 Solution

Part (d) makes the residual and therefore $\hat{\nu }=X^T(Y-X\hat{\beta })$ unique, so $E$ is unique. The Karush-Kuhn-Tucker conditions from part (c) imply that every nonzero block satisfies $\Vert {\hat{\nu }}^{(k)}{\Vert }_2=n\lambda \sqrt{m}$. Hence ${\hat{\beta }}^{(k)}=0$ for $k\notin E$.

Any two minimizers have the same fitted value and vanish outside $E$. Their difference $h$ is therefore supported on $E$ and satisfies $\tilde{X}{h}_E=0$. If $\tilde{X}$ has full column rank, then $h_E=0$, proving uniqueness of $\hat{\beta }$.