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A nonzero $u$ satisfying this relation, with $Ku\ne 0$, generalizes the eigenfunction of an absolutely one-homogeneous functional to the quadratic geometry induced by the forward map. Pairing with $u$ yields $\lambda =J(u)/\Vert Ku{\Vert }^2$. With unit data norm it is a generalized singular vector. If $\lambda >0$, $0\le \alpha \lambda \le 1$ and the exact data are $Ku$, the vector $(1-\alpha \lambda )u$ minimizes the quadratic variational regularization objective. Substitute its raywise subgradient into the subgradient optimality condition. Every minimizing branch has the same predicted data, and injectivity of $K$ gives equality of vectors. The Bregman distance along this ray is zero even though the reconstruction is shrunk. For nonnegative $J$ the branch continues as $(1-\alpha \lambda {)}_{+}u$; nonnegativity supplies $0\in \partial J(0)$ for the threshold argument.

Background on these generalized singular vectors: Benning and Burger, Ground States and Singular Vectors of Convex Variational Regularization Methods.

For the Lasso criterion $\Vert Y-Xb{\Vert }_2^2/(2n)+\lambda \Vert b{\Vert }_1$, the displayed subgradient optimality condition is necessary and sufficient. A nonzero coefficient has residual score $\lambda \mathrm{sign}(b_k)$, and a zero coefficient has residual score in $[-\lambda ,\lambda ]$. The subdifferential of the L1 norm supplies both cases.

This stronger source condition in variational regularization requires a subgradient $w=K^{\ast }Kv$ at $u^{†}$. For each $\alpha >0$ it is equivalent to making $u^{†}$ minimize $\Vert Ku-K\bar{u}{\Vert }^2/2+\alpha J(u)$ for artificial input $\bar{u}=u^{†}+\alpha v$. The subgradient optimality condition proves both implications. The source vector may be zero: $K=I$ and $J\equiv 0$ satisfy the condition but admit no nonzero source vector.

The forward subgradient step and backward subgradient step are the set-valued mapsThus $u\in B_{\tau f}(z)$ means $z-u\in \tau \partial f(u)$. They are respectively the explicit and implicit time discretizations of subgradient flow $\dot{x}\in -\partial f(x)$. At a differentiable point the forward update is $x-\tau \nabla f(x)$, while the backward update evaluates the gradient at the new point. For proper closed convex data the latter is the proximal operator

To prove at most one output, suppose $u,v\in B_{\tau f}(z)$. Then $(z-u)/\tau \in \partial f(u)$ and $(z-v)/\tau \in \partial f(v)$. The two subgradient inequalities imply monotonicity of a convex subdifferential, givingSince $\tau >0$, $u=v$. Convexity supplies this monotonicity; lower semicontinuity is not needed for this at-most-one argument.

Under the full printed assumptions the output actually exists at every $z$. Properness ensures a finite point and excludes $-\infty$. Proper lower-semicontinuous convex $f$ has an affine minorant, so the quadratic proximal objective is coercive. Lower semicontinuity makes its minimum attained on a compact sublevel set. Convexity and the positive quadratic curvature make it strongly convex, hence its minimizer is unique. The subgradient optimality condition, with the differentiable quadratic term, is exactly $z-u\in \tau \partial f(u)$. ThusThis also identifies precisely which assumptions provide existence, uniqueness and the update's optimality interpretation.

With the centred data, no intercept is needed. Use the Lasso normalizationWrite $r=Y-Xb$ and $c=X^{\mathsf{T}}r/n$. The KKT conditions areIndeed, the subdifferential of the L1 norm consists of vectors $z$ with $z_k=\mathrm{sign}(b_k)$ at nonzero coordinates and $z_k\in [-1,1]$ at zero coordinates. The subgradient optimality condition is $0=-X^{\mathsf{T}}(Y-Xb)/n+\lambda z$. convexity makes these conditions sufficient as well as necessary. These are the Karush-Kuhn-Tucker conditions for the Lasso.

Here is a correlation-parameter version of least angle regression. Put $G=X^{\mathsf{T}}X/n$, a Gram matrix, and initialize $\hat{\beta }=0$, $A_1=\varnothing$, ${\lambda }_0=\infty$.

The two hitting formulas come from $c_j(t)-h{a}_j=\pm (t-h)$. This is LAR hitting time. Active variables are not dropped if a coefficient crosses zero: that is the distinction between this algorithm and the modified Lasso path algorithm.

The active Gram matrix is an invertible matrix under the assumed unique positive hits. Inductively, a candidate column in the span of the current active columns would have $c_j(\lambda )$ equal to a fixed linear combination of their correlations, hence proportional to $\lambda$ on the whole segment. Its ratio $|c_j(\lambda )|/\lambda$ could not first reach one at a positive interior knot. Thus every uniquely entering column adds a new independent direction. This also allows more predictors than observations; the algorithm stops before a dependent direction needs to be added at zero correlation.

We now prove the requested invariant. At the first positive knot the newly active correlation has magnitude ${\lambda }_1$. At a later knot the old active correlations have that knot's magnitude by induction, and the entering correlation has the same magnitude by the hitting rule. Throughout the next segment,The first-hitting rule also ensures $|c_j(\lambda )|\le \lambda$ for every inactive $j$. If no positive hit remains, this inequality continues to zero, where all correlations vanish. Consequently, on every printed closed segment,This is the LAR active correlation invariant.

If active nonzero coefficients have the same signs as their residual correlations, the active equalities are precisely the nonzero-coordinate Lasso KKT conditions. An active coefficient which is zero also satisfies the zero-coordinate KKT conditions, because its correlation is within $[-\lambda ,\lambda ]$. Inactive coefficients are zero and satisfy those same inequalities by construction. Thus every point of the least angle regression path is a Lasso minimizer for $\lambda >0$. Under the stated uniqueness assumption,This proves the desired implication, in fact under the weaker sign compatibility of LAR and Lasso condition requiring agreement only at nonzero coefficients.

There is an endpoint convention in the printed sufficient condition. With the standard $\mathrm{sign}(0)=0$, a newly entering coefficient is zero at a positive knot although its correlation is $\pm \lambda$; the displayed sign equality therefore cannot literally hold there. At the final zero knot the correlation signs are zero as well. Interpret sign compatibility on positive open segments, or in the nonzero-coefficient/subgradient sense just proved. The KKT conditions already handle zero coefficients at knots, and no equality of solution paths at $\lambda =0$ is needed for the requested conclusion.

Fix the normalizationThe Lasso penalty uses the L1 norm. All arguments below apply to every minimizer; uniqueness of the coefficient vector is not assumed.

The Karush-Kuhn-Tucker conditions are the subgradient optimality conditionThese conditions are necessary and sufficient because the objective is a convex function. They use the subgradient of the absolute value, coordinate by coordinate.

Put $\delta =\hat{\beta }-{\beta }^0$. Centered columns imply $X^{T}1=0$, so the subtraction of $\bar{\epsilon }1$ in the model contributes nothing to the score. The Karush-Kuhn-Tucker conditions becomeMultiply by ${\delta }^T$. Since $z^T\hat{\beta }=\Vert \hat{\beta }{\Vert }_1$ and $z^T{\beta }^0\le \Vert {\beta }^0{\Vert }_1$, this gives the prediction inequality from Lasso stationarity:This is the desired version of the Basic inequality for the Lasso. Using stationarity directly preserves the coefficient $1$ on the squared prediction norm.

For the Gaussian score event for the Lasso, write $X_j$ for column $j$. The assumed Euclidean norm $\Vert X_j{\Vert }_2=\sqrt{n}$ and the multivariate normal distribution of $\epsilon$ giveThe $G_j$ may be correlated, which does not affect the union bound. Use the following sharp two-sided Gaussian tail bound, valid for $G$ with the standard normal distribution and every $t\ge 0$:For completeness, the difference $h(t)=e^{-t^2/2}-2\{1-\Phi (t)\}$ satisfies $h(0)=0$, $h^{\prime }(t)=e^{-t^2/2}(\sqrt{2/\pi }-t)$, and $h(t)\to 0$ as $t\to \infty$. It first increases and then decreases to zero, so it is nonnegative. This bound avoids introducing an unnecessary factor two.

Here $\Phi$ is the standard normal distribution function. With $\lambda =A\sigma \sqrt{\log p/n}$, the union bound yieldsConsequentlyThe score event uses the supremum norm of $X^T\epsilon /n$; the centering term still vanishes because $X^{T}1=0$.

We next work deterministically on $\Omega$. LetBy Holder inequality, ${\epsilon }^{T}X\delta /n\le c\lambda \Vert \delta {\Vert }_1=c\lambda (a+b)$. Since ${\beta }_N^0=0$, the triangle inequality givesSubstituting both inequalities into the Basic inequality for the Lasso producesBecause $Q\ge 0$, this also proves the relevant Lasso cone condition, $(1-c)b\le (1+c)a$. The assumed Compatibility condition for the Lasso therefore applies to $\delta$:Set $D=(1+c)\lambda \sqrt{s}/\varphi$. Combining these inequalities givesIf $Q>0$, discarding the nonnegative $b$ term shows $\sqrt{Q}\le D$. If $Q=0$, compatibility gives $a=0$, and the preceding inequality then gives $b=0$. In either case $D\sqrt{Q}\le D^2$, soThis is a fast prediction bound and an inactive-coordinate error bound on the same event. It is the fast-rate Lasso prediction bound under compatibility.

Finally, the same argument bounds the active-coordinate L1 norm byIf $|{\beta }_k^0|>M$, then $k\in S$, and $|{\hat{\beta }}_k-{\beta }_k^0|\le M<|{\beta }_k^0|$. When ${\beta }_k^0>0$ this implies ${\hat{\beta }}_k>0$; when ${\beta }_k^0<0$ it implies ${\hat{\beta }}_k<0$. Thus the sign function obeysThis is Lasso sign recovery under compatibility. It concerns the sufficiently large nonzero coefficients; it does not assert that every estimated inactive coefficient is zero.

Write $\Omega ={\Sigma }^{-1}$ for a positive-definite matrix. Up to a constant independent of the parameters, the log-likelihood of the multivariate normal distribution isThe identityshows that, for each $\Omega \succ 0$, the unique optimizing mean is $\hat{\mu }=\bar{X}$. Using the matrix trace to rewrite the first quadratic sum, the profile likelihood isTherefore maximum likelihood estimation of the precision matrix, whenever the maximum exists, is exactlyThis is the profile likelihood for a Gaussian precision matrix. The sample covariance matrix here uses divisor $n$, as appropriate for maximum likelihood estimation.

There is an important rank qualification in the printed assumptions. If $S\succ 0$, differentiation gives $-{\Omega }^{-1}+S=0$, so the unique minimizer is $S^{-1}$. Uniqueness follows because the negative log-determinant is a strictly convex function on the positive-definite matrices. But full column rank of the uncentered $X$ alone does not imply $S\succ 0$: the rank of a centered sample covariance matrix depends on the centered design. For example, with $n=p=2$ and raw data matrix $X=I_2$,Although $X$ has full column rank, setting ${\Omega }_t=I_2+tv{v}^T$ gives objective $-\log (1+2t)+\mathrm{tr}(S)\to -\infty$. Thus an unpenalized maximizer need not exist under the literal raw-rank assumption. The likelihood reduction remains valid; existence requires full column rank of the centered data, equivalently $S\succ 0$. For the nonsingular Gaussian model this holds almost surely when $n>p$.

Use the entrywise-penalty convention for the Graphical Lasso:The sum is the entrywise matrix L1 norm. Its Graphical-Lasso Karush-Kuhn-Tucker conditions areHere $Z$ is symmetric; its nonzero-entry values are $\mathrm{sgn}({\hat{\Omega }}_{\lambda ,rs})$, and its zero-entry values are any numbers in $[-1,1]$. These are the subgradient optimality condition for this convex optimization problem. Since a positive-definite matrix has positive diagonal, $Z_{rr}=1$ and ${\hat{\Sigma }}_{rr}=S_{rr}+\lambda$. Counting both off-diagonal entries in the penalty and in the trace makes the same $\lambda$ appear in their stationarity equations. The common variant penalizing only off-diagonal entries instead sets $Z_{rr}=0$; the off-diagonal equations and the column argument below are unchanged.

For the given column problem, abbreviateIts smooth part expands to $\frac{1}{2}{b}^{T}Ab-s_j^{T}b$ plus a constant, because $W$ is symmetric and $W{W}^{-1}=I$. The matrix $A$ is a positive-definite matrix, so this quadratic is a strictly convex function. Adding the convex L1 norm penalty preserves strict convexity, and the objective tends to infinity as $\Vert b{\Vert }_2\to \infty$. The minimizer $b^{\ast }$ exists and is unique. Its Karush-Kuhn-Tucker conditions are

We identify that minimizer directly from the precision matrix. Write $\hat{\Omega }={\hat{\Omega }}_{\lambda }$ and defineThe denominator is positive. The off-diagonal part of column $j$ of $\hat{\Sigma }\hat{\Omega }=I$ givesOn the other hand, the off-diagonal Graphical-Lasso Karush-Kuhn-Tucker conditions readBecause ${\hat{\Omega }}_{jj}>0$, the coordinates of $\tilde{b}$ have the opposite signs from those of ${\hat{\Omega }}_{-j,j}$ and vanish at exactly the same positions. The subgradient of the absolute value therefore satisfiesIt follows thatThese are exactly the Lasso column problem's Karush-Kuhn-Tucker conditions. Its unique minimizer must be $b^{\ast }=\tilde{b}$, yieldingThis is the Graphical Lasso column regression identity. The minus sign is essential: it converts the precision-entry subgradient into the regression-coefficient subgradient. The argument uses the inverse identity directly, so it does not require proving the supplied block inverse formula.

Use the conventionThis proximal map is also the resolvent of a monotone operator $(I+\tau \partial f{)}^{-1}$. A proper lower semicontinuous convex function has an affine minorant, so the quadratic term makes this minimization coercive and strongly convex. A unique minimizer exists for every $x$. Its subgradient optimality condition isThe Moreau–Yosida regularisation is $f_{\tau }=f\square g^{\tau }$. The conjugate of an infimal convolution and the quadratic conjugate giveThe factor $\tau$ here is essential. Apply subgradient inversion under convex conjugacy, followed by the subdifferential sum rule with the everywhere differentiable quadratic:The last equivalence is precisely the unique proximal minimization condition. ThusThis proves both existence and uniqueness of the subgradient, rather than only identifying a possible element. The finite convex function $f_{\tau }$ is therefore differentiable, with $\nabla f_{\tau }=(I-P_{\tau })/\tau$, the gradient of a Moreau envelope.

For completeness, monotonicity of $\partial f$ applied to the two proximal conditions givesHence $P_{\tau }$ is firmly nonexpansive. Expanding the same inequality shows that $I-P_{\tau }$ is firmly nonexpansive too. In particular, $\nabla f_{\tau }$ is $1/\tau$-Lipschitz continuous. None of this requires a bounded effective domain; the result applies to the next example as well.

Set $r^k=A{x}^k-c$. Completing the square in the $y$-minimization givesThus, with $\sigma =\tau /(1+2\tau )$, the steps in terms of the proximal map of $\varphi$ areThe factor $2$ comes from the squared residual without a $1/2$ prefactor. The proximal map can equivalently be denoted $(I+\sigma \partial \varphi {)}^{-1}$. Since $\varphi$ is a finite convex function, it is continuous, and the quadratic makes the proximal minimizer unique.

A general convex $\varphi$ may be nondifferentiable, so this last example need not satisfy the full $L$-smooth hypothesis of the preceding part. The formulas nevertheless remain well-defined: only the $x$ derivative is used explicitly. If $\varphi$ has an $L_{\varphi }$-Lipschitz gradient, one may take $L=2(\Vert A{\Vert }_2^2+1)+L_{\varphi }$ in the smooth proof, where $\Vert A{\Vert }_2$ is the operator norm.

There is also a useful nonsmooth extension. Put $h(x,y)=\Vert Ax+y-c{\Vert }_2^2$, whose Lipschitz gradient constant is $L_h=2(\Vert A{\Vert }_2^2+1)$. For two updates, retain the mixed-point differences $(p,q)$ above, let $(a,b)=\nabla h(w)-\nabla h(\tilde{w})$, and let $d$ be the difference of the two selected subgradients of $\varphi$ at $y^{+},{\tilde{y}}^{+}$. The input difference is $(p,q+\tau (b+d))$ and the output difference is $(p-\tau a,q)$. Monotonicity of $\partial \varphi$ gives $⟨q,d⟩\ge 0$. The same norm expansion now yieldsFor $0<\tau L_h\le 1$, discard the nonpositive $b$ term and use $2\tau /L_h-{\tau }^2\ge {\tau }^2$. The result is the firmly nonexpansive mapping inequality for the full residual $\tau (a,b+d)$. Fixed points satisfy ${\nabla }_{x}h=0$ and $0\in {\nabla }_{y}h+\partial \varphi$, exactly the subgradient optimality condition for the convex objective. Thus the stated algorithm also converges for nonsmooth $\varphi$ under this sufficient step bound, provided a minimizer exists; this extension uses monotonicity, not an unavailable gradient of $\varphi$.

Yes: the solution map is locally affine, hence has the Aubin property. Keep the regularization parameter $\alpha >0$ fixed and define the active hinge setSince none of the margins vanishes, the optimality condition at the reference data has no fractional weights:For nearby $a$, define the candidateEach map $a\mapsto 1-a_i{x}_I(a)$ is continuous and is nonzero at $\hat{a}$. Because there are finitely many samples, all their signs remain unchanged on a common neighborhood $U$ of $\hat{a}$. Therefore the same set $I$ is active at $x_I(a)$ for every $a\in U$. The hinge-loss optimality weights are still $1$ on $I$ and $0$ off $I$, and the defining equation for $x_I(a)$ proves the subgradient optimality condition. Strong convexity makes this candidate the unique global minimizer. Consequently the solution map of a parametric optimization problem satisfiesThis argument establishes stability of the active pattern without presupposing continuity of the unknown optimizer; continuity is used only for an explicit candidate and then optimality is checked.

For $a,b\in U$, the Cauchy-Schwarz inequality givesThis is the Aubin property with $\kappa =\sqrt{|I|}/(\alpha n)$. In fact this constant is the exact local Lipschitz continuity modulus of the affine branch, since its derivative isThe active-set dependence of this derivative is the active-set sensitivity of hinge-loss minimization. At a vanishing margin the affine-branch argument no longer applies, although failure of this particular argument does not by itself prove failure of the Aubin property.

For the illustration, take $n=2$, $\alpha =1$, and $a=(s,1)$ with $s>0$. The hinge-loss optimality weights giveOn the middle branch the first margin is exactly zero, so that branch lies outside the strict-margin hypothesis. The example makes clear why the local affine conclusion is tied to that hypothesis, and also shows that a zero margin need not destroy local Lipschitz stability.

Interpret the normal-operator source condition as the displayed intersection condition: there is $v\in U$ with $w=K^{\ast }Kv\in \partial J(u^{†})$. Fix any $\alpha >0$ and put $\bar{u}=u^{†}+\alpha v$. The subgradient optimality condition for the artificial-data objective isIt holds because its first term equals $-\alpha K^{\ast }Kv=-\alpha w$. By convexity, this is sufficient for global minimality, proving the forward implication. Conversely, if $u^{†}$ minimizes that objective, the same optimality condition givesTaking $v=(\bar{u}-u^{†})/\alpha$ proves the intersection condition. Thus the two conditions are equivalent when the source vector is allowed to be zero.

The printed additional requirement $v\ne 0$ is not equivalent in general. For $K=I$ and $J\equiv 0$, the displayed source condition holds, and choosing $\bar{u}=u^{†}$ makes $u^{†}$ the minimizer, but $\partial J(u^{†})=\{0\}$ forces $v=0$. A nonzero-source version needs an additional hypothesis, such as requiring $w\ne 0$; it cannot be inferred from the printed intersection alone.

The relevant nonlinear generalized singular vector relation, without imposing unit-data normalization, isThis is a generalized eigenfunction in the forward-operator metric, rather than a distributional generalized eigenfunction of a linear operator. It supplies the normal-operator source condition with $v=\lambda u^{†}$. For a convex positively one-homogeneous functional, the subgradient inequality applied at zero and at $2u^{†}$ gives $⟨p,u^{†}⟩=J(u^{†})$. If $K{u}^{†}\ne 0$, thenAssume the usual positive eigenvalue $\lambda >0$ and $0<\alpha \lambda \le 1$. By positive homogeneity,The shifted-comparator Bregman bound therefore yieldsThe vector $(1-\alpha \lambda )u^{†}$ is itself a minimizer: $p$ remains a subgradient along the nonnegative ray, and substitution satisfies the subgradient optimality condition. If $K$ is injective, this also proves $u_{\alpha }=(1-\alpha \lambda )u^{†}$. Otherwise distinct minimizers can differ in the kernel while sharing the same predicted data.

For a nonnegative convex positively one-homogeneous functional, the branch continues as $(1-\alpha \lambda {)}_{+}{u}^{†}$ above the threshold, since $p/(\alpha \lambda )\in \partial J(0)$ there. Nonnegativity is needed for that clipping argument; bare positive homogeneity allows signed linear functionals.

Zero Bregman distance does not imply exact recovery. For $K=I$, $J(s)=|s|$, $u^{†}=a>0$ and $0<\alpha <a$, the minimizer is $u_{\alpha }=a-\alpha \ne a$, but $D_J^1(a-\alpha ,a)=0$. Both points touch the same supporting line, as described by zero Bregman distance and supporting faces.

Use the scaled proximal operator ${\mathrm{prox}}_{\gamma E}$; take $\gamma =1$ for each unscaled resolvent.

For the quadratic case, write $D^{T}D=\mathrm{diag}(d_1,\dots ,d_n)$ and $\xi =Qx$. Since $Q$ is an orthogonal matrix, the quadratic proximal term becomes $\Vert \xi -Qz{\Vert }^2/2$. The subgradient optimality condition givesAll $d_j\ge 0$, so each diagonal entry is positive, including when $d_j=0$. Multiplying by the inverse diagonal matrix givesThe proximal operator of Poisson data fidelity separates pointwise. At each $t$, the scalar optimality equation on $x>0$ isSince $y>0$, the two roots have opposite signs. Only the positive root is admissible in the logarithm:The negative root is outside the effective domain. The second derivative $1+\gamma y/x^2$ of the proximal objective is positive, establishing the unique pointwise minimizer. The integral formula is understood on the function spaces where these quantities are admissible.

Finally, for $E(x)=|x|$, positive and negative minimizers obey $x-z+\gamma =0$ and $x-z-\gamma =0$, respectively. At zero the subdifferential condition is $z\in [-\gamma ,\gamma ]$. Combining these cases yields soft thresholding:Thus $\gamma =1$ supplies all three requested closed forms.

For a proper convex function $E$ that is lower semicontinuous on a Hilbert space, its proximal operator at scale $\gamma >0$ isThe minimizer is unique because the objective is strongly convex; existence follows from the closed proper convex functional and the coercive quadratic, using an affine lower bound. The subgradient optimality condition is $0\in x-z+\gamma \partial E(x)$, which is exactly the displayed resolvent of a monotone operator relation. Setting $\gamma =1$ gives the unscaled proximity or resolvent operator requested here. This is a nonlinear resolvent of the subdifferential, distinct from the spectral resolvent of a linear operator.

Work in the real Hilbert space $L^2({\mathbb{R}}^2)$, with $g\in L^2$ and $\alpha >0$. The penalty $J$ is a proper convex function: it is finite at zero, and the bounded-variation space domain is convex. Its subdifferential at a finite-penalty $u$ consists of $q\in L^2$ satisfying $J(v)\ge J(u)+⟨q,v-u⟩$ for every $v\in L^2$.

If $q=(g-u)/\alpha \in \partial J(u)$, expand the quadratic term and use the subgradient inequality:Thus $u$ is the unique minimizer.

Conversely, let $u$ be a minimizer. Its penalty is finite since comparison with zero gives a finite objective. For any $v$ with $J(v)<\infty$, set $w_t=u+t(v-u)$, where $0<t\le 1$. Convexity gives $J(w_t)-J(u)\le t(J(v)-J(u))$. Minimality and quadratic expansion then implyLetting $t\downarrow 0$ yields $J(v)\ge J(u)+⟨(g-u)/\alpha ,v-u⟩$. The inequality is automatic if $J(v)=\infty$. HenceThis also follows from the subdifferential sum rule, since the quadratic term is everywhere continuous and differentiable, and the subgradient optimality condition. The direct proof above needs no unproved existence theorem.

On the entire plane, the printed global BV domain is not closed in the L2 geometry. The usual $BV({\mathbb{R}}^2)$ definition includes an $L^1$ condition. For $1<a\le 2$, $f(x)=(1+|x|{)}^{-a}$ belongs to $L^2$, has finite distributional variation $2\pi a{\int }_0^{\infty }r(1+r{)}^{-a-1}dr$, and fails to belong to $L^1$. The truncated $f_T=f{\chi }_{B(0,T)}$ lies in $BV\cap L^2$ and tends to $f$ in $L^2$. Its variation is the interior variation plus $2\pi T(1+T{)}^{-a}$, and tends to a finite limit. Thus $J(f_T)$ stays bounded but the literal $J(f)=\infty$, establishing failure of L2 closure of the global BV domain. One must not infer universal minimizer existence from an inapplicable closed-penalty proximal operator theorem. The standard closed extension uses the homogeneous bounded-variation space, allowing finite distributional variation without global $L^1$. The optimality equivalence just proved is valid for the literal penalty whenever a minimizer exists; the next datum has an explicit certified minimizer in its domain.

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.