then $r$ itself is smooth. The simplest argument I could construct was more delicate than I'd originally expected, so I've included it in some detail. It's entirely possible there is a more elegant argument.

Lemma: If $R$ is a smooth, real-valued function on $\Reals$ such that $R(x) = O(|x|)$ near $0$, and if $R(x) = xr(x)$ for all real $x$, then $r$ is smooth.

Proof: Because $R$ is smooth, $r$ is smooth except possibly at $0$. Let $D$ denote the derivative operator and write, e.g., $D^{k}(x^{n})$ to denote the $k$ th derivative of the $n$th power function. If $n \geq 0$ is an integer, a straightforward induction (compare the identity $R^{(n)}(x) = xr^{(n)}(x) + nr^{(n-1)}(x)$, or $r^{(n)}(x) = \frac{1}{x}\bigl[R^{(n)}(x) - nr^{(n-1)}(x)\bigr]$, which holds for $x \neq 0$), gives the formula $$ r^{(n)}(x) = \frac{1}{x^{n+1}} \sum_{k=0}^{n} (-1)^{k} D^{k}(x^{n}) D^{n-k}R(x). \tag{1} $$ For example, \begin{align*} r'(x) &= \frac{xR'(x) - R(x)}{x^{2}}, \\ r''(x) &= \frac{x^{2} R''(x) - 2x R'(x) + 2R(x)}{x^{3}}, \\ r'''(x) &= \frac{x^{3} R'''(x) - 3x^{2} R''(x) + 6x R'(x) - 6R(x)}{x^{4}}. \end{align*} The numerator (i.e., the sum) in (1) is a smooth function, and clearly vanishes at $0$. (The terms containing a positive power of $x$ vanish because $R$ is smooth, while $R(0) = 0$ by hypothesis.)

Upon differentiating this numerator and shifting the index of summation, everything cancels except one term: \begin{align*} D\sum_{k=0}^{n} (-1)^{k} D^{k}(x^{n}) D^{n-k}R(x) &= \sum_{k=0}^{n} (-1)^{k} \bigl[D^{k+1}(x^{n}) D^{n-k}R(x) + D^{k}(x^{n}) D^{n-k+1}R(x)\bigr] \\ &= -\sum_{k=1}^{n+1} (-1)^{k} D^{k}(x^{n}) D^{n-k+1}R(x) + \sum_{k=0}^{n} (-1)^{k} D^{k}(x^{n}) D^{n-k+1}R(x) \\ &= x^{n} R^{(n+1)}(0). \end{align*} Consequently, applying l'Hôpital's rule to (1) gives $$ \lim_{x \to 0} r^{(n)}(x) = \lim_{x \to 0} \frac{1}{x^{n+1}} \sum_{k=0}^{n} (-1)^{k} D^{k}(x^{n}) D^{n-k}R(x) = \lim_{x \to 0} \frac{x^{n} R^{(n+1)}(0)}{(n+1)x^{n}} = \frac{R^{(n+1)}(0)}{n+1},\quad n\geq 0. $$

To show $r$ is of class $C^{1}$, expand $R$ to third order at $0$: $$ R(x) = R'(0)x + \frac{R''(0)}{2!}x^{2} + O(|x|^{3}),\qquad r(x) = R'(0) + \frac{R''(0)}{2!}x + O(|x|^{2}). $$ It follows at once that $r'(0)$ exists and is equal to $\frac{1}{2}R''(0) = \lim(r', 0)$.

Using this fact as a base case, assume inductively that $r$ is of class $C^{n}$ for some $n \geq 0$. The function $r^{(n)}$ is continuous on $\Reals$, differentiable except possibly at $0$, and $\lim(r^{(n+1)}, 0)$ exists. It follows that $r^{(n)}$ is differentiable at $0$, and that $$ r^{(n+1)}(0) = \lim_{x \to 0} r^{(n+1)}(x). $$ This establishes that $r$ is of class $C^{n+1}$ at $0$. By induction on $n$, $r$ is smooth.

Theorem: If $R$ is a smooth function such that $R(x) = O(|x|^{N})$ near $0$ for some integer $N \geq 1$, and if $R(x) = x^{N}r(x)$, then $r$ is smooth.

Proof: Immediate from the lemma. (We have $R(x) = O(|x|)$, so $x^{N-1}r(x) = O(|x|^{N-1})$ is smooth, etc.)

Armed with the theorem (which I'd blithely been assuming), we turn to the question at hand.

Suppose $f$ is smooth, and that $f^{(n)}(0) = 0$ when $n$ is not a multiple of $3$. For each positive integer $N$, $f$ has an order-$3N$ Taylor expansion $$ f(x) = \sum_{k=0}^{N} \frac{f^{(3k)}(0) x^{3k}}{(3k)!} + R_{3N}(x) \tag{2} $$ with $R_{3N}$ smooth, and $R_{3N}(x) = O(|x|^{3N + 3})$ at $0$. Write $R_{3N}(x) = x^{3N + 3} r(x)$. By the theorem, $r$ is smooth.

Define $g$ by $g(x) = f(x^{1/3})$ for all real $x$. Fix an integer $N \geq 1$, and use (2) to write $$ g(x) = \sum_{k=0}^{N} \frac{f^{(3k)}(0) x^{k}}{(3k)!} + R_{3N}(x^{1/3}) = \sum_{k=0}^{N} \frac{f^{(3k)}(0) x^{k}}{(3k)!} + x^{N+1} r(x^{1/3}). $$ To show $g$ is smooth, it suffices to show the remainder $R(x) := R_{3N}(x^{1/3}) = x^{N+1} r(x^{1/3})$ is of class $C^{N}$ (since $N \geq 1$ is arbitrary).

Since $r$ is smooth, $R$ is smooth except possibly at $0$. For $x \neq 0$, \begin{align*} R'(x) &= (N + 1)x^{N} r(x^{1/3}) + x^{N+1} r'(x^{1/3}) \tfrac{1}{3} x^{-2/3} \\ &= x^{N}\bigl[(N + 1) r(x^{1/3}) + \tfrac{1}{3} x^{1/3} r'(x^{1/3})\bigr] \\ &= x^{N} r_{1}(x^{1/3}), \end{align*} with the final equality serving to define $r_{1}$. Clearly, $r_{1}$ is smooth, which (i) proves $R$ is of class $C^{1}$, and (ii) expresses $R'(x) = x^{N} r_{1}(x^{1/3})$, allowing us to continue differentiating inductively, with the result that $R^{(N)}(x) = x r_{N}(x^{1/3})$ for some smooth function $r_{N}$. Particularly, $R$ is of class $C^{N}$.

Solution 2:

Not an entirely different proof, just the way I see it.

1. Let $f$ smooth function on $\mathbb{R}$. Take $n\ge 0$ a natural number. We have the Taylor expansion of order $n$ at $0$ with integral remainder

$$f(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k + R_n(x)$$ where $$R_n(x) = \int_{0}^x \frac{(x-t)^n}{n!} f^{(n+1)}(t) dt= x^{n+1} \int_0^1\frac{(1-t)^n}{n!}f^{(n+1)}(x t)dt= x^{n+1} g_{n+1}(x)$$ with $g_{n+1}$ smooth.

2. Take now $n = 3m$ and assume that $f$ satisfies the conditions. Then we get $$f(\sqrt[3]{x}) = \sum_{k=0}^m a_k x^k + x^{m+\frac{1}{3}}g_{3m+1}(\sqrt[3]{x})$$

3. Now it's enough to prove the following lemma: let $h$ a function of class $C^m$. Then the function $x^{m+\frac{1}{3}} h(\sqrt[3]{x})$ is of class $C^m$. The proof is by induction on $m$. For $m=0$ it's clear. Assume it's true for $m-1$. The function is clearly continuous. The derivative of the function $x^{m+\frac{1}{3}} h(\sqrt[3]{x})$ for $x\ne 0$ is $$x^{(m-1) +\frac{1}{3}} \left(\frac{1}{3}h'(\sqrt[3]{x}) + (m+\frac{1}{3})h( \sqrt[3]{x}) \right )$$ and since this extends at $0$ with value $0$, it's true for all $x$. By the induction hypothesis, we get that the derivative is of class $C^{m-1}$. Hence the function is of class $C^m$.

Solution 3:

Just wanted to cover up the $\implies$ case, where the other answers doesn't addressed, and the OP just explained vaguely "differentiating and using the chain rule gives that the required derivatives of $f$ vanish". As stated in a similar question, it is not just that simple.

Let's show that if $f \circ \psi^{-1}\colon \widetilde{\mathbb{R}}\to \mathbb{R}$ is smooth, where $\psi^{-1}(x) = x^{\frac{1}{3}} $, then $f^{(n)} (0) = 0$ for all $n \nmid 3$. We can express $f = (f \circ \psi^{-1}) \circ \psi)$, and apply the combinatorial form of the Faà di Bruno's formula to express the $n$-th derivative: $$ \frac{d ^n}{d x^n} (\varphi \circ \psi) (x)= \sum_{\pi \in \Pi} \frac{d^{|\pi|} \varphi}{d x^{|\pi|} } \left( \psi(x) \right) \cdot \prod_{B \in \pi} \frac{d^{|B|} \psi}{d x^{|B|} } (x)\,, $$ where $\varphi = f \circ \psi^{-1}$, $\pi$ runs throught $\Pi$, which is the collection of all partitions of the set $\{i\}_{i=1}^n$, and $B$ runs through all blocks of partition $\pi$. The values of $\psi^{(n)} (0)$ are all $0$, except for $n = 3$. So, the product is positive only if $|B| = 3$ for all $B \in \pi$. It follows that the formula above can only give non-zero values if $n$ is divisible by $3$.