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 $f$.



Gaussians at lattice points
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 $\epsilon < 1$.


