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Let $0 \neq \phi \in L^2(\mathbb R^n)$ be a square-integrable function and $\mathcal F \subset \mathbb R^n$ a finite set. If we are in the one-dimensional setting $n=1$ then the set of translates of $\phi$ by $\mathcal F$, i.e. $$ \{ \phi(\cdot - x) : x \in \mathcal F \} \subset L^2(\mathbb R) $$ is linear independent. I found this result in the book Christensen, In introduction to frames and Riesz bases, p. 228.

I was wondering if this statement holds in arbitrary dimensions, i.e. without restriction of $n$ to $n=1$. If yes, does somebody knows a reference for this?

Thanks in advance!

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    $\begingroup$ This is the same as asking if finitely many exponentials $e^{ix\cdot t}$, $x\in F$, are linearly independent (take Fourier transforms). The answer is yes, and for example the second answer here can be adapted (consider the exponent with largest real part; if there are several, we have reduced the dimension by one, so an induction on the dimension takes care of that): math.stackexchange.com/questions/1451281/… $\endgroup$ Apr 3 at 17:38
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    $\begingroup$ Not really important, but asking whether every finite subfamily of a family $(f_i)_{i\in I}$ is linearly independent is exactly the same as asking whether $(f_i)_{i\in I}$ is so: linear combinations by definition only involve finitely many elements, so linear independence likewise. $\endgroup$
    – Gro-Tsen
    Apr 3 at 19:02

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I do not know what Christensen's proof is but here is a simple proof that works in $R^n$. Suppose that translates are linearly dependent: $$\sum c_j\phi(x-t_j)\equiv 0,$$ where $t_j$ are all distinct. Take Fourier transform; shift correspnds to multilication on an exponential: $$\sum c_j e^{-it_j\cdot s}\hat{\phi}(s)\equiv 0.$$ But the multiplier $$m(s):=\sum c_j e^{-it_js}$$ is a non-zero entire function, since all $t_j$ are distinct, so its zeros make a proper analytic subset of $R^n$, and such a set must be of zero measure. Therefore $\hat{\phi}(s)=0$ almost everywhere, so $\phi=0.$

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  • $\begingroup$ As I see it correctly the argument here is that since the Fourier transform of $\phi$ is not the zero function, it must be non-zero on a set of positive Lebesgue measure. Are we then using that an entire function of several complex variables which vanishes on a set in $\mathbb R^n$ of positive Lebesgue measure must be the zero function? If yes, is there a reference for this? $\endgroup$
    – J. Swail
    Apr 4 at 10:21
  • $\begingroup$ A reference is any book on several complex variables. $\endgroup$ Apr 4 at 13:21
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A standard proof of the independence of the exponentials $f_j(x)=e^{ix\cdot \xi_j}$ is the following. Take a vector $\omega \in \mathbb R^n$ such that $\omega\cdot (\xi_k-\xi_j)\neq 0$ for $k \neq j$ (this follows by induction on the numbers of vectors) and let $D=\omega \cdot \nabla$. Then $D(f_j)=i\omega \cdot \xi_j f_j$ so that the $f_j$ are eigenvectors associated to distinct eigenavalues of a linear operator.

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    $\begingroup$ A more straightforward argument which I use in my undergraduate linear algebra class is to compute the Wronskian determinant. $\endgroup$ Apr 3 at 23:48
  • $\begingroup$ @Eremenko I agree that it is easier to use the Wronskian if $n=1$. For general $n$ I do not see how to use it easily and induction on the dimension is puzzzling. $\endgroup$ Apr 4 at 6:50
  • $\begingroup$ Wronskian of any number of exponentials is easily compute: it is an exponential times Vandermonde. If functions are linearly independent on a line then they are surely linearly independent in the whole space. Take a line $x=\{ x_0t:t\in R\}$ such that $(x_0,\xi_j)\neq 0$ for all $j$. $\endgroup$ Apr 4 at 15:15

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