Gaussians at lattice points

Question

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 $f$.

Answer 1

For the question as it is currently stated, we only need to consider finitely many dimensions, since $||N_0||_2$ goes to $0$ as $d$ goes to infinity, and the inequality trivially holds when $||N_0||_2 < \epsilon$.

However, I’ve discussed this with the author of the post privately over email and he intended for $N_0$ to be an $L^2$-normalized gaussian so that $||N_0||_2 = 1$. I'll use a different but equivalent formulation: we can keep $N_0$ as the usual d-dimensional standard gaussian (whose $L^2$ norm will be less than $1$) and require that the norm of the projection is less than $\epsilon||N_0||_2$ rather than less than $\epsilon$.

For this modified problem, no such $c$ exists. For the ball-indicator version, I’ve made no progress.

I’ll use the following notation for my solution:

$N_{d,x}$ : the $d$-dimensional standard Gaussian centered at $x$.

$M_d$ : $d$-dimensional Gaussian with double the covariance as standard, centered at $0$. In other words, the convolutional square of $N_{d, 0}$. Note that $\langle N_{d, x}, N_{d, y} \rangle = M_d(x - y)$

$\mathcal{V}_{d, c}$: The same as $\mathcal{V}$ as defined in the question, but with a subscript to indicate the dimension and the value of $c$ used.

The norm of the projection of $N_{d,0}$ onto $\mathcal{V}_{d, c}$ is equal to the sup of the norm of the projection of $N_{d,0}$ onto each one-dimensional subspace of $\mathcal{V}_{d, c}$. Applying this fact, and making a substitution for $\epsilon$, we have that the problem is equivalent to the following:

Does there exist a $c$ such that $$\frac{\langle v, N_{d,0} \rangle ^2}{\langle v,v\rangle \langle N_{d, 0}, N_{d, 0} \rangle} < \epsilon$$ for all $v \in \mathcal{V}_{c, d}$, uniformly for all $d$?

Let $S_{d, n}$ be the set of all $x \in \mathbb{Z}^d$ with a $1$ in exactly $n$ coordinates and a $0$ in the remaining $d - n$ coordinates. Let $f_{d, n} = \sum_{x \in S_{d ,n}}N_{d, x}$.

We can explicitly compute the values $\langle f_{d,n} , N_{d,0} \rangle$ and $\langle f_{d,n} , f_{d,n} \rangle$. For this, let $G_d(r)$ be the value of $M_d$ at a point of norm $r$, so $M_d(x)$ = $G_d(||x||_2)$. Then we have $$\langle f_{d,n} , N_{d,0} \rangle = \binom{d}{n}G_d(\sqrt{n})$$ $$\langle f_{d,n} , f_{d,n} \rangle=\binom{d}{n} \sum_{m = 0}^{n} \binom{n}{m} \binom{d - n}{n - m} G_d(\sqrt{2(n - m)})$$

Here $m$ basically parameterizes the size of the overlap of nonzero coordinates between a pair of the terms comprising $f_{d,n}$.

If we fix $n$, and let $d$ go to infinity, the term with $m = 0$ dominates and we have: $$\frac{\langle f_{d,n} , f_{d,n} \rangle}{\binom{d}{n}^2 G_d(\sqrt{2n})} = 1 + O(\frac{1}{d})$$

We therefore have, from the computations above, and the fact that $\langle N_{d, 0}, N_{d, 0} \rangle = G_d(0)$:

$$\lim_{d \to \infty}{\frac{\langle f_{d, n}, N_{d,0} \rangle ^2}{\langle f_{d, n}, f_{d, n} \rangle \langle N_{d, 0}, N_{d, 0} \rangle}}$$ $$= \lim_{d \to \infty}{\frac{G_d(\sqrt{n})^2}{G_d(\sqrt{2n})G_d(0)}}$$ $$= 1$$

So no matter how large $c$ is, we can always choose a sequence of functions $f_{d,n}$ with a fixed $n > c^2$ which eventually breaks the inequality, as long as $\epsilon < 1$.

Answer 2

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.