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The Beckmann potential integrates each link's delay function. Its gradient with respect to route flows is the corresponding vector of route delays, making Wardrop equilibrium a convex optimization problem.

This finite-image convex optimization model combines robust L1 norm fidelity with discrete isotropic total variation and pixel-range constraints. An exact second-order cone program uses two orthant fidelity slacks per pixel, two box slacks per pixel and a three-dimensional Lorentz cone for each gradient magnitude. The product canonical barrier has parameter $6n$ for $n$ pixels.

A constraint qualification is a regularity hypothesis on optimization constraints that guarantees a specified optimality or duality conclusion. For the necessity of Karush-Kuhn-Tucker conditions, an appropriate constraint qualification ensures that a local minimum admits suitable Lagrange multipliers; linear independence of the gradients of equality optimization constraints and active constraints is one sufficient hypothesis in KKT regularity at a local minimum.^[ref] Slater's condition provides a different constraint qualification for convex optimization, yielding strong duality and dual attainment.^[ref]

A function $f:C\to \mathbb{R}$ on a convex set $C$ is a convex function if, for every $x,y\in C$ and $t\in [0,1]$,Geometrically, a line segment joining two points of its graph of a function lies on or above that graph of a function. The convex set condition ensures that the intermediate input stays in the domain of a function. An affine function satisfies the defining inequality with equality.

For the quadratic function $f(x)=x^2$, the inputs $1$ and $3$ have equally weighted arithmetic mean $2$. Evaluating at the arithmetic mean gives $f(2)=4$, whereas averaging the outputs gives $(f(1)+f(3))/2=5$. Averaging first saves one unit in this example. More generally, direct expansion givesThis identity proves convexity and also shows exactly when its defining inequality is strict.

For a differentiable convex function, the tangent line in one dimension, or the corresponding supporting affine function in several dimensions, lies below the entire graph of a function. This is the subgradient inequality. The equivalent geometric statement is that the epigraph, consisting of points on or above the graph of a function, is a convex set. These characterizations are developed in Boyd and Vandenberghe's convex-function slides, sections 3.1–3.4.

Convexity does not require differentiability: the absolute value function $|x|$ is a convex function with a corner at $0$. When a function is twice continuously differentiable on an open interval, it is a convex function exactly when its second derivative is nonnegative throughout that interval. In several dimensions, the corresponding criterion on an open convex set is that its Hessian matrix is positive semidefinite everywhere. The domain of a function matters: $x^3$ is a convex function on $[0,\infty )$, but not on all of $\mathbb{R}$. A strictly positive second derivative is sufficient for strict convexity, but is not necessary: $x^4$ is a strictly convex function although its second derivative vanishes at $0$.

The usefulness of convexity is that local information can certify a global conclusion. Every local minimum of a convex function on a convex set is a global minimum: if a lower value existed elsewhere, the defining inequality would produce lower values arbitrarily close to the proposed local minimum. For a differentiable convex function on an open convex set, a zero gradient therefore certifies a global minimum. Strict convexity gives uniqueness when a global minimum exists; it does not supply existence by itself. These facts explain the role of convex functions in convex optimization.

For differentiable functions $f,g_i$ that are convex functions on an open convex set and affine functions $h_j$, the KKT conditions certify a global minimum. Once Lagrange multipliers satisfy them, no extra Slater condition is needed.^[ref]

The certificate is a short chain of inequalities. Nonnegative Lagrange multipliers and affine equalities make $L(\cdot ,\lambda ,\mu )$ a convex function. Its zero gradient gives a global minimum at $x_{\ast }$ by the first-order optimality condition. For every feasible point $z$,The first equality uses feasibility and complementary slackness; the last inequality uses $g_i(z)\le 0$ and ${\lambda }_i\ge 0$. Equivalently, the Lagrange dual function attains the same value as the primal objective function, proving strong duality. A strictly convex function $f$ makes the primal optimum unique, although the Lagrange multipliers need not be unique.

Affine equality optimization constraints matter: an arbitrary signed multiple of a nonlinear equality can destroy convexity of the optimization Lagrangian. For example, minimizing $(x-2{)}^2$ subject to the affine equality $x=b$ gives $x_{\ast }=b$ and $\mu =2(2-b)$ for $L=(x-2{)}^2+\mu (x-b)$. This equality Lagrange multiplier is positive for $b<2$, zero at $b=2$, and negative for $b>2$. Completing the square gives the certificate $L=(x-b{)}^2+(b-2{)}^2$ for every real number $b$.

Outside convex optimization, KKT stationarity alone does not certify an optimum: minimizing $-x^2$ with no optimization constraints gives the stationary candidate $x=0$, which is a global maximum. The Lagrangian sufficiency theorem remains applicable whenever a global minimum of the optimization Lagrangian can be proved directly.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

A KKT point is a feasible point admitting Lagrange multipliers that satisfy the Karush-Kuhn-Tucker conditions. Some authors use the name for the full primal and multiplier pair; specify the multipliers when this distinction matters. For a convex optimization problem with a differentiable convex function as its objective function, convex inequality functions and affine equality constraints, the KKT conditions certify a global minimizer. Their necessity requires an appropriate constraint qualification.

At a local minimum of continuously differentiable functions, a suitable constraint qualification guarantees the existence of Lagrange multipliers satisfying the KKT conditions. One sufficient local regularity assumption is that the gradients of the equality optimization constraints and the active constraints are jointly linearly independent. This assumption concerns the description of the feasible set, not just the smoothness of the objective function.^[ref]

To see why a hypothesis is needed, minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The feasible set is the singleton $\{0\}$, so $x_{\ast }=0$ is a global minimum. Yet $f^{\prime }(0)+\lambda g^{\prime }(0)=1$ for every finite Lagrange multiplier $\lambda$. Thus this convex optimization problem has an attained optimum without a KKT multiplier. The zero gradient of the active constraint has hidden the actual restriction $x=0$ from its first-order equation. Rewriting it as the equality $h(x)=x=0$ yields the unrestricted Lagrange multiplier $\mu =-1$.

For convex optimization, Slater's condition requires strict inequalities and affine equalities at a point in the relative interior of the common domain of a function. A finite optimum value then gives strong duality and dual attainment. Primal attainment and differentiability yield KKT multipliers. Affine inequalities may be nonstrict in the refined version.^[ref]

Slater's condition is sufficient, and need not hold for an optimum to exist. The singular example above also shows that mere primal attainment is not a substitute for a constraint qualification.

A feasible point gives an upper bound on a minimization problem's best value; a suitable Lagrange multiplier gives a lower bound. Lagrangian duality searches for the strongest bound obtainable this way. Matching bounds certify a global minimum.

Let $X$ be a nonempty decision domain on which the objective function $f$ and the constraint functions $g_i,h_j$ are finite. Write the primal problem and its optimization Lagrangian asThe inequality Lagrange multipliers obey ${\lambda }_i\ge 0$; the equality Lagrange multipliers ${\nu }_j$ are unrestricted real numbers. Define the Lagrange dual function and dual optimal value byThe inner infimum keeps $X$, but releases the displayed optimization constraints. It can be $-\infty$, giving an uninformative lower bound. A different domain changes the bound.^[ref]

Here is the entire weak duality argument. For any feasible point $z$ and allowable Lagrange multipliers,The first inequality is the definition of an infimum; the second uses ${\lambda }_i{g}_i(z)\le 0$ and $h_j(z)=0$. Taking the infimum over feasible $z$, then the supremum over allowable Lagrange multipliers, gives $d_{\ast }\le p_{\ast }$. No convexity is used.

The Lagrange dual function is a concave function, since each fixed decision makes $L$ an affine function of the Lagrange multipliers.^[ref] Explicitly, write $y=(\lambda ,\nu )$ and choose two points where $q$ is finite. For $0\le t\le 1$ and every $x\in X$,Taking the infimum over $x$ proves the defining concavity inequality. Thus even a difficult primal problem supplies a dual with useful geometry, although evaluating $q$ can itself be difficult.

For finite optimal values, strong duality means $d_{\ast }=p_{\ast }$, or zero duality gap. It need not imply dual attainment. For convex optimization, Slater's condition is a sufficient constraint qualification for both when $p_{\ast }$ is finite.^[ref] The Strong Lagrangian property requires an actual finite Lagrange multiplier achieving equality. Convex perturbation duality interprets such a bound as a supporting slope of the perturbed value function.

Use Lagrange multipliers ${\alpha }_i\ge 0$ for $1-{\xi }_i-y_i(w\cdot x_i+b)\le 0$ and ${\mu }_i\ge 0$ for $-{\xi }_i\le 0$. The optimization Lagrangian isThe Lagrange dual function is the infimum over the now unconstrained primal variables. The first-order optimality conditions in $w$ and $b$ giveIf the latter equality fails, the infimum over $b$ is $-\infty$. The first-order optimality condition in ${\xi }_i$ givesSubstitution, or completing the square separately in $w$ and each ${\xi }_i$, producesprovided ${\sum }_i{\alpha }_i{y}_i=0$. For fixed nonnegative ${\alpha }_i$, this expression is maximized over ${\mu }_i\ge 0$ at ${\mu }_i=0$. Thus the simplest Lagrangian dual problem isIn particular, there is no upper box constraint ${\alpha }_i\le C$. That constraint belongs to the linear-slack penalty; squared slack instead contributes the displayed diagonal quadratic penalty. With $Y=\mathrm{diag}(y_1,\dots ,y_m)$ and Gram matrix $G_{ij}=x_i\cdot x_j$, the objective can also be writtenSince $G$ is positive semidefinite and $C>0$, the matrix in parentheses is positive-definite. The objective of the Lagrangian dual problem is therefore strictly concave, and the feasible maximizer is unique.

The primal problem is a convex optimization problem, and $w=0,b=0,{\xi }_i=2$ is strictly feasible for both families of inequalities. Slater's condition gives strong duality, so an optimum of the Lagrangian dual problem yields a primal optimum. The Karush-Kuhn-Tucker conditions supplyFor any $i$ with ${\alpha }_i>0$, the intercept is consequentlyAll such choices agree at an exact optimum. If all labels have one sign, the optimal coefficients can all be zero: choose $w=0$, $\xi =0$, and any constant intercept of that sign with magnitude at least one.

Taking the natural logarithm of the positive product constraint gives $h(x)={\sum }_i{a}_i\log x_i=0$. Its gradient has entries $a_i/x_i>0$, so the equality constraint qualification holds everywhere on the feasible set. Use the optimization LagrangianStationarity requires $a_i-\lambda a_i/x_i=0$, so every $x_i=\lambda$. Substitution into the constraint, using ${\sum }_i{a}_i=1$, gives $\lambda =1$. Thus the only candidate for a local minimum is $x=(1,\dots ,1)$.

To prove global minimality without assuming the inequality to be established, set $u_i=\log x_i$. The constraint becomes the affine hyperplane ${\sum }_i{a}_i{u}_i=0$, and the objective function becomes $F(u)={\sum }_i{a}_i{e}^{u_i}$. Its Hessian matrix $\mathrm{diag}(a_i{e}^{u_i})$ is positive-definite, so $F$ is strictly convex. The elementary tangent inequality $e^t\ge 1+t$, with equality only at zero, givesEquality forces every $u_i=0$. ThereforeThis point is the unique global minimizer and the only local minimizer. The logarithm change of variables also explains why a multiplicative constraint leads to a convex optimization argument.

If $x,y$ are feasible points and $0\le t\le 1$, concavity givesfor every $i$. Thus the feasible set is a convex set. Since the objective function is convex, this is a convex optimization problem.

Use closest points on two line segments as a small constrained convex optimization problem. Write the two line segments asand put $w=P_1-P_2$. Minimize the squared Euclidean distance $D(s,t)=\Vert w+sd-te{\Vert }^2$; minimizing its square also minimizes the distance. Compute the dot productsFor nonzero, nonparallel directions, an interior minimum must satisfy the two derivative equationsTheir determinant is $\Delta =\alpha \gamma -{\beta }^2>0$, so the stationary candidate isRetain this candidate only when both parameters lie in the unit interval.

The other possibilities are on the boundary. Define $\mathrm{clip}(x)=\max (0,\min (1,x))$. Minimizing the same quadratic on each of the four edges gives these candidates by orthogonal projection onto a line segment:Evaluate $D$ at every retained candidate and choose the smallest. The resulting $R_1(s),R_2(t)$ are the required closest points. The quadratic is convex because it is the squared norm of an affine map. A minimum in the interior is stationary, and a minimum on the boundary minimizes one of these edge restrictions, so the list is exhaustive. Ties correspond to multiple equally valid answers.

Handle degenerate line segments before dividing: if $\alpha =0$ and $\gamma >0$, set $s=0$ and $t=\mathrm{clip}(\epsilon /\gamma )$; if $\gamma =0$ and $\alpha >0$, set $t=0$ and $s=\mathrm{clip}(-\delta /\alpha )$; if both vanish, use the two endpoints. For exactly parallel directions, $\Delta =0$. The distance depends on a single linear combination of $s,t$; any interior minimizing level line reaches the boundary, so the four edge candidates still include a minimum.

For nearly parallel directions, compare $\Delta$ with the scale $\alpha \gamma$, and solve the interior least-squares problem using QR decomposition or singular value decomposition if necessary. A small determinant should trigger a stable solve, not arbitrary rejection of a possibly valid interior solution. In particular, do not merely clip both coordinates of the infinite-line solution: the dot product $\beta$ couples the two parameters, and changing one changes the optimum for the other.

A useful starting model treats a road network as a directed flow network and travelers as a divisible population in a nonatomic congestion game: an individual traveler has negligible effect on link congestion. For each source-sink pair $k$, fix a demand $d_k$ and a finite set ${\mathcal{P}}_k$ of allowed routes. Let $f_p\ge 0$ be the flow on route $p$, with ${\sum }_{p\in {\mathcal{P}}_k}{f}_p=d_k$. The link-route incidence matrix $A$ gives link loads $v=Af$. If link $e$ has delay ${\ell }_e(v_e)$, the route travel time isA Wardrop equilibrium is a feasible route-flow vector for which every used route has minimum travel time among the routes serving the same source-sink pair. Equivalently, there are numbers ${\lambda }_k$ such thatThe condition says that a traveler cannot lower their own delay by changing route. It does not assert that their choice minimizes the total delay experienced by all travelers. At a Wardrop equilibrium, unused routes may have exactly the same delay as used ones.

This static flow network model requires delay to depend on present link load, with each source-sink demand fixed. Models from queueing theory instead describe vehicles or packets as discrete customers whose delay depends on random arrivals, service and accumulated queues; a time-dependent model must also track changes of flow and travel-time propagation. Those models answer different questions from a static Wardrop equilibrium. The static model is nevertheless valuable because its route-choice condition has a precise convex optimization formulation, as developed next.

Let $R_s$ be the routes serving source-sink pair $s$. The source-sink route incidence matrix and link-route incidence matrix areThus $Hx=f$ fixes the demands, and $y=Ax$ gives link throughputs. Assume a finite route set with at least one route for each positive demand. A Wardrop equilibrium is a feasible flow for which every used route has minimum delay among the routes for its source-sink pair. Writingthis means that, for each $s$, there is a number ${\lambda }_s$ with $c_r(x)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$. No individual user can reduce its delay by switching routes while treating the aggregate throughputs as fixed.

The feasible set $K=\{x\ge 0:Hx=f\}$ is a nonempty compact convex set: within each source-sink pair its route flows lie in a simplex of total $f_s$. The Beckmann potentialis continuous and convex, and has $\partial \Phi /\partial x_r=c_r(x)$. It therefore attains a minimum on $K$. For a differentiable convex function, $x$ minimizes $\Phi$ exactly whenIndeed necessity follows by differentiating along the feasible segment from $x$ to $z$, and sufficiency follows from the supporting-tangent inequality. This condition is equivalent to Wardrop equilibrium. Necessity of the route condition follows by moving a small amount of positive flow from a used route to any other route of the same pair. Conversely, under the route condition,because the terms involving $x_r$ vanish and both flows have the same pair totals. This proves both existence and the claimed convex optimization characterization, with $y=Ax$.

Since each $D_j$ is strictly increasing, its primitive is strictly convex. Thus ${\sum }_j{\int }_0^{y_j}{D}_j(u)du$ is strictly convex in $y$. If two minimizing flows had different link throughputs, their midpoint would have strictly smaller potential. Hence the equilibrium link throughputs are unique.

The route flows need not be unique. Consider one unit of demand through two successive stages, each with two parallel links, labelled $a,b$ in the first stage and $c,d$ in the second. The four routes are $ac,ad,bc,bd$, and all link delays are $D_j(u)=u$. For any $0\le t\le 1/2$, chooseEvery link then carries $1/2$ and every route has delay one, so each vector is a Wardrop equilibrium. They all produce the same link throughputs but give distinct route decompositions.

For the toll construction, put $F={\sum }_s{f}_s>0$. The total delay is $L(x)={\sum }_j{y}_j{D}_j(y_j)$ and the average delay is $L(x)/F$, so they have the same minimizers. Compactness of $K$ gives a global minimizer $x^{\ast }$, with $y^{\ast }=A{x}^{\ast }$. Its differentiable objective satisfies the necessary route-flow inequalityThis inequality is necessary even when total delay is not convex, since it follows from the right derivative along each feasible segment.

Use optimal-flow calibrated congestion tolling:Here $(v{)}_{+}=\max (v,0)$. The toll is nonnegative, continuously differentiable and nondecreasing, because $D_j^{\prime }\ge 0$. The perceived cost ${\hat{D}}_j(u)=D_j(u)+T_j(u)$ is strictly increasing. At $y_j^{\ast }$ it equals $D_j(y_j^{\ast })+y_j^{\ast }{D}_j^{\prime }(y_j^{\ast })$, so the displayed necessary inequality makes $x^{\ast }$ a Wardrop equilibrium for these perceived costs, by the preceding Beckmann potential argument. Their strictly increasing costs give unique equilibrium link throughputs. Therefore every tolled equilibrium has link loads $y^{\ast }$ and minimizes average delay. Constant tolls $T_j(u)=y_j^{\ast }{D}_j^{\prime }(y_j^{\ast })$ also suffice; the added positive-part term makes them genuinely traffic-dependent without moving the optimum. Zero total demand is the trivial empty-flow case.

The usual marginal external cost toll $T_j(u)=u{D}_j^{\prime }(u)$ makes its Beckmann potential equal $L$, but concluding that all its equilibria minimize $L$ requires convexity of $L$. Strict increase of delays alone does not supply this. For example, on two parallel links with $D(u)=1-e^{-u}$ and demand six, the equal split $(3,3)$ is an equilibrium under marginal-cost tolls, while its total delay $6(1-e^{-3})$ exceeds $4(1-e^{-4})+2(1-e^{-2})$, the delay at $(4,2)$. The calibrated construction avoids adding a hypothesis absent from the printed question.

Let $f(x)=-{\sum }_i\log ({\alpha }_i+x_i)$. The feasible probability simplex is a nonempty compact convex set, so the extreme value theorem guarantees a minimizer. Its Hessian matrix is diagonal with positive entries $({\alpha }_i+x_i{)}^{-2}$, making $f$ a strictly convex function. Hence the minimizer is unique.

Use the Lagrange multipliers $\lambda$ for the sum constraint and ${\nu }_i\ge 0$ for $-x_i\le 0$. The Lagrangian isThe KKT conditions areA strictly positive feasible allocation exists, so the Slater condition holds. These conditions are necessary and sufficient for this convex optimization. At least one coordinate is positive, giving $\lambda >0$. Put $\tau =1/\lambda$. For a positive coordinate, complementary slackness gives ${\alpha }_i+x_i=\tau$. At a zero coordinate, stationarity gives ${\nu }_i=1/\tau -1/{\alpha }_i\ge 0$, or ${\alpha }_i\ge \tau$. ThusHere $(u{)}_{+}=\max (u,0)$. This logarithmic water filling raises the smaller baseline values to a common level. The scalar left side is a continuous function and strictly increasing above ${\min }_i{\alpha }_i$, starts at zero, and tends to infinity, so exactly one $\tau$ solves it. One can find it by interval bisection, or sort the baseline values increasingly and find an index $k$ for whichusing ${\alpha }_{(n+1)}=+\infty$. Equality at the next baseline simply gives a zero allocation. Sorting and a running sum implement the water-filling algorithm in $O(n\log n)$ time.

For a minimization problem over a domain $D$ with inequality constraints $g_j(x)\le 0$ and equality constraints $h_{\ell }(x)=0$, letThe Lagrangian sufficiency theorem states: if $x_{\ast }$ is feasible, ${\lambda }_j\ge 0$, ${\lambda }_j{g}_j(x_{\ast })=0$, and $x_{\ast }$ globally minimizes $\mathcal{L}(\cdot ,\lambda ,\mu )$ on $D$, then $x_{\ast }$ globally minimizes the constrained objective. Equality multipliers have no sign restriction.

For any feasible $x$, the sign of the inequality multipliers gives $\mathcal{L}(x,\lambda ,\mu )\le f(x)$. Feasibility and complementary slackness give $\mathcal{L}(x_{\ast },\lambda ,\mu )=f(x_{\ast })$. ThereforeThis proves the theorem. No convexity is necessary if the global minimization of the Lagrangian is already known. In a differentiable convex optimization problem with convex inequalities and affine equalities, a stationary point of this convex Lagrangian is automatically a global minimizer, which gives the familiar sufficiency of the Karush-Kuhn-Tucker conditions.

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.

For the Euclidean norm, rotational symmetry and minimizing the radial objective give the radial soft thresholding formulaIndeed, if $r=\Vert x{\Vert }_2$, an optimal $u$ is a nonnegative scalar multiple of $x$; minimizing $s+(r-s{)}^2/(2\tau )$ over $s\ge 0$ gives $s=(r-\tau {)}_{+}$. The gradient of a Moreau envelope is consequentlyAt $x=0$ this is $\{0\}$, and the two expressions agree when $\Vert x{\Vert }_2=\tau$. The Moreau envelope of the Euclidean norm itself isThus a quadratic core replaces the nondifferentiable tip while the outer gradient remains the normalized radial direction. The Euclidean norm has an unbounded effective domain, unlike the earlier bounded-domain hypothesis; the preceding argument explicitly shows that this restriction is unnecessary here.