\begin{comment}
\begin{multline*}
    \nabla^2 \mathcal{L}(\theta) =
    \frac{1}{|\mathcal{T}|}\sum_{t\in \mathcal{T}}x^t(x^t)^T\bigg[
        x_i^{t+1}\frac{f''f - f'^2}{f^2}(\inprod{\theta_i}{x^t})\\
    - (1-x_i^{t+1})\frac{f''(1-f) + f'^2}{(1-f)^2}(\inprod{\theta_i}{x^t})\bigg]
\end{multline*}
\end{comment}


\subsection{Proposition~\ref{prop:irrepresentability}}
In the words and notation of Theorem 9.1 in \cite{vandegeer:2009}: 
\begin{lemma}
\label{lemm:irrepresentability_proof}
Let $\phi^2_{\text{compatible}}(L,S) \defeq \min \{ \frac{s \|f_\beta\|^2_2}{\|\beta_S\|^2_1} \ : \ \beta \in {\cal R}(L, S) \}$, where $\|f_\beta\|^2_2 \defeq \{ \beta^T \Sigma \beta \}$ and ${\cal R}(L,S) \defeq \{\beta : \|\beta_{S^c}\|_1 \leq L \|\beta_S\|_1 \neq 0\}$. If $\nu_{\text{irrepresentable}(S,s)} < 1/L$, then $\phi^2_{\text{compatible}}(L,S) \geq (1 - L \nu_{\text{irrepresentable}(S,s)})^2 \lambda_{\min}^2$. 
\end{lemma}

Since ${\cal R}(3, S) = {\cal C}$, $\|\beta_S\|_1 \geq \|\beta_S\|_2$, and $\|\beta_S\|_1 \geq \frac{1}{3} \|\beta_{S^c}\|_1$ it is easy to see that $\|\beta_S\|_1 \geq \frac{1}{4} \|\beta\|_2$ and therefore that: $\gamma_n \geq \frac{n}{4s}\phi^2_{\text{compatible}}(3,S)$

Consequently, if $\epsilon > \frac{2}{3}$, then $\nu_{\text{irrepresentable}(S,s)} < 1/3$ and the conditions of Lemma~\ref{lemm:irrepresentability_proof} hold.