Distance from vectors in $\mathbb{Z}^d$ to the cube $[-1/2,1/2]^d$

Question

Let $\|\;\|_2$ be the Eucliden norm on $\mathbb{R}^d$.

Problem: Suppose $x\in Q:=\big[-\tfrac12,\tfrac12\big]^d$. Is $$ \|x-k\|_2\geq \frac{1}{2\sqrt{d}}\|k\|_2 $$ for all $k\in\mathbb{Z}^d$?

Notice that for all $x\in Q$, $\|x-k\|_2\geq d(k,Q):=\inf_{x\in Q}\|x-k\|_2$. So it is to enough to show that $$d(k,Q)\geq \frac{\|k\|_2}{2\sqrt{d}},\quad k\in\mathbb{Z}^d$$

---

I think this holds but I can't reproduce a proof at the moment. The relevance of this simple geometric result is that it provides a criteria for absolute and uniform convergence of the Poisson summation $$\sum_{k\in\mathbb{Z}^d}f(x+k)=\sum_{k\in\mathbb{Z}^d}\widehat{f}(k)e^{2\pi ik\cdot x}$$ in the $\mathbb{T}^d$ torus, where $f\in L_1(\mathbb{R}^d)$. If $f$ can be bounded poitwise by an integrable decreasing radial function, i.e. $|f(x)|\leq \phi_0(\|x\|_2)$, where $\phi_0$ is monotone non increasing and $\phi_0\circ\|\;\|_2\in L_1(\mathbb{R}^d)$

A proof or a good hint will be appreciated.

Answer 1

I think we can prove even sharper bound: for all $x\in Q=[-1/2,1/2]^d$ and $k\in\mathbb{Z}^d$ we have $$ \|x-k\|_{2}\ge\frac{1}{2}\|k\|_{2}. $$ Indeed, let $x=(x_1,\ldots,x_d)\in Q$ and $k=(k_1,\ldots,k_d)\in\mathbb{Z}^d$. We need to prove that $$ \|x-k\|_{2}\ge\frac{1}{2}\|k\|_{2}~\text{or}~\sum_{i=1}^{d}|x_i-k_i|^2\ge\frac{1}{4}\sum_{i=1}^{d}|k_i|^2. $$ Clearly, it's sufficient to prove that $|x_i-k_i|\ge\frac{1}{2}|k_i|$ for all $i\in\{1,\ldots,d\}$.

Hence, we need to prove that for all $x\in[-1/2,1/2]$ and $k\in\mathbb{Z}$ we have $$ |x-k|\ge\frac{1}{2}|k|. $$ If $k=0$, then inequality is trivial and if $k\neq 0$, then $|k|\ge 1$, so by triangle inequality $$ |x-k|\ge|k|-|x|\ge|k|-\frac{1}{2}\ge\frac{1}{2}|k|. $$ Thus, $|x-k|\ge\frac{1}{2}|k|$ for all $x\in[-1/2,1/2]$ and $k\in\mathbb{Z}$, as desired.

Remark. The constant $\frac{1}{2}$ is sharp: just consider $x=(1/2,\ldots,1/2)$ and $k=(1,\ldots,1)$.

Answer 2

After giving it more thought, I came up with the following solution below. I am still interested in accepting a more elegant solution which may make clever use of the law of cosines and/or the triangle inequality without resorting to induction.

---

My solution:

As stated, $Q=\big[-\tfrac12,\tfrac12\big]^d$. For any $\mathbf{k}=(k_1,\ldots,k_d)\in\mathbb{Z}^d$, define $$\Delta(\mathbf{k})=\{1\leq j\leq d: k_j\neq0\}$$ By compactness, there is $\mathbf{q}\in\partial Q$ such that $$d(\mathbf{q},\mathbf{k})=d(\mathbf{k},Q)$$ Since $0<\big|m-\operatorname{sign}(m)\frac12\big|\leq |m-t|$ for all integer $m\neq0$ and $|t|\leq\frac12$ $$ d(\mathbf{k},Q) =\sum_{j\in\Delta(\mathbf{k})}\big(|k_j|-\tfrac12\big)^2 $$ That is, $\mathbf{q}=(q_1,\ldots,1_d)$ where $q_j=\operatorname{sign}(k_j)\frac12$ when $j\in\Delta(\mathbf{k})$ and $q_j=0$ otherwise.

We proceed to show

$$\begin{align} d(\mathbf{k},Q)\geq\frac{\|\mathbf{k}\|_2}{2\sqrt{d}}\tag{1}\label{one} \end{align} $$ or equivalently $$\begin{align} 4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq\sum_{j\in\Delta(\mathbf{k})}k^2_j\tag{1'}\label{onep} \end{align}$$ by induction on the dimension $D$ of the space.

• For $D=1$, \eqref{onep} holds trivially for $k=0$, and also for $k\neq0$ since $2\big||k|-\tfrac12\big|=2|k|-1\geq|k|$ since $|k|\geq1$.

• Asume \eqref{onep} holds for $D=1,\ldots,d-1$, where $d-1\geq1$. Let $\mathbf{k}\in\mathbb{Z}^d$ and $\alpha_\mathbf{k}=\#\Delta(\mathbf{k})$. If $\alpha_\mathbf{k}<d$, then $$ \begin{align} 4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq4\alpha\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq \sum_{j\in\Delta(\mathbf{k})}k^2_j \end{align} $$ by the indiction hypothesis. If $\alpha=d$, then $|k_j|>0$ for all $1\leq j\leq d$ and so, \begin{align} 4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2 &=4d\sum^{d-1}_{j=1}(|k_j|-\tfrac12)^2 + 4d(|k_d|-\tfrac12)^2\\ &\geq 4(d-1)\sum^{d-1}_{j=1}(|k_j|-\tfrac12)^2 + 4(|k_d|-\tfrac12)^2\\ &\geq \sum^{d-1}_{j=1}k^2_j + k^2_d=\|\mathbf{k}\|^2_2 \end{align}

This completes the induction argument.

---