 Solution

Let $\delta =\hat{\beta }-{\beta }^0$. Comparing the Lasso objective at $\hat{\beta }$ and ${\beta }^0$ gives the Basic inequality for the Lasso. On $\mathcal{T}$ it impliesSincewe obtain the Lasso cone condition

The Karush-Kuhn-Tucker conditions also giveso on $\mathcal{T}$,The assumed cone invertibility condition therefore yieldsIf ${\min }_{j\in S}|{\beta }_j^0|>3\lambda /(2\psi )$, then every active coefficient remains nonzero and retains its sign. Substituting $\lambda =A\sqrt{\log p/n}$ proveson an event of the required probability. This is Lasso sign recovery from cone invertibility.