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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.

One Answer

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 $0<big|m-operatorname{sign}(m)frac12big|leq |m-t|$ 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 $alpha_mathbf{k}<d$, 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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