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Suppose I have a function $f:\mathbb{R}^d \to \mathbb{R}$ and I know that $f$ is smooth ($C^\infty$) along each ray $t \mapsto f(td)$ on $t \in [-\epsilon, \epsilon]$ and all directions $d \in \mathbb{R}^d$.

Is smoothness along these rays sufficient for $f$ to be smooth around $0$ as a multivariate function (all partial derivatives exist)?


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No. Just take a non-smooth function $g \colon S^{d-1} \to \mathbb{R}$ with $g(-p) = -g(p)$ and define $f(v) = \|v\|g(v/\|v\|)$. This is smooth since on each ray it is just $t \mapsto k t$ for some $k$, but the non-smoothness of $g$ means that $f$ is not smooth. The closest to this that I know of is if you take all smooth curves (not just rays). Then $f$ is smooth. This is due to Jan Boman. – Loop Space Aug 24 '11 at 10:13
That is indeed a nice counterexample. Would you happen to have a reference of this result of Jan Boman? In addition, would smoothness along all real analytic curves be sufficient too? – Bart Aug 24 '11 at 12:15
The result Andrew Stacey refers to is this paper of Boman (MathSciNet link): The article doesn't look like it is available online. In the MathSciNet review the following counterexample is mentioned: there exists a non-continuous function $f$ from $\mathbb{R}^2\to\mathbb{R}$ such that $f\circ u \in C^{\infty}(\mathbb{R},\mathbb{R})$ for every quasianalytic $u$. (Though it is not clear which quasianalytic class the example applies to.) This should give at least a partial answer to your follow-up question. – Willie Wong Aug 24 '11 at 13:12
Thanks Willie. Looking at the Boman paper, demanding smoothness along only analytic curves is indeed not sufficient. It turns out that when $f \circ u$ is real analytic for every real analytic curve $u$, that $f$ is real analytic. See "An Ontology of Directional Regularity Implying Joint Regularity" published in Real Analysis Exchange, available at . – Bart Aug 24 '11 at 13:35
Apparently, in the published version there is the condition that the k-th partial derivative of $f \circ u$ needs to be smaller than $C k! / r^k$ for some $r>0$. – Bart Aug 24 '11 at 14:19

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