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I am trying to understand an argument used and not proved in a paper, which is:

To see if a certain vector $v$ is close to the lattice, we check $\langle v , w \rangle$ for a short vector $w$ in the dual lattice and see if it is close to an integer. Writing $v=x+e$ where $x$ is in the lattice and $e$ is the perturbation vector, this method is not effective when $||e||$ is much bigger than $1/||w||$.

In other words, we can state the argument as:

Let $L \subseteq \mathbb{R}^n$ be a lattice and $v \in \mathbb{R}^n$.

Let $w$ be a short vector of $L^*$ (the dual lattice).

Then, $\langle v , w \rangle$ is close to an integer if and only if $v$ is close to $L$.

Let's try to prove this.

Recall the definition of $L^*$, $$L^* = \{z \in \text{span}\{b_1,\ldots,b_n\} : \langle x,z \rangle \in \mathbb{Z} \text{ for all } x \in L\}$$ where $b_1,\ldots,b_n$ is a basis of $L$, and the fact that $(L^*)^*=L$.

What I got so far:

$v$ is close to $L$ if and only if $v=x+e$ for $x \in L$ and $e$ small perturbation vector.

Let $w \in L^*$ be a small vector of $L^*$.

$\langle v,w \rangle = \langle x+e,w \rangle = \underbrace{\langle x,w \rangle}_{integer} + \langle e,w \rangle$

I don't see how to proceed from here, I need some help here.

Thanks in advance.

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So $\def\<#1>{\left<#1\right>}\$ is close to an integer, namely the integer $\$, if $\$ is small. Now we have by Cauchy-Schwarz $$ |\| \le \|e\|\|w\| $$ If now $w$ is a small vector from $L^*$, this is small for small $e$, hence $\$ is close to an integer iff $e$ is small, hence $v$ close to $L$.

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    Thanks for answering; I think that this proves the implication $v$ close to $L$ $\Rightarrow$ $\langle v,w \rangle$ close to an integer. The other implication is not very clear, and I think that we need the extra hypothesis that $||e||$ is sufficiently smaller than $1/||w||$, right?2017-01-08