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Assume $\psi_n(x),\;x\in\mathbb R$ is an $L_2(\mathbb R)$ complete orthonormal series.

Let $f:\mathbb R^d\rightarrow\mathbb R$ be smooth enough.

Is it true that if $$\sum_{i=1}^d \frac {\partial f(x)} {\partial x_i}=1,\;\forall x\in\mathbb R^d$$ then $$\psi_n(f(x)),\;x\in\mathbb R^d$$ is an $L_2(\mathbb R^d)$ complete orthonormal series?

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Of course not. Take $f(x)=x_1$, the first coordinate. The condition holds, but all you get from $\psi_n\circ f$ are functions in $x_1$ only.

Informally, the elements of an ONB in $L_2(\mathbb R^d)$ should have $d$ indices; they are not naturally numbered by a single integer $n$.

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Thanks! Let me ask you this way: If all partial derivatived were 1, couldn't we use Change of Variable in the relevant integrals? – Troy McClure Jan 14 '13 at 10:22
@Troy No. Take $f=\sum x_i$ and it's the same problem: everything in the span of $\psi_n \circ f$ is a function of $f$ only. The change of variables formula does not work between spaces of different dimensions. // Try $\psi_n\circ f_k$ where $k$ runs from $1$ to $d$. – user53153 Jan 14 '13 at 12:32
ok so it's a span of $f$, cannot it be dense? – Troy McClure Jan 14 '13 at 12:38
@Troy Every function in the closed span of $\{\psi_n\circ f\}$ is constant on each level set of $f$. There are functions in $L^2(\mathbb R^d)$ that are not constant on level sets of $f$. – user53153 Jan 14 '13 at 13:03
thanks! i studied a new term: "level set" – Troy McClure Jan 14 '13 at 13:25

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