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Set $y_j=-\log (1-B_j)$ and let $e_C(a)=E(a,C)$. For positive integer capacities,The function $U$ is the carried load of an Erlang loss resource evaluated at its inverse blocking parametrization. It is continuous and strictly increasing from zero to $C$. Hence $F$ is a coercive function and a strictly convex function on the nonnegative orthant. Its first-order conditions are the fixed-routing Erlang fixed point equations, including zero prices at unused links.

A convex function satisfies $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$ for all $x,y$ in its domain and $0\le t\le 1$. If the second derivative exists on an interval, this is equivalent to $g^{\prime \prime }\ge 0$ there. The equivalence follows from the nondecreasing first derivative, or from the supporting-line inequalities obtained by integration. HereThus it is a strictly convex function, and in particular convex.

The exact normalizing sum can be expensive on a large loss network. The Erlang fixed point approximation replaces joint resource acceptance by a product of marginal acceptances. Here consider unit requirements $A_{jr}\in \{0,1\}$, positive integer capacities and finitely many fixed routes. Let $B_j$ be resource blocking and $q_j=1-B_j$. A route-$r$ call contributes to the offered traffic at resource $j$ after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is ${\prod }_{j\in r}{q}_j$. This is an independence approximation, not an alternative exact factorization of the stationary law in part (i).

Uniqueness follows from a convex potential for the Erlang fixed point. For a single resource, the carried load of an Erlang loss resource is $m_C(v)=v[1-E(C,v)]$. It increases strictly from zero to $C$ as $v$ increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to $\log v$ gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.

Use $p_j=-\log q_j\ge 0$. Let $v_C(p)$ be the unique offered load with $1-E(C,v_C(p))=e^{-p}$, and set $g_C(p)=v_C(p)e^{-p}=m_C(v_C(p))$, with $g_C(0)=0$. This function is strictly increasing and tends to $C$. DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore $F$ is a strictly convex function. It is also a coercive function on the nonnegative orthant: an unbounded coordinate makes its integral grow asymptotically linearly with positive slope $C_j$. A unique minimizer exists.

If resource $j$ carries some positive offered route, its inward derivative at $p_j=0$ is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by $q_j$ gives precisely $v_j={\sum }_{r:j\in r}{\rho }_r{\prod }_{k\in r\setminus \{j\}}{q}_k$. A resource with no positive offered route uniquely has $p_j=0$, hence $B_j=0$. Thus the minimizer and the fixed point coincide, proving existence and uniqueness of the Erlang fixed point for fixed unit-resource routing. This does not by itself guarantee convergence of every simultaneous substitution algorithm; the uniqueness claim concerns the solution of the equations.

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 the Erlang loss formula, assume a Poisson process of arrivals with rate $\lambda$, independent holding times with exponential distribution of mean $1/\beta$, unit capacity usage per accepted call, and rejection without waiting when all $C$ units are occupied. Here $C$ is a positive integer and the offered load is $a=\lambda /\beta$. The call population on $\{0,\dots ,C\}$ is a birth-death process; detailed balance givesBy Poisson arrivals see time averages, the fraction of arrivals rejected is the time probability of full occupancy. ThereforeThe paper's $\nu$ is the offered load, rather than necessarily an arrival rate. Exponential holding times suffice for this derivation; no unproved insensitivity of loss networks assumption is needed.

A loss network with fixed routing has capacities $C_j$, call types $r$, and a link-route incidence matrix $A$ with $A_{jr}=1$ exactly when a type-$r$ call requires link $j$. An accepted call holds one unit on every link of its route for its holding time; if any required link is full, the call is rejected everywhere. Take independent Poisson processes of calls, independent holding times, and route offered loads ${\nu }_r$. For the finite-variable formulation below, take $C_j\ge 1$; a zero-capacity link and all routes using it may be removed first.

The Erlang fixed point approximation treats blocking at different links as independent and thins the traffic offered to link $j$ by acceptance at the other links. Its reduced offered load isThe equality uses $A_{jr}\in \{0,1\}$: only routes with $A_{jr}=1$ contribute. Treating this reduced stream as Poisson leads to $B_j=E(a_j(B),C_j)$. Independence and Poisson thinning here are the approximation, not exact properties of a general loss network.

To construct the convex potential for the Erlang fixed point, write $e_C(a)=E(a,C)$ and $m_C(a)=a[1-e_C(a)]$, the carried load of an Erlang loss resource, equivalently the mean occupancy in the single-link stationary distribution. Differentiating its finite sums gives, for $a>0$,The variance is positive because the finite occupancy law is nondegenerate for $a>0$. Furthermore $e_C$ maps $[0,\infty )$ increasingly onto $[0,1)$, while $m_C$ increases from zero to $C$. SetIn particular $U(0,C)=0$, $U(\cdot ,C)$ is continuous and strictly increasing, and $U(z,C)\to C$ as $z\to \infty$.

Forthe first sum is convex, and each integrated strictly increasing function is strictly convex. Thus $F$ is a strictly convex function on $y\ge 0$. Since $U$ tends to a positive limit on every coordinate, $F$ is a coercive function there; it has a unique minimizer. Its coordinate derivative isThe Erlang fixed point equation is exactly ${\partial }_{j}F=0$. For an unused link, $a_j=0$ and the optimum is $y_j=0$; for a used link with positive offered load, the derivative at zero is negative and the optimum is positive. Hence every fixed point gives the unique minimizer, and the minimizer gives a fixed point. The Erlang fixed point is unique under these fixed-routing assumptions.

A convergent cyclic substitution for the Erlang fixed point updates one coordinate at a time, repeatedly cycling through all links:Always use the most recently available values of the other coordinates. Start, for example, with $B=0$. In $y$ coordinates this is exact coordinate descent for $F$: with other coordinates fixed, the displayed update is the unique coordinate minimizer.

Here is a convergence proof. All iterates stay in the initial compact sublevel set of the coercive function $F$, and their potential values decrease to a limit. Each single-coordinate update is continuous, since its reduced load is continuous and finite and $E(a,C)<1$. Thus a whole-sweep map $T$ is continuous. If a full-sweep subsequence converges to $z$, continuity gives $F(Tz)=F(z)$, because both consecutive potential values converge to the same limit. Strict coordinate convexity means equality can occur only when no coordinate changes during that sweep. Thus $z$ minimizes every coordinate, satisfies the nonnegative-orthant first-order conditions, and is the unique global minimizer of $F$. Every subsequential limit is therefore the same point, proving convergence of the full-sweep iterates and, by continuity of the updates, all intermediate iterates. This argument applies to sequential substitution; it does not assume simultaneous updates converge.

Let $R_s$ be the finite nonempty set of routes for source-sink pair $s$, with fixed nonnegative demand $f_s$. A Wardrop equilibrium is a feasible route-flow vector such that every route carrying positive flow has minimum total delay among the routes for its pair. Ifthis means that, for each $s$, some ${\lambda }_s$ satisfies $c_r(y)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$.

The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$ and zero otherwise, or the traversal multiplicity if repeated traversals are permitted. The source-sink route incidence matrix has $H_{sr}=1$ when route $r$ serves pair $s$ and zero otherwise. Thus $Ax=y$ gives link throughputs, and $Hx=f$ gives the fixed demands.

The feasible route-flow set is a product of finite simplexes, hence nonempty and both a compact set and a convex set. The Beckmann potentialis continuous and convex, so it attains a minimum. Its route derivative is ${\partial }_r\Phi =c_r(Ax)$. At a minimum, moving a small amount of flow from any used route to any other route for the same pair cannot decrease the potential. Hence every used route has minimum route delay, exactly the Wardrop equilibrium condition. Conversely, at a Wardrop equilibrium $x$, for any feasible $\tilde{x}$,because the used flow has cost ${\lambda }_s$ and competing routes cost at least ${\lambda }_s$. The first-order inequality for a convex function makes $x$ a global minimum. This proves both existence and the specified optimization characterization.

Since each $D_j$ is strictly increasing, its primitive is a strictly convex function. Therefore the objective is strictly convex as a function of link loads $y$, and the equilibrium throughputs $y$ are unique. The mapping $x\mapsto Ax$ need not be injective on $Hx=f$, so route flows need not be unique.

For a concrete route-flow nonuniqueness at a Wardrop equilibrium example, use one unit-demand pair with two successive stages, each containing two parallel links. There are four routes, indexed by their two link choices. Give every link delay $D(u)=u$. For every $0\le t\le 1/2$,produces load $1/2$ on all four links. Every route has delay one, so each vector is a Wardrop equilibrium with the same unique link loads.

To implement minimum average delay, minimize the total delayThe average differs only by the fixed total demand ${\sum }_s{f}_s$, provided it is positive. Choose any global minimizer $x^{\ast }$, which exists by compactness, and write $y^{\ast }=A{x}^{\ast }$. Its first-order route-exchange conditions are the Wardrop equilibrium conditions for marginal link costsA genuinely traffic-dependent choice guaranteed under exactly the printed hypotheses is the calibrated toll functionHere $v_{+}=\max (v,0)$. The calibrated base charges are nonnegative because differentiable increasing delays have nonnegative derivatives, and the added penalties are nonnegative and nondecreasing in $u$. The perceived link costs $D_j(u)+T_j(u)$ remain strictly increasing, and at $y^{\ast }$ they equal $m_j^{\ast }$. Hence $x^{\ast }$ is a tolled Wardrop equilibrium. The same strictly convex Beckmann potential argument makes all its equilibrium link loads equal to $y^{\ast }$; every tolled equilibrium therefore has the globally minimum average delay. This is optimal-flow calibrated congestion tolling.

The familiar marginal external cost toll is $T_j(u)=u{D}_j^{\prime }(u)$, whose perceived-cost primitive is exactly $u{D}_j(u)$. It implements the optimum directly when the total-delay objective is convex, for example when each $u{D}_j(u)$ is convex. That extra hypothesis does not follow from the printed strict increase of $D_j$: for two identical parallel links with $D(u)=1-e^{-u}$ and total demand six, equal loads of three form a tolled Wardrop equilibrium, but the second derivative of $uD(u)$ at three is $-e^{-3}<0$, so an unequal nearby split lowers total delay. The calibrated traffic-dependent toll above establishes the requested existence without silently assuming this additional convexity.

For a twice differentiable function, convexity on ${\mathbb{R}}^2$ is equivalent to its Hessian matrix being positive semidefinite everywhere. Here the Hessian matrix is constant:Its diagonal entries are positive, so positive semidefiniteness is equivalent to $\det H=8-{\alpha }^2\ge 0$. The required parameter range isThe endpoints give convex functions with one flat direction; strict inequalities give a Hessian matrix defining a positive-definite quadratic form and a strictly convex function.

Use the normalizationThis is ridge regression; the penalty is the squared Euclidean norm. The centering removes the need for an intercept. Differentiating gives the normal equationFor every nonzero $v$, $v^T(X^{T}X+\lambda I_p)v=\Vert Xv{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. The coefficient matrix is therefore a positive-definite matrix, even if the design matrix has deficient matrix rank. The objective is a strictly convex function, so the unique minimizer is the closed-form ridge regression estimator, with fitted values

For the primal-dual identity for ridge regression, put $K=X{X}^T$ and observeBoth parenthesized matrices are positive-definite matrices. Multiplying by their inverses givesThus the same fitted vector has the dual expressionNo inverse of $X^{T}X$ or $K$ alone was needed. This is also the kernel-ridge hat matrix for the linear kernel.

Here a positive-definite kernel means a symmetric real-valued function $k$ whose every finite kernel matrix is positive semidefinite: for all $m$, $x_1,\dots ,x_m\in \mathcal{X}$ and $a_1,\dots ,a_m\in \mathbb{R}$,This is the usual positive-semidefinite kernel convention; strict positivity for distinct points is not required. In particular $k(x,x)\ge 0$. Write $a=k(x,x)$, $b=k(x^{\prime },x^{\prime })$ and $r=k(x,x^{\prime })$. Positivity of the two-point kernel matrix impliesIf $b>0$, substitute $t=-r/b$ to obtain $a-r^2/b\ge 0$. If $b=0$, the affine expression $a+2tr$ can be nonnegative for every $t$ only if $r=0$. Both cases give the Cauchy-Schwarz inequality for positive-semidefinite kernels:The zero-diagonal case is essential when the positive-definite kernel is not strictly positive.

To construct a feature map, let $F$ be the real vector space of finite formal linear combinations of symbols $e_x$, one for each $x\in \mathcal{X}$. Define a symmetric bilinear form byThe positive-definite kernel property makes this a positive semidefinite bilinear form. It need not yet be an inner product: distinct formal vectors can have zero squared length. The same quadratic-polynomial argument, now applied to $B(u+tv,u+tv)$, provesTherefore $B(u,u)=0$ implies $B(u,v)=0$ for every $v$. The setis the radical of a bilinear form, hence a linear subspace. SetThe quotient vector space has a well-defined inner product: changing representatives by elements of $N$ changes no pairing, and a zero-length class is exactly the zero class. It is therefore an inner-product space.

Define $\varphi (x)=e_x+N$. By construction,Every positive-definite kernel thus has a feature representation, including degenerate kernels and the zero kernel. This is its canonical feature space of a positive-semidefinite kernel. The construction already gives the requested inner-product space.

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.