Let $\epsilon > 0$. I would like to know if there exists $c < \infty$ such that for all $d \in \mathbb{N}$ the following holds. If $x \in \mathbb{R}^d$ let $N_x$ be the standard Gaussian centered at $x$. Let $\mathcal{V}$ be the subspace of $L^2(\mathbb{R}^d)$ spanned by the vectors $\{N_x: x \in \mathbb{Z}^d,||x||_2 \geq c\}$. Then the $L^2$ norm of the orthogonal projection of $N_0$ onto $\mathcal{V}$ is less than $\epsilon$.

I am also interested in the version of this question where the functions $N_x$ are the normalized indicators of balls of radius $\sqrt{d}$.

An idea that might be useful is as follows. In order to obtain the desired bound, it suffices to bound the norm of the projection onto the closed span $\mathcal{W}$ of $\{N_x :x \in \mathbb{R}^d,||x||_2 \geq c \}$. Since both $N_0$ and $\mathcal{W}$ are rotation-invariant, the projection of $N_0$ onto $\mathcal{W}$ must be rotation-invariant. Intuitively, this suggests that the projection should be a positive scalar multiple of

$f = \int_{\mathrm{O}(d)} u \cdot N_{\overline{x}} \, \mathrm{d} \sigma(u)$

where $\sigma$ is the Haar measure on the orthogonal group $\mathrm{O}(d)$ and $\overline{x} \in \mathbb{R}^d$ is any fixed point with $||\overline{x}||_2 = c$. Now,

$\langle N_0,f \rangle = \int_{\mathbb{R}^d} N_0(x) f(x) \, \mathrm{d}x = \int_{\mathbb{R}^d} N_0(x) \left( \int_{\mathrm{O}(d)} u \cdot N_{\overline{x}} \, \mathrm{d} \sigma(u) \right)(x) \, \mathrm{d}x \\ \int_{\mathrm{O}(d)} \int_{\mathbb{R}^d} N_0(x)( u \cdot N_{\overline{x}})(x)\, \mathrm{d}x \, \mathrm{d} \sigma(u) \\ = \int_{\mathrm{O}(d)} \int_{\mathbb{R}^d} (u \cdot N_0)(x) N_{\overline{x}}(x) \, \mathrm{d}x \,\mathrm{d}\sigma(u) \\ = \int_{\mathbb{R}^d} N_0(x) N_{\overline{x}}(x) \, \mathrm{d}x \leq \delta(c)$

where $\delta:[0,\infty) \to [0,\infty)$ is a function independent of $d$ with $\lim_{c \to \infty} \delta(c) = 0$.

Let $p$ be the projection of $N_0$ on $\mathcal{W}$ and let $\alpha > 0$ be such that $p = \alpha f$. We have $||p||^2 = \langle N_0,p \rangle = \langle N_0,\alpha f \rangle \leq \alpha \delta(c)$. Since $||p||_2 \leq 1$, we have $\alpha \leq 1/||f||_2$. Thus one would need to show that $||f||_2 > \beta$ for some absolute constant $\beta > 0$, as well as confirming the hypothesis about $f$.

  • $\begingroup$ Do you mean that $c$ is uniform in $d$? $\endgroup$ – Asaf May 14 '18 at 21:46
  • $\begingroup$ Yes, otherwise it's obvious that the case of indicators of balls has a positive answer. $\endgroup$ – burtonpeterj May 14 '18 at 22:01

Here is a suggestion for the one d case. First, look on the fourier transform side. Let $\phi_a$ etc. be the Gaussian shifted by a. On the the fourier transform side $\hat{\phi_a} \rightarrow e^{2 \pi a \xi} \phi_0$, and the question is can you approximate $\phi_0 $ by a linear combination of $e^{2 \pi n \xi} \phi_0$ where n is an integer and $|n| > c$ in $L^2$. This would entail that there is a linear combination so that $$ \int |\sum c_n e^{2 \pi n \xi} - 1 | ^ 2 \phi^2(\xi) d\xi $$ is small, where is sum is over $|n| > c$. But it cannot be small because $\phi_0^2 > m > 0$ on $(0, 2 \pi)$ so if it were small $$ \int_0^{2 \pi} |\sum c_n e^{2 \pi n \xi} - 1 | ^ 2 d \xi$$ would also be small, which it cannot be. This can be turned into a bound by considering what $m$ actually is, and getting the constants straight, which I have not done.

  • $\begingroup$ If we try to apply this to the $d$-dimensional case, we need to argue that $\int_{\sqrt{d} - 10 \leq |\xi| \leq \sqrt{d}+10} |1- \sum c_x e^{2 \pi i x \cdot \xi}|^2 \, \mathrm{d}\xi$ cannot be small if the sum is over $x$ with $|x|> c$. This seems very plausible ... $\endgroup$ – burtonpeterj May 14 '18 at 16:19
  • $\begingroup$ If I understand correctly, this bound doesn't get any better as c is increased, and therefore can't lead to a solution of the problem on its own. That being said, it is still interesting. $\endgroup$ – Tom Price Nov 11 '18 at 5:46

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