Let $n$ be a natural number.
Let $U \subseteq \mathbb{F}\_2^n$ be a linear subspace of dimension $k$.
What is the maximum number of vectors in $U$ of Hamming weight $\ell$?


The case I am specifically interested in is $k \approx 0.99n$, $\ell = n/2$, where $n$ is a large enough even number. Is this number at most $C \cdot 2^k \cdot \frac{\binom{n}{n/2}}{2^n}$ for some constant $C$, independent of $n$?



Number of binary vectors of a given Hamming weight in a subspace of the Hypercube
I just found out that the question (at least with the set of parameters written above) was asked by Ben-Or and answered by Linial and Samordinsky.
Their paper proves that for any constant $r>1/2$ a subspace of dimension $k = rn$ has at most $C\_r \cdot 2^k \cdot \frac{\binom{n}{n/2}}{2^n}$ vectors of weight $n/2$.
<http://www.cs.huji.ac.il/~nati/PAPERS/lin_codes.pdf>


