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Use the normalizationfor the Lasso. A centered sub-Gaussian random variable with parameter $\sigma$ satisfies $\mathbb{E}{e}^{tW}\le e^{{\sigma }^2{t}^2/2}$ for all $t\in \mathbb{R}$; for a noncentered variable apply this definition to $W-\mathbb{E}W$. With fixed $X$, independence of the errors givesThe exponential Markov bound, optimized over $t$ separately for the two signs, gives $\mathrm{Pr}(|X_j^T\epsilon |/n>u)\le 2e^{-n{u}^2/(2{\sigma }^2)}$. Take $u=\lambda /2=\sigma A\sqrt{\log p/n}$ and apply the union bound over columns:The specified positive tuning parameter requires $p\ge 2$; at $p=1$ its formula is zero. For small $A$ the probability lower bound can be negative and hence uninformative. Although centering makes the centered errors dependent, $X^{T}1=0$ means $X^T(\epsilon -\bar{\epsilon }1)=X^T\epsilon$, so the preceding proof uses the original independent errors, not independence after centering.

The Karush-Kuhn-Tucker conditions, using the subdifferential of the L1 norm, areConsequently, for nonempty $B\subseteq \hat{S}$, on $\Omega$,This establishes the required active-set inequality without any rank condition on $X$.

For $s>0$, a sufficient Compatibility condition for the Lasso isIndeed, for $\delta =\hat{\beta }-{\beta }^0$, comparison of the two Lasso objective values and the score bound on $\Omega$ give the Basic inequality for the LassoIt follows that $\delta$ satisfies the Lasso cone condition and $\Vert {\delta }_S{\Vert }_1\le \sqrt{s{P}_E}/\varphi$. Thus $P_E\le 3\lambda \sqrt{s{P}_E}/\varphi$, including the trivial $P_E=0$ case. This proves the slightly stronger $P_E\le 9{\lambda }^{2}s/{\varphi }^2$, and in particular the requested $16{\lambda }^{2}s/{\varphi }^2$. If $s=0$, the same basic inequality forces $\hat{\beta }=0$ on $\Omega$; no nonempty-support compatibility constant is needed.

For the support-size argument, use the stated prediction bound with constant $16$. Set $b=|B|$. By the Cauchy-Schwarz inequality and the definition of the sparse maximum eigenvalue,Combining with the lower bound and squaring gives $\boxed{b\le 64{\kappa }_b^{2}s/{\varphi }^2}$ for every nonempty $B\subseteq \hat{S}$. If $m_{\ast }<\infty$ and $\hat{s}\ge m_{\ast }$, choose $B$ of size $m_{\ast }$; this contradicts the strict inequality defining $m_{\ast }$. If $m_{\ast }=\infty$, $\hat{s}\le p<\infty$ proves the assertion directly. Hence $\boxed{\hat{s}<m_{\ast }}$ in both cases.

When $1<m_{\ast }\le p$, minimality and monotonicity of the sparse maximum eigenvalues giveWhen $m_{\ast }=1$, the first result gives $\hat{s}=0$, so the same final bound holds. Thus the finite-index conclusion is $\boxed{\hat{s}\le 64{\kappa }_{m_{\ast }}^{2}s/{\varphi }^2(m_{\ast }<\infty )}$.

There is a genuine domain omission in the printed last assertion: the definition of ${\kappa }_m$ only makes sense for $1\le m\le p$, so ${\kappa }_{m_{\ast }}$ is undefined when $m_{\ast }=\infty$. This case can occur: take centered orthogonal columns with $X^{T}X/n=I_p$, $n\ge p+1$, $s=p$, and ${\varphi }^2=1$. Then ${\kappa }_m^2=1$ for every available $m$ and none exceeds $64p$, so $m_{\ast }=\infty$. A universally defined replacement, from $B=\hat{S}$ and monotonicity, isAlternatively, explicitly extend the definition by ${\kappa }_m={\kappa }_p$ for $m>p$, including infinity. Under that added convention the printed final expression is meaningful and follows from this replacement bound. Neither convention nor finiteness should be silently assumed.

For a convex function $f$, its subdifferential at $x$ is the set of supporting slopesThe subgradient optimality condition isIndeed, the defining inequality with $g=0$ is exactly the global minimum inequality. For the subdifferential of the L1 norm, the coordinates vary independently:

Let $r=Y-X{\hat{\beta }}_{\lambda }$. The Karush-Kuhn-Tucker conditions for Lasso give a vector $z\in \partial \Vert {\hat{\beta }}_{\lambda }{\Vert }_1$ such that $X^{T}r/n=\lambda z$. If ${\hat{\sigma }}_{\lambda }=\Vert r{\Vert }_2/\sqrt{n}>0$, the unsquared residual norm is differentiable at this fit, with gradientUsing the subdifferential sum rule, choose $\gamma (\lambda )=\lambda /{\hat{\sigma }}_{\lambda }$ to obtainConvexity now proves the tuning relation between Lasso and square-root Lasso:The positive residual assumption is needed both for this gradient and for the quotient defining the tuning parameter.

For the final score calculation, use the usual fixed-design interpretation of the normal linear model: regard $X,Z$ and the tuning parameter $\gamma$ in the regression of $Z$ as fixed, and choose that regression's minimizer using only those inputs. Its nonzero residual $R=R_{\gamma }$ then has the Square-root Lasso optimality conditionExpanding the response residual givesSince $R/\Vert R{\Vert }_2$ is a fixed unit vector and $\epsilon \sim N_n(0,{\sigma }^2{I}_n)$, a linear image of a multivariate normal vector gives $W\sim N(0,{\sigma }^2)$. Finally, Holder inequality and the residual score bound yieldThis is the residual-score decomposition from square-root Lasso. The normality statement also holds conditionally on a design independent of the noise. If $\gamma$ were chosen from the response noise, for example through the preceding $\gamma (\lambda )$ formula, the direction $R$ would require a separate independence argument; the final calculation treats $\gamma$ as fixed.