Proof

In order to prove a lower bound for the distance of a code ${\displaystyle C^{*}}$ we prove that the Hamming distance of an arbitrary but distinct pair of codewords has a lower bound. So let ${\displaystyle \Delta (c^{1},c^{2})}$ be the Hamming distance of two codewords ${\displaystyle c^{1}}$ and ${\displaystyle c^{2}}$. For any given

${\displaystyle m_{1}\neq m_{2}\in \left(\mathbb {F} _{q^{k}}\right)^{K},}$

we want a lower bound for ${\displaystyle \Delta (C^{*}(m_{1}),C^{*}(m_{2})).}$

Notice that if ${\displaystyle C_{out}(m)=(c_{1},\cdots ,c_{N})}$, then ${\displaystyle C^{*}(m)=(C_{in}^{1}(c_{1}),\cdots ,C_{in}^{N}(c_{N}))}$. So for the lower bound ${\displaystyle \Delta (C^{*}(m_{1}),C^{*}(m_{2}))}$, we need to take into account the distance of ${\displaystyle C_{in}^{1},\cdots ,C_{in}^{N}.}$

Suppose

${\displaystyle {\begin{aligned}C_{out}(m_{1})&=\left(c_{1}^{1},\cdots ,c_{N}^{1}\right)\\C_{out}(m_{2})&=\left(c_{1}^{2},\cdots ,c_{N}^{2}\right)\end{aligned}}}$

Recall that ${\displaystyle \left\{C_{in}^{1},\cdots ,C_{in}^{N}\right\}}$ is a Wozencraft ensemble. Due to "Wonzencraft ensemble theorem", there are at least ${\displaystyle (1-\varepsilon )N}$ linear codes ${\displaystyle C_{in}^{i}}$ that have distance ${\displaystyle H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k.}$ So if for some ${\displaystyle 1\leqslant i\leqslant N,c_{i}^{1}\neq c_{i}^{2}}$ and the code ${\displaystyle C_{in}^{i}}$ has distance ${\displaystyle \geqslant H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k,}$ then

${\displaystyle \Delta \left(C_{in}^{i}\left(c_{i}^{1}\right),C_{in}^{i}\left(c_{i}^{2}\right)\right)\geqslant H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k.}$

Further, if we have ${\displaystyle T}$ numbers ${\displaystyle 1\leqslant i\leqslant N}$ such that ${\displaystyle c_{i}^{1}\neq c_{i}^{2}}$ and the code ${\displaystyle C_{in}^{i}}$ has distance ${\displaystyle \geqslant H_{q}^{-1}({\tfrac {1}{2}}-\varepsilon )\cdot 2k,}$ then

${\displaystyle \Delta \left(C^{*}(m_{1}),C^{*}(m_{2})\right)\geqslant H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k\cdot T.}$

So now the final task is to find a lower bound for ${\displaystyle T}$. Define:

${\displaystyle S=\left\{i\$ :\ 1\leqslant i\leqslant N,c_{i}^{1}\neq c_{i}^{2}\right\}.}

Then ${\displaystyle T}$ is the number of linear codes ${\displaystyle C_{in}^{i},i\in S}$ having the distance ${\displaystyle H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k.}$

Now we want to estimate ${\displaystyle |S|.}$ Obviously ${\displaystyle |S|=\Delta (C_{out}(m_{1}),C_{out}(m_{2}))\geqslant (1-R)N}$.

Due to the Wozencraft Ensemble Theorem, there are at most ${\displaystyle \varepsilon N}$ linear codes having distance less than ${\displaystyle H_{q}^{-1}({\tfrac {1}{2}}-\varepsilon )\cdot 2k,}$ so

${\displaystyle T\geqslant |S|-\varepsilon N\geqslant (1-R)N-\varepsilon N=(1-R-\varepsilon )N.}$

Finally, we have

${\displaystyle \Delta (C^{*}(m_{1}),C^{*}(m_{2}))\geqslant H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k\cdot T\geqslant H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right)\cdot 2k\cdot (1-R-\varepsilon )\cdot N.}$

This is true for any arbitrary ${\displaystyle m_{1}\neq m_{2}}$. So ${\displaystyle C^{*}}$ has the relative distance at least ${\displaystyle (1-R-\varepsilon )H_{q}^{-1}\left({\tfrac {1}{2}}-\varepsilon \right),}$ which completes the proof.