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

Mathematics Asked by Oliver Diaz on December 2, 2020

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

Problem: Suppose $$xin Q:=big[-tfrac12,tfrac12big]^d$$. Is
$$|x-k|_2geq frac{1}{2sqrt{d}}|k|_2$$
for all $$kinmathbb{Z}^d$$?

Notice that for all $$xin Q$$, $$|x-k|_2geq d(k,Q):=inf_{xin Q}|x-k|_2$$. So it is to enough to show that
$$d(k,Q)geq frac{|k|_2}{2sqrt{d}},quad kinmathbb{Z}^d$$

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
$$sum_{kinmathbb{Z}^d}f(x+k)=sum_{kinmathbb{Z}^d}widehat{f}(k)e^{2pi ikcdot x}$$
in the $$mathbb{T}^d$$ torus, where $$fin L_1(mathbb{R}^d)$$. 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 $$phi_0circ|;|_2in L_1(mathbb{R}^d)$$

A proof or a good hint will be appreciated.

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, $$Q=big[-tfrac12,tfrac12big]^d$$. For any $$mathbf{k}=(k_1,ldots,k_d)inmathbb{Z}^d$$, define $$Delta(mathbf{k})={1leq jleq d: k_jneq0}$$ By compactness, there is $$mathbf{q}inpartial Q$$ such that $$d(mathbf{q},mathbf{k})=d(mathbf{k},Q)$$ Since $$0 for all integer $$mneq0$$ and $$|t|leqfrac12$$ $$d(mathbf{k},Q) =sum_{jinDelta(mathbf{k})}big(|k_j|-tfrac12big)^2$$ That is, $$mathbf{q}=(q_1,ldots,1_d)$$ where $$q_j=operatorname{sign}(k_j)frac12$$ when $$jinDelta(mathbf{k})$$ and $$q_j=0$$ otherwise.

We proceed to show

begin{align} d(mathbf{k},Q)geqfrac{|mathbf{k}|_2}{2sqrt{d}}tag{1}label{one} end{align} or equivalently begin{align} 4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geqsum_{jinDelta(mathbf{k})}k^2_jtag{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 $$kneq0$$ since $$2big||k|-tfrac12big|=2|k|-1geq|k|$$ since $$|k|geq1$$.

• Asume eqref{onep} holds for $$D=1,ldots,d-1$$, where $$d-1geq1$$. Let $$mathbf{k}inmathbb{Z}^d$$ and $$alpha_mathbf{k}=#Delta(mathbf{k})$$. If $$alpha_mathbf{k}, then begin{align} 4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geq4alphasum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geq sum_{jinDelta(mathbf{k})}k^2_j end{align} by the indiction hypothesis. If $$alpha=d$$, then $$|k_j|>0$$ for all $$1leq jleq d$$ and so, begin{align} 4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2 &=4dsum^{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.

Correct answer by Oliver Diaz on December 2, 2020

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