McMillan inequality for finite binary uniquely decodable codes

Put $K:={\sum }_{i=1}^m{2}^{-{\mathrm{\ell }}_i}$ and $\lambda :={\max }_i{\mathrm{\ell }}_i$. For each integer $k\ge 1$, expand $K^k={\sum }_{(i_1,\dots ,i_k)\in [m{]}^k}{2}^{-({\mathrm{\ell }}_{i_1}+\cdots +{\mathrm{\ell }}_{i_k})}$. For each total length $s$, let $N_{k,s}$ count the tuples with ${\mathrm{\ell }}_{i_1}+\cdots +{\mathrm{\ell }}_{i_k}=s$. Their concatenated codewords are distinct by [F1] and each is a binary word of length $s$, of which there are exactly $2^s$ because each of the $s$ positions has two choices; hence $N_{k,s}\le 2^s$.

Since every ${\mathrm{\ell }}_i\ge 1$, the only possible totals satisfy $k\le s\le k\lambda$, so there are $k(\lambda -1)+1$ possible integer values of $s$. Grouping step 1.1 by $s$ gives $K^k={\sum }_{s=k}^{k\lambda }{N}_{k,s}{2}^{-s}\le k(\lambda -1)+1$. Therefore $K\le (k(\lambda -1)+1{)}^{1\mathrm{/}k}$ for every $k$.

Suppose $K>1$, set $ϵ=K-1>0$, and put $a=\lambda -1\ge 0$. By [F2], the integer $k:=\lfloor 1+2a\mathrm{/}{ϵ}^2\rfloor +1$ satisfies $k>1+2a\mathrm{/}{ϵ}^2$ and $k\ge 2$, so $k(k-1){ϵ}^2\mathrm{/}2>ak$. The binomial theorem gives $K^k=(1+ϵ{)}^k\ge 1+kϵ+\left(\genfrac{}{}{0px}{}{k}{2}\right){ϵ}^2>1+ak$, contradicting step 2.1, which gives $K^k\le ak+1$. Hence $K\le 1$.