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Fix vectors $x,y\in\mathbb{R}^d$ and a smooth function $\phi:\mathbb{R}\rightarrow \mathbb{R}$. Define $\phi^d: \mathbb{R}^d \rightarrow \mathbb{R}^d$ as applying $\phi$ entrywise (i.e. $\phi^d(x_1, x_2, \dots) = (\phi(x_1),\phi(x_2), \dots)$). Under what conditions do we have the equality

$$\mathbb{R}^d=\text{span}(\{\phi^d(\alpha x+y)~\big |~\alpha\in \mathbb{R}\})$$

Clearly if $\phi$ is the identity operator $\phi(x)=x$ or if any two entries of $x$ are equal, this equality does not hold. Main claim is equality holds for generic (i.e. almost all) $x,\phi$ choices.

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    $\begingroup$ Suppose the span is not the full $\mathbb{R}^d$, then there exists a normal vector to the span which we can call $\nu$. This means that $\sum \nu_i \phi(\alpha x_i) = 0$ for all $\alpha$. Take $k$ derivatives with respect to $\alpha$ you still get 0. Evaluate at $\alpha = 0$ you get that $\phi^{(k)}(0) \sum \nu_i (x_i)^k = 0$. So provided not too many derivatives of $\phi$ vanish at zero (generically true), such $\nu$ cannot exist for generic $x_i$. $\endgroup$ – Willie Wong Jan 17 at 22:22
  • $\begingroup$ For the final claim, see en.wikipedia.org/wiki/Vandermonde_matrix $\endgroup$ – Willie Wong Jan 17 at 22:28
  • $\begingroup$ There could be some typo around, because the question does not make any sense, as it is. $\endgroup$ – Pietro Majer Jan 18 at 0:10
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    $\begingroup$ @Pietro Majer I think that $\phi (x)=(\phi (x_1),\ldots, \phi (x_d)) $. $\endgroup$ – Jochen Wengenroth Jan 18 at 8:44
  • $\begingroup$ So maybe it is a map $\phi:\mathbb{R}^d\to\mathbb{R}^d$? $\endgroup$ – Pietro Majer Jan 18 at 8:55

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