 Solution

Let $\delta =\hat{\beta }-{\beta }^0$. Standard sub-Gaussian random variable and Gaussian-product concentration gives, for $n>\log p$,Thus $A=4$ makes this probability at most $2/p$. On the complementary event, the basic inequality and Holder inequality giveThe parenthesis is at most $2\Vert \delta {\Vert }_1$ by the triangle inequality and at most $2\Vert {\beta }^0{\Vert }_1$ by $\Vert \delta {\Vert }_1\le \Vert \hat{\beta }{\Vert }_1+\Vert {\beta }^0{\Vert }_1$. Substituting $\lambda =A\sigma v\sqrt{\log p/n}$ proves ${\Omega }_1$, whose probability tends to one.