Consider the sparse linear regression model $y=X\theta^\*+w$, where $w\sim N(0, \sigma^2 I\_{n\times n})$ and $\theta^\*\in R^d$ is supported on a subset $S$. Suppose that the sample covariance matrix $\hat{\Sigma}=X^TX/n$ has its diagonal entries uniformly upper bounded by 1, and that for some parameter $\gamma>0$, it also satisfies an $\ell\_{\infty}$-curvature condition of the from
$$\|\hat{\Sigma}\Delta\|\_\infty\ge \gamma\|\Delta\|\_\infty, \, \Delta\in C\_3(S)
$$
where $C\_3(S):=\{\Delta\in R^d: \|\Delta\_{S^c}\|\_1\le 3 \|\Delta\_{S}\|\_1\}$



Lasso of sparse linear regression model
The proof you are looking for should be in



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> Theorem 1 of *Sup-norm convergence rate and sign concentration property of Lasso and Dantzig estimators* by Karim Lounici (<https://arxiv.org/abs/0801.4610>).
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Check the relationship of your curvature condition with Assumption 2 of that paper. The proof there should give you the main ideas, in particular the standard argument to prove that $\hat \theta-\theta^\*\in C\_3(S)$.


This proof fails for random design matrices when $|S|>>\sqrt n$ because the curvature condition in $\ell\_\infty$ norm cannot hold. It is possible to overcome this difficult for Gaussian random design matrices, see



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> Theorem 5.1 in *De-Biasing The Lasso With Degrees-of-Freedom Adjustment*
> by Bellec and Zhang (<https://arxiv.org/abs/1902.08885>)
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