[proofplan] We prove the result by working on the high-probability event on which the supremum norm error is at most $r_n$. On that event, inactive coordinates cannot exceed the threshold because their true coefficients are zero, while active coordinates must exceed the threshold because the beta-min condition leaves a margin larger than $r_n$. The same margin also prevents any active coefficient from crossing zero, which gives sign recovery. Finally, the probability lower bound follows because the support and sign recovery event contains the supremum norm control event. [/proofplan]

[step:Define the supremum norm control event and work pointwise on it] Define the event $E\in\mathcal F$ by \begin{align*} E:=\{\omega\in\Omega:\|\hat{\beta}(\omega)-\beta^*\|_\infty\le r_n\}. \end{align*} By hypothesis, \begin{align*} \mathbb P(E)\ge 1-\delta_n. \end{align*} Fix $\omega\in E$. By the definition of the supremum norm on $\mathbb R^p$, for every $j\in\{1,\dots,p\}$, \begin{align*} |\hat{\beta}_j(\omega)-\beta^*_j|\le r_n. \end{align*} It is enough to prove that $\hat S_\tau(\omega)=S$ and that the signs agree on $S$, because then $E\subseteq\{\hat S_\tau=S\}$ and $E$ is also contained in the sign recovery event. [/step]

[step:Exclude every inactive coordinate from the thresholded support]Let $j\in\{1,\dots,p\}\setminus S$. By the definition of $S$, $\beta^*_j=0$. Since $\omega\in E$, \begin{align*} |\hat{\beta}_j(\omega)| = |\hat{\beta}_j(\omega)-\beta^*_j| \le r_n \le \tau. \end{align*} The thresholded support $\hat S_\tau(\omega)$ contains exactly those indices $k$ for which $|\hat{\beta}_k(\omega)|>\tau$. Therefore $j\notin\hat S_\tau(\omega)$. Since $j\notin S$ was arbitrary, \begin{align*} \hat S_\tau(\omega)\subseteq S. \end{align*}[/step]

[guided]We first show that thresholding creates no false positives. Take an index $j\notin S$. The definition of the support $S$ says exactly that the true coefficient at this coordinate is zero: \begin{align*} \beta^*_j=0. \end{align*} Because $\omega$ belongs to the event $E$, the coordinatewise error is bounded by $r_n$: \begin{align*} |\hat{\beta}_j(\omega)-\beta^*_j|\le r_n. \end{align*} Substituting $\beta^*_j=0$ gives \begin{align*} |\hat{\beta}_j(\omega)| = |\hat{\beta}_j(\omega)-0| = |\hat{\beta}_j(\omega)-\beta^*_j| \le r_n. \end{align*} The threshold assumption includes $r_n\le\tau$, so \begin{align*} |\hat{\beta}_j(\omega)|\le\tau. \end{align*} But $\hat S_\tau(\omega)$ was defined using the strict inequality $|\hat{\beta}_k(\omega)|>\tau$. Hence an index with $|\hat{\beta}_j(\omega)|\le\tau$ is not selected. Thus every inactive coordinate is excluded, and we have proved \begin{align*} \hat S_\tau(\omega)\subseteq S. \end{align*}[/guided]

[step:Keep every active coordinate after thresholding] Let $j\in S$. By the [reverse triangle inequality](/theorems/2300) and the coordinatewise bound on $E$, \begin{align*} |\hat{\beta}_j(\omega)| \ge |\beta^*_j|-|\hat{\beta}_j(\omega)-\beta^*_j| \ge |\beta^*_j|-r_n. \end{align*} Since \begin{align*} \tau<\min_{k\in S}|\beta^*_k|-r_n\le |\beta^*_j|-r_n, \end{align*} we obtain \begin{align*} |\hat{\beta}_j(\omega)|>\tau. \end{align*} Therefore $j\in\hat S_\tau(\omega)$. Since $j\in S$ was arbitrary, \begin{align*} S\subseteq\hat S_\tau(\omega). \end{align*} Combining this inclusion with $\hat S_\tau(\omega)\subseteq S$ gives \begin{align*} \hat S_\tau(\omega)=S. \end{align*} [/step]

[step:Use the same margin to preserve the signs on the active coordinates] Let $j\in S$. The beta-min condition gives \begin{align*} |\beta^*_j|\ge \min_{k\in S}|\beta^*_k|>2r_n, \end{align*} and therefore $r_n<|\beta^*_j|$. Since $\omega\in E$, \begin{align*} |\hat{\beta}_j(\omega)-\beta^*_j|\le r_n<|\beta^*_j|. \end{align*} If $\beta^*_j>0$, then \begin{align*} \hat{\beta}_j(\omega)\ge \beta^*_j-r_n>0. \end{align*} If $\beta^*_j<0$, then \begin{align*} \hat{\beta}_j(\omega)\le \beta^*_j+r_n<0. \end{align*} Thus in both cases \begin{align*} \operatorname{sgn}(\hat{\beta}_j(\omega))=\operatorname{sgn}(\beta^*_j). \end{align*} Since $j\in S$ was arbitrary, signs are recovered on all active coordinates. [/step]

[step:Take probabilities to obtain the recovery guarantee] The preceding steps prove that every $\omega\in E$ satisfies both $\hat S_\tau(\omega)=S$ and sign agreement on $S$. Hence \begin{align*} E\subseteq\{\omega\in\Omega:\hat S_\tau(\omega)=S\}. \end{align*} Taking probabilities and using the assumed lower bound for $\mathbb P(E)$ gives \begin{align*} \mathbb P(\hat S_\tau=S) \ge \mathbb P(E) \ge 1-\delta_n. \end{align*} The same containment with the sign agreement event gives sign recovery with probability at least $1-\delta_n$. [/step]