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A standard sufficient form of Brenier theorem assumes $\mu ,\nu \in \mathcal{P}({\mathbb{R}}^d)$ have finite second moments and $\mu \ll {\mathcal{L}}^d$, that is, absolute continuity of measures with respect to Lebesgue measure. For the cost $|x-y{|}^2$, there exists a unique optimal transport plan, and it is induced by a transport map:Here $\Phi$ is a proper convex function, which may be chosen sequentially lower semicontinuous, and its gradient exists $\mu$-almost everywhere. The map $\nabla \Phi$ is unique $\mu$-almost everywhere and is the unique minimizer of the Monge optimal transport problem; its cost equals the Kantorovich optimal transport problem minimum. Equivalently, it is the unique gradient of a convex function transporting $\mu$ to $\nu$.

The uniqueness claim concerns the map and the transport plan, not a globally unique potential. The potential may be shifted by a constant, and additional nonuniqueness away from the source can occur. No density assumption is required on $\nu$. A primary reference is Brenier's Polar factorization and monotone rearrangement of vector-valued functions.

Define $T^{†}(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu (\{0\})=0$. In polar coordinates, the two probability density functions giveBoth angular distributions are uniform, with independence of angle and radius. The proposed transport map preserves the angle and sends $r$ to $s=\sqrt{r}$. Thereforeand preservation of the angle proves $(T^{†}{)}_{\#}\mu =\nu$. As a local check using the Jacobian determinant, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt{r})$ and $1/\sqrt{r}$, respectively, so $\det D{T}^{†}=1/(2r)$ and $g(T^{†}(x))\det D{T}^{†}(x)=(2r/\pi )/(2r)=f(x)$.

Now use the convex functionIt is convex because the Euclidean norm is convex and $s\mapsto (2/3)s^{3/2}$ is increasing and convex on $[0,\infty )$. It is differentiable, including at zero, andIts graph transport plan lies in the graph of $\partial \Phi$, so the Knott–Smith optimality criterion proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the Monge optimal transport problem as well. The minimum provides a useful independent check:

The empirical risk isFor $g(z)={\log }_2(1+e^{-z})$,so $g$ is a convex function. Each signed margin $y_i{x}_i^T\beta$ is an affine function of $\beta$, and composition with an affine function preserves convexity. Equivalently,is a positive semidefinite matrix. Therefore $f$ is convex.

For $0\le t\le 1$ and $x,y\in {\mathbb{R}}^d$, convexity givesTaking the maximum over $i$ and then bounding each term by the corresponding endpoint maxima yieldsThus the finite pointwise maximum of convex functions is convex. Summing the original inequalities over $i$ proves that ${\sum }_i{f}_i$ is convex as well.

Taking $\Theta =0$ gives $\Vert \Sigma \Theta -I{\Vert }_{\max }=1$, so every matrix is $1$-invertible. The smallest admissible value isThe map inside the norm is affine in $\Theta$, and a norm composed with an affine map is a convex function. The space of matrices is a convex feasible set, so this is a convex optimization problem.

A $J$-minimizing solution is an exact solution $u_J^{†}$ satisfyingThe source condition in variational regularization says that some $p^{†}\in Y$ satisfieswhere $\partial J$ is the subdifferential.

Fix $\alpha >0$. The subgradient optimality condition forat $u_J^{†}$ isIf the source condition holds, choose $g=f+\alpha p^{†}$. Since $A{u}_J^{†}=f$, the displayed inclusion holds. Because the objective is a convex function, this condition is necessary and sufficient for global minimality.

Conversely, if the stated range condition holds for some $g$, optimality supplies $\xi \in \partial J(u_J^{†})$ withThus $\xi =A^{\ast }((g-f)/\alpha )$, which is precisely the source condition. This proves the equivalence.

For a convex function $f$, its subdifferential isFor $f(x)=\gamma |x|$,The defining inequality shows directly that $0\in \partial f(x)$ exactly when $x$ is a global minimizer.

Strict convexity meansfor distinct $x,y$ and $0<\lambda <1$. Two distinct minimizers would make the midpoint have a strictly smaller value, so a minimizer is unique.

A function $f$ on an interval is a convex function ifJensen inequality states that for an integrable random variable $X$, when all terms are defined,

The function $\varphi (x)=x\log x$ is convex on $(0,\infty )$ becauseApplying Jensen's inequality to the uniform distribution on the positive numbers $x_1,\dots ,x_n$ givesWriting $S={\sum }_i{x}_i$ and multiplying by $n/S$ yields

A vector $g$ is a subgradient of the convex function $f$ at $x$ whenfor every $y$. The set of all such vectors is the subdifferential $\partial f(x)$. The proximal operator satisfiesMore generally, $u={\mathrm{prox}}_{tf}(x)$ exactly when $(x-u)/t\in \partial f(u)$.

Each map $x\mapsto ⟨a_i,x⟩+b_i$ is an affine function. For $0\le t\le 1$, the pointwise maximum of convex functions satisfiesso $f$ is convex.

A vector $g$ is a subgradient of a convex function $f$ at $x$ whenfor every $y$. Choose any active index $j\in I(x):=\{i:f(x)=⟨a_i,x⟩+b_i\}$. Thenand hence $\boxed{a_j\in \partial f(x)}$. More generally, every convex combination of the active vectors is a subgradient, and in fact

The first-order subgradient inequality for a differentiable convex function givesand, after interchanging $v$ and $w$,Adding and rearranging yieldsThus the gradient $F=\nabla f$ is a monotone operator.

The function $U(s)=V_0{e}^{-2s}$ is strictly convex. Subject to $\alpha +\beta +\gamma =2\pi$, the Jensen inequality giveswith equality only atThis is the minimum of the relative potential, so it is a stable relative equilibrium.

Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and putThe gap perturbations arewhose sum vanishes. A Taylor expansion gives the quadratic energiesThus the linearized equations areThe matrix in parentheses is the Graph Laplacian of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three normal modes and their angular frequencies is thereforeThe zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general normal modes of three equal masses with a symmetric gap potential.

Coordinate descent cycles through the intercept and coefficient coordinates, minimizing the convex ridge objective in one coordinate while holding the others fixed. Given current coefficients, updateFor coordinate $j$, form the partial residualand update it by the exact one-dimensional minimizerRepeated sweeps converge to the unique fitted value because the objective is a convex function; with $\lambda >0$ it is strictly convex in $\beta$.

The Euclidean projection onto a convex set is nonexpansive, and $x^{\ast }=P_C(x^{\ast })$ because the optimum is feasible. Therefore the projected subgradient method satisfiesThe subgradient inequality gives $⟨g_i,x_i-x^{\ast }⟩\ge f(x_i)-f^{\ast }$, while Lipschitz continuity of the finite convex function gives $\Vert g_i{\Vert }_2\le G$. HenceSumming this telescoping inequality for $0\le i<k$, and then bounding the smallest term by the average, yieldsWriting $D=\Vert x_0-x^{\ast }{\Vert }_2$, the right-hand side is minimized by the constant step sizeSubstitution givesIf $D=0$, the initial point is already optimal and the result is immediate.

Write $a_i^T$ for row $i$ of $A$, and define affine functionsThenThe pointwise maximum of convex functions is convex, so $f$ is a convex function. Moreover,so $f$ has Lipschitz continuity. The simpler bound $L\le \Vert c{\Vert }_2+M{\max }_i\Vert a_i{\Vert }_2$ is also valid.

Let $I(x)=\{i:{\ell }_i(x)=f(x)\}$ be the active set. The subdifferential isthe convex hull of all active slopes. In particular, choosing any active index gives the subgradientAt a tie, every convex combination of the tied slopes is also valid.

For a proper lower-semicontinuous convex function $f$, its proximal operator isThe squared norm is strongly convex, so the minimizer is unique. The subdifferential sum rule gives the necessary and sufficient conditionMore generally,The subgradient inversion rule for the convex conjugate says $u\in \partial f(x)$ exactly when $x\in \partial f^{\ast }(u)$. Hencewhich is precisely the proximal optimality conditionSince $x=y-tu$, this proves the generalized Moreau decomposition

For a convex function $f:{\mathbb{R}}^d\to \mathbb{R}$, its subdifferential at $x$ isIts elements are the subgradients of $f$ at $x$.

Suppose ${\hat{\beta }}_1$ and ${\hat{\beta }}_2$ minimize $Q$. Their midpoint is also a minimizer because $Q$ is a convex function. The group penalty is convex, while the squared Euclidean norm is strictly convex in the fitted value. If $X{\hat{\beta }}_1\ne X{\hat{\beta }}_2$, strict convexity would make the midpoint objective strictly smaller than the minimum. Hence $X{\hat{\beta }}_1=X{\hat{\beta }}_2$, so the fitted values are unique.

Let $0\le \lambda \le 1$ and put $P=\lambda P_1+(1-\lambda )P_2$ and $Q=\lambda Q_1+(1-\lambda )Q_2$. For each alphabet symbol $x$, apply the log-sum inequality to $a_1=\lambda P_1(x)$, $a_2=(1-\lambda )P_2(x)$ and the corresponding $b$ values. Summing over $x$ giveswhich is joint convexity in $(P,Q)$.

An element $u^{†}$ is a $J$-minimizing solution whenIt satisfies the source condition in variational regularization when there is a $w^{†}\in Y^{\ast }$ such thatwhere $\partial J(u^{†})$ is the subdifferential of the convex functional $J$.