$xf$ smooth $\Rightarrow$ $f$ smooth?

Question. Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be a continuous function such that $$x\mapsto xf(x)$$ is smooth. Is $$f$$ necessarily smooth?

Here are some thoughts on the problem:

Clearly $$f$$ is smooth on $$\mathbb{R}\backslash 0$$ so one only needs to show that the derivatives $$f^{(j)}(x)$$ (for $$x\neq 0$$) have the same limit as $$x\rightarrow 0$$ from left and right.

If $$xf$$ is analytic, then also $$f$$ is analytic. This is clear as its power series around $$0$$ is obtained from the one of $$xf$$ by factoring out one $$x$$. For smooth functions one would need to know something about the regularity of $$x\mapsto x^{-1}R_kf(x)$$, where $$R_kf$$ is the remainder of the order $$k$$ Taylor expansion. Showing smoothness of this is essentially equivalent to the original question.

One can without loss of generality assume that $$f$$ has compact support. Then the Fourier transform satisfies $$\hat f\in C_0$$ and $$\frac{d}{d\xi}\hat f \in \mathcal{S}$$ (Schwartz functions). If this would imply that $$\hat f\in \mathcal{S}$$, then the question would be solved.

• To clarify: For me smooth means of class $C^\infty$. – Jan Bohr May 27 at 13:15
• All you need to use is Taylor theorem. Let $g(x)=xf(x)$. The zeroth derivative of $f$ is $f(0)$. Assume that you have already proven that $f$ has derivatives up to order $n-1$. Write $g(x)=xf(x)=0+g'(0)x+...+g^{(n-1)}(0)\frac{x^{n-1}}{(n-1)!}+g^{(n)}(0)\frac{x^n}{n!}+g^{(n+1)}(0)\frac{x^{n+1}}{(n+1)!}+x^{n+1}h(x)$, with $h(x)\to0$ as $x\to0$ (Taylor therorem). – logarithm May 27 at 13:16
• Then, divide by $x$ and take $n-1$ derivatives to get $f^{(n-1)}(x)=g^{(n-1)}(0)+g^{(n)}(0)x+g^{(n+1)}(0)\frac{x^2}{2}+xH(x)$, for some $H$ satisfying $H(x)\to0$ as $x\to0$. Now, $f^{(n)}(0)=\lim_{h\to0}\frac{f^{(n-1)}(h)-g^{n-1}(0)}{h}=\lim_{h\to0}(g^{n}(0)+g^{(n+1)}(0)\frac{h}{2}+H(h))=g^{(n)}(0)$. – logarithm May 27 at 13:16
• I think you implicitly use that $h(x)$ is differentiable of some order. This is the problem I address in my second remark and it is not clear to me why it should be true. – Jan Bohr May 27 at 13:20
• @kesa Relax, this result is a well known theorem. You won't find any counterexample. In your example, $xf(x)=x|x|$ is not smooth and it is $\neq x^2$. – logarithm May 27 at 13:36

There is well-known trick exploiting integration to solve this division problem which avoids the more cumbersome Taylor series approach. Given a smooth function $$f$$ of a single variable $$x$$ which vanishes at $$0$$, one would like to be able to "divide out $$x$$" and write $$f=xg \tag{1}$$ where $$g$$ is also a smooth function. More generally, if $$f$$ is a smooth function of $$n$$ variables $$x_1,\ldots,x_n$$ which vanishes at $$(0,\ldots,0)$$ one would like to be able to write $$f = x_1 g_1+ \ldots + x_n g_n \tag{2}$$ where the $$g_i$$ are also smooth functions of $$n$$ variables. In other words, the coordinate functions $$x_i$$ generate the ideal of functions vanishing at $$(0,\ldots,0)$$ inside the ring of smooth functions. This more general result is useful when setting up a good theory of tangent spaces on a smooth manifold. Note that by taking $$\frac{\partial}{\partial x_i}$$ and evaluating at $$(0,\ldots,0)$$, one gets $$g_i(0 ,\ldots,0)=\frac{\partial f}{\partial x_i}(0,\ldots,0)$$ must hold, so the $$g_i$$ are at least uniquely determined at the origin. In general, for $$n \geq 2$$, the $$g_i$$ are not uniquely determined away from the origin. However, for $$n =1$$, there is obviously only one choice for a $$g$$ satisfying (1), we must define $$g(x) = \begin{cases} f'(0) && \text{ if } x = 0 \\ \frac{f(x)}{x} && \text{ if } x \neq 0 \\ \end{cases} \tag{3}$$ So the key lemma to prove is
Proposition: Suppose that $$f: \mathbb{R} \to \mathbb{R}$$ is a $$C^\infty$$ function with $$f(0)=0$$. Then, the function $$g : \mathbb{R} \to \mathbb{R}$$ defined by (3) above is also $$C^\infty$$.
Proof: Observe that another formula for $$g$$ is $$g(x) = \int_0^1 f'(xt) \ dt \tag{4}.$$ Since $$(x,t) \mapsto f'(xt)$$ is a $$C^\infty$$ function of two variables, we are done by the general fact that, if $$H : \mathbb{R}^2 \to \mathbb{R}$$ is a $$C^\infty$$ function of two variables, then $$h(x) = \int_0^1 H(x,t) \ dt$$ defines a $$C^\infty$$ function of one variable. The latter can be proven one derivative at a time; check that $$h'$$ exists and is given by $$h'(x) = \int_0^1 \frac{\partial H}{\partial x} (x,t)\ dt$$ by justifying the interchange of differentiation and integration there.