[proofplan] The Dantzig selector constraint measures the maximum absolute component of the empirical score vector $X^\top(Y-X\beta)/n$. To prove feasibility of the true parameter, we substitute $\beta=\beta^*$ into this constraint. The model equation $Y=X\beta^*+\varepsilon$ turns the residual $Y-X\beta^*$ into exactly $\varepsilon$, so the feasibility condition becomes precisely the assumed score bound. [/proofplan] [step:Evaluate the Dantzig residual at the true parameter] Define the residual map \begin{align*} r: \mathbb{R}^p \to \mathbb{R}^n, \qquad r(\beta) := Y - X\beta. \end{align*} Evaluating this map at $\beta^*$ and using the model equation gives \begin{align*} r(\beta^*) = Y - X\beta^* = (X\beta^* + \varepsilon) - X\beta^* = \varepsilon. \end{align*} [guided] The Dantzig feasibility condition is written in terms of the residual vector $Y-X\beta$. Therefore the first task is to compute that residual for the specific candidate $\beta=\beta^*$. Define \begin{align*} r: \mathbb{R}^p \to \mathbb{R}^n, \qquad r(\beta) := Y - X\beta. \end{align*} This is a well-defined map because $X \in \mathbb{R}^{n \times p}$ and $\beta \in \mathbb{R}^p$, so $X\beta \in \mathbb{R}^n$, while $Y \in \mathbb{R}^n$. Now substitute the true parameter $\beta^*$. Since the model equation states that $Y=X\beta^*+\varepsilon$, we obtain \begin{align*} r(\beta^*) = Y - X\beta^* = (X\beta^* + \varepsilon) - X\beta^* = \varepsilon. \end{align*} Thus, at the true parameter, the residual is exactly the noise vector. [/guided] [/step] [step:Substitute the residual into the Dantzig constraint] By the previous step, \begin{align*} \frac{X^\top(Y-X\beta^*)}{n} = \frac{X^\top \varepsilon}{n}. \end{align*} Taking the $\ell^\infty$ norm on $\mathbb{R}^p$ gives \begin{align*} \left\| \frac{X^\top(Y-X\beta^*)}{n} \right\|_\infty = \left\| \frac{X^\top \varepsilon}{n} \right\|_\infty. \end{align*} By the assumed score bound, the right-hand side is at most $\lambda$. Therefore \begin{align*} \left\| \frac{X^\top(Y-X\beta^*)}{n} \right\|_\infty \leq \lambda. \end{align*} [/step] [step:Conclude that the true parameter belongs to the feasible set] The defining condition for membership in $\mathcal{F}_\lambda$ is \begin{align*} \left\| \frac{X^\top(Y-X\beta)}{n} \right\|_\infty \leq \lambda. \end{align*} The previous step proves this condition for $\beta=\beta^*$. Hence $\beta^* \in \mathcal{F}_\lambda$, as required. [/step]