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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.

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  • $\begingroup$ To clarify: For me smooth means of class $C^\infty$. $\endgroup$ – Jan Bohr May 27 at 13:15
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    $\begingroup$ 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). $\endgroup$ – logarithm May 27 at 13:16
  • $\begingroup$ 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, $f^{(n)}(0)=\lim_{h\to0}\frac{f^{(n-1)}(h)-g^{n-1}(0)}{h}=\lim_{h\to0}(g^{n}(0)+g^{(n+1)}(0)\frac{h}{2}+H(h))=g^{(n)}(0)$. $\endgroup$ – logarithm May 27 at 13:16
  • $\begingroup$ 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. $\endgroup$ – Jan Bohr May 27 at 13:20
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    $\begingroup$ @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 $\neq x^2$. $\endgroup$ – logarithm May 27 at 13:36
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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.

The case of multivariable smooth functions proceeds along similar lines, the only difference being that there is more than one reasonable choice for a path from the origin to a given point along which to do the integral. You can find the full argument in most references on smooth manifolds close by to the definition of "tangent spaces", at least in references which make use of the "tangent vectors are derivations" approach to things.

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