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

A: 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 $k=(1,\ldots,1)$.
