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Use the Lagrange multiplier with the sign conventionFor every $\lambda >0$, this is a strictly convex quadratic in $x$, with unique global minimizerThese formulas follow by setting the three partial derivatives of the optimization Lagrangian to zero; its Hessian is $\mathrm{diag}(2\lambda ,2\lambda ,4)$, which is positive-definite.

Choose $\lambda >0$ to make this minimizer feasible. Its constraint value is $5/(2{\lambda }^2)-\lambda /4$, so the scalar equation isThe left side of the first equation is strictly decreasing on $(0,\infty )$, with limits $+\infty$ and $-\infty$. Thus exactly one positive multiplier exists and can be obtained by a bracketed scalar root search. The Lagrangian sufficiency theorem proves that its associated feasible point is the global optimum; strict Lagrangian convexity makes that optimum unique.

Consequently the unique solution isIts objective value is $-5/\lambda +{\lambda }^2/8$. This scalar multiplier certificate for a quadratic equality constraint does not require finding the multiplier in radicals, and does not mistake the nonconvex equality surface for a convex set.

The stated portfolio has initial wealth $5$ and terminal payoffs $W_u=4\cdot 6-3\cdot 7=3$ and $W_d=4\cdot 6-3\cdot 4=12$. For CRRA utility, $U^{\prime }(w)=w^{-R}$. The market is complete, and an interior optimum in its positive payoff domain has marginal utility proportional to the unique state-price density. ThereforeThe Lagrange multiplier is $y^{\ast }=9/(10\sqrt{3})$, and both state equations $U^{\prime }(W)=y^{\ast }Z$ are satisfied. Part (b) verifies sufficiency and strict concavity gives uniqueness of the optimal payoff. This illustrates risk aversion recovered from an optimal two-state payoff.

The ungauge-fixed Polyakov action has worldsheet diffeomorphism and Weyl transformation symmetries. They must survive quantization so that conformal gauge remains a valid gauge choice. In a curved target, the string nonlinear sigma model has a field-dependent coupling $g_{ab}(X)$. Its worldsheet Weyl anomaly is controlled by the renormalization of this coupling, the sigma-model beta function.

Restore the Euclidean normalization $(4\pi {\alpha }^{\prime }{)}^{-1}\int g_{ab}(X)\partial X^a\partial X^b$. Use a covariant background field expansion of a string sigma model, with geodesic fluctuations about a slowly varying embedding. The quadratic fluctuation operator contains the target Riemann curvature tensor coupled to two background embedding derivatives. The ultraviolet coincident propagator is $⟨{\xi }^a{\xi }^b{⟩}_{\mathrm{UV}}=\alpha \prime g^{ab}\log (\Lambda /\mu )$: the kinetic operator supplies $2\pi \alpha \prime$, and the two-dimensional momentum integral supplies $(2\pi {)}^{-1}\log (\Lambda /\mu )$. Contracting the curvature vertex with this propagator gives a logarithmic metric counterterm proportional to $\alpha \prime R_{ab}$. The resulting one-loop metric counterterm givesThe trace of the worldsheet stress tensor contains this coefficient multiplying $\partial X^a\partial X^b$, together with the curvature anomaly controlled by total central charge. Thus, with zero antisymmetric background, constant dilaton, and the critical matter/ghost system,The central-charge condition also requires $d=26$ if there is no extra internal conformal theory. This is a leading-order equation: higher-curvature terms enter the beta function at higher orders in ${\alpha }^{\prime }$, so a Ricci-flat metric alone is not a general all-orders quantum consistency criterion.

To derive the local T-duality, work where the spacelike Killing vector has nonzero norm $V>0$. Use dimensionless coordinates and ${\alpha }^{\prime }=1$ units for the duality formulas. The isometry makes the action depend on $z$ only through its derivatives. Replace those derivatives by an independent worldsheet one-form $A_{\mu }$, and enforce its flatness with a Lagrange multiplier $\tilde{z}$. With ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$ and ${ϵ}^{\tau \sigma }=1$, the relevant first-order worldsheet duality action isVarying $\tilde{z}$ sets $dA=0$. Locally $A=dz$, recovering the original theory. For the other elimination, integrate the multiplier term by parts. Varying $A_{\nu }$ gives $V{A}^{\nu }={ϵ}^{\mu \nu }{\partial }_{\mu }\tilde{z}$, orSubstituting into the entire first-order worldsheet duality action, including the multiplier term, givesThe multiplier contribution is necessary for the sign of the dual kinetic term. This is the first-order action for Abelian worldsheet duality, with Buscher rulesThe coordinate map is ${\partial }_{\tau }\tilde{z}=V{\partial }_{\sigma }z$, ${\partial }_{\sigma }\tilde{z}=V{\partial }_{\tau }z$. The original $z$ equation is precisely the integrability condition for $\tilde{z}$; conversely the original identity $d(dz)=0$ becomes the dual equation. Thus this transformation exchanges an equation of motion with the Bianchi identity for an Abelian p-form. It reverses one chiral derivative and preserves the other.

For full closed-string equivalence, the local derivation must include the global data. For a free compact circle isometry, the period of $\tilde{z}$ and the allowed gauge-field holonomies are chosen so that the Polyakov path integral exchanges momentum and winding modes. A circle of radius $R$ becomes a circle of radius ${\alpha }^{\prime }/R$ when dimensions are restored. A spacelike Killing field alone does not fix these periods or ensure a free global circle action; the global qualifications of Abelian T-duality are additional to the local field transformation.

At the quantum level integrating out $A_{\mu }$ also gives a regulated functional determinant. The Buscher dilaton shift isin the stated units and conventional Euclidean dilaton normalization. It generates the dilaton curvature coupling even though the original dilaton vanished. The dilaton curvature coupling is defined before fixing conformal gauge; its metric variation improves the worldsheet stress tensor even when the reference worldsheet is flat. As a normalization check, $e^{-2\tilde{\Phi }}\sqrt{|\tilde{g}|}=e^{-2\Phi }\sqrt{|g|}$ for this block-diagonal background. The local dual metric by itself is therefore only part of the equivalent quantum background.

With a nonconstant dilaton and no antisymmetric field, the leading metric-dilaton Weyl condition istogether with the scalar dilaton anomaly condition. It does not demand ${\tilde{R}}_{ab}=0$ separately. The new Ricci tensor can be nonzero while the Hessian of the shifted dilaton compensates it.

For a direct example, take flat polar coordinates away from the origin, with $d{s}^2=d{r}^2+r^{2}d{z}^2$ and flat spectator directions. The polar-coordinate T-dual background isHereBoth metric equations vanish. In critical dimension the scalar equation also holds: $4|\tilde{\nabla }\tilde{\Phi }{|}^2-4\tilde{\square }\tilde{\Phi }-\tilde{R}=4/r^2-8/r^2+4/r^2=0$. The example is local on $r>0$; the circle degenerates at the excluded origin.

Nonzero dual Ricci curvature is consistent because quantum T-duality transforms the dilaton as well as the metric. The displayed curvature equations and classical Buscher rules are interpreted at their stated leading derivative order; higher-order renormalized descriptions require the corresponding corrections and field-redefinition conventions. A constant $V$ is a special case in which the dual metric may also remain Ricci-flat.

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.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

Use $x^{\pm }=(x^0\pm x^{D-1})/\sqrt{2}$, transverse coordinates $x^i$, $i=1,\dots ,d=D-2$, and the Minkowski metricThe relativistic particle phase-space action becomesIn the light-cone gauge $x^{+}=t$, solve the mass-shell condition for $p_{+}$, assuming $p_{-}\ne 0$. The reduced phase-space action is $\int dt({\dot{x}}^{-}{p}_{-}+{\dot{x}}^i{p}_i-H)$, withThe last equality selects the future-directed momentum sector and makes positivity transparent. With $\hslash =1$ and $p_a=-i{\partial }_a$, the Schrodinger equation isThe inverse acts only on Fourier modes with nonzero $p_{-}$. Multiplication by $2p_{-}$ gives $2{\partial }_{-}{\partial }_{+}\Psi =({\Delta }_{\perp }-{\mu }^2)\Psi$. ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.

For the massive two-form field, take $\mu \ne 0$. Apply ${\partial }^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes ${\partial }^n{\partial }^m{F}_{mnp}=0$, soExpanding $F=dA$, the other divergence terms vanish by this condition, leaving $({\square }_D-{\mu }^2)A_{np}=0$. The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking $n=-$ and $n=i$, respectively, givesThe $n=+$ equation follows from these expressions: the two terms containing ${\partial }_{+}{\partial }_i{A}_{-i}$ cancel and ${\partial }_i{\partial }_j{A}_{ji}=0$. Consequently $A_{-i}$ and $A_{ij}$ are independent, each satisfying the Klein-Gordon equation with mass $\mu$. The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group $SO(D-1)$. In the analogous Proca equation, ${\partial }^m{A}_m=0$ determines $A_{+}$ from $A_{-}$ and $A_i$, leaving $D-1$ components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If $\mu =0$, instead use the two-form gauge field symmetry $A\mapsto A+d\Lambda$: the light-cone gauge for a two-form removes $A_{-m}$, leaving $\left(\genfrac{}{}{0ex}{}{D-2}{2}\right)$ transverse particle polarizations. The massive and massless counts are different.

In the closed-string mode expansion, $x^m,p_m$ are center-of-mass canonical variables, while ${\alpha }_k^i,{\tilde{\alpha }}_k^i$ are independent left- and right-moving transverse string oscillators. Their complex conjugates are ${\alpha }_{-k}^i,{\tilde{\alpha }}_{-k}^i$. The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey ${\alpha }_n^{i†}={\alpha }_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor $k>0$, $a_k^i={\alpha }_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j†}]={\delta }_{kl}{\delta }^{ij}$. HenceStarting with $N|0;p⟩=0$, a finite product with $r_{ki}$ creation operators of mode $k$ has eigenvalue ${\sum }_{k,i}k{r}_{ki}$. The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces $N=\tilde{N}$.

There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string $a$, these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts $d(d+1)/2-1$, $d(d-1)/2$, and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes $a=1$. Equivalently, regularized transverse zero-point energy gives $a=(D-2)/24$, and Lorentz consistency fixes the critical dimension of the bosonic string $D=26$.

It follows that the bosonic string mass spectrum isThe ground state has $M_0^2=-8\pi T$ and is a tachyon; level one is massless; for $N\ge 2$ the mass is $M_N=\sqrt{8\pi T(N-1)}$. The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical $L_0$.

A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the symmetric traceless square $S_0^{2}V$ of the massive little group vector space $V={\mathbb{R}}^{D-1}$. The full closed-string level is $S_0^{2}V\otimes S_0^{2}V$. For two symmetric trace-free matrices $S,\tilde{S}$, the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\tilde{S}$ with only its symmetric $ab$ entry nonzero. Their commutator gives the $ab$ antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly $\left(\genfrac{}{}{0ex}{}{D-1}{2}\right)$ particle polarizations and is described by the massive field equation with ${\mu }^2=8\pi T$. At $D=26$ this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components $A_{-i}$ and 276 components $A_{ij}$.

For the relativistic particle phase-space action, the first-class constraint $\phi =(p^2+m^2)/2$ generatesIndeed the integrand varies by $\frac{1}{2}d[ϵ(p^2-m^2)]/dt$. The canonical gauge transformation is an invariance when the gauge parameter vanishes at fixed temporal endpoints, or when all fields and the parameter are periodic. The boundary restriction matters for the proper-time modulus.

Normalize the worldline interval to $[0,1]$. Thenis invariant because $\delta s=ϵ(1)-ϵ(0)=0$. Every allowed $e$ in its orbit can be written $e(t)=s+\dot{ϵ}(t)$: set $ϵ(t)={\int }_0^t[e(u)-s]du$. Thus $s$ remains a gauge-invariant integration variable, not another removable nonconstant mode. For a worldline circle the constant gauge parameter is a residual zero mode. Without the endpoint restriction, the assertion that $s$ is invariant would not hold.

The worldline gauge-orbit determinant is the Jacobian from gauge-orbit coordinates $ϵ$ to the nonconstant part of $e$ is the Faddeev-Popov determinant of ${\partial }_t$. Equivalently, the gauge-fixing identity has the formThe determinant is taken between the appropriate boundary-condition spaces, with the modulus removed; on a circle the prime also removes the constant parameter. Gauge fixing therefore leaves a factor $\det {}^{\prime }{\partial }_t$ and a modulus measure, after division by any residual gauge volume. Even though this determinant is field independent in the present Abelian example, it is the required change-of-variables Jacobian. A Grassmann integral over the Faddeev-Popov ghosts exponentiates it:The overall determinant phase depends on the integration convention and can be absorbed into normalization. Zero modes and the same endpoint restrictions must be treated separately rather than included in an invertible determinant.

For the free-ended open string, take $0\le \sigma \le \pi$. A canonical cosine expansion at a fixed time isIt implements the Neumann boundary conditions and has ${\alpha }_n^{†}={\alpha }_{-n}$. With ${\alpha }_0^m=p^m/\sqrt{\pi T}$, its Nambu-Goto phase-space action, up to a total time derivative, isReality requires ${\lambda }_{-n}={\lambda }_n^{\ast }$. Numerical factors can be absorbed into these Lagrange multipliers. In this covariant quantization of the bosonic string the oscillators retain all $D$ spacetime components, in contrast to the transverse oscillators in the preceding solution. The canonical commutation relations areThe oscillator vacuum $|0;p⟩$ is annihilated by ${\alpha }_n^m$ for $n>0$. Its momentum label will sometimes be suppressed.

Define the matter Virasoro algebra generators using normal ordering:No additive intercept is included in this definition of $L_0$. For $n>0$ the indices of the two factors in each term add to $n$; they cannot both be negative. After normal ordering there is a positive-mode annihilation operator on the right, possibly accompanied by the zero mode. Hence $L_n|0;p⟩=0$ for every $n>0$. In $L_0$, commuting positive modes past negative modes formally adds $\frac{D}{2}{\sum }_{k>0}k$. This divergent constant needs a prescription, and a finite shift is an ordering ambiguity. Our convention instead puts the physical string intercept into the constraint $L_0-a$.

With this convention the matter Virasoro algebra isA different additive constant in $L_0$ would change the linear-in-$m$ central term, so stating the convention is essential.

For the worldsheet ghost fields, useand choose a ghost oscillator vacuum withThen $c_0|0{⟩}_{\mathrm{gh}}\ne 0$ since $\{b_0,c_0\}=1$. This choice specifies the ghost zero-mode doublet; it is not a claim that both zero modes annihilate one state. With the printed ghost Virasoro zero-mode convention, $L_0^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ and $L_m^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ for $m>0$. The latter follows by putting positive ghost modes on the right; a possible contraction requires $m=0$ and is absent here.

Apply the supplied BRST charge to the matter state times this ghost oscillator vacuum. Terms with a rightmost $c_m$, $m>0$, vanish, as do the positive-mode ghost generators. ThusThese one-ghost states are independent, as can also be seen by applying $b_0,b_m$. Therefore the BRST physical-state constraints of an open string areFor the matter oscillator vacuum, $L_0={\alpha }^{\prime }{p}^2$, so $p^2=1/{\alpha }^{\prime }$ and $M^2=-1/{\alpha }^{\prime }=-2\pi T$: the physical ground state is a tachyon. The momentum must satisfy this equation; the zero-momentum oscillator vacuum by itself would not be BRST-closed.

Matter and ghost generators commute with one another. Add their two algebras and write ${\mathcal{L}}_m=L_m+L_m^{\mathrm{gh}}$. The given ghost constant must be retained:Define the shifted generators ${\hat{\mathcal{L}}}_m={\mathcal{L}}_m-{\delta }_{m,0}$. ThenBRST nilpotence requires cancellation of the anomalous central term in this shifted constraint algebra, together with the intercept one already present in the charge. At $m=2,n=-2$ the remaining anomalous coefficient is $(D-26)/2$, so $D=26$. At this value the shifted total generators obey the Witt algebra. The unshifted ${\mathcal{L}}_m$ still have the displayed linear zero-mode shift; it must not be silently discarded.

For a maximization problem with $g_i(u)\le 0$ and $h_j(u)=0$, use the optimization LagrangianThe Lagrangian sufficiency theorem says: if $u^{\ast }$ is feasible, $u^{\ast }$ globally maximizes $L(\cdot ,{\lambda }^{\ast },{\mu }^{\ast })$ over its original domain, and complementary slackness holds, ${\lambda }_i^{\ast }{g}_i(u^{\ast })=0$, then $u^{\ast }$ globally maximizes $f$ over the feasible set. Equality Lagrange multipliers have no sign restriction. For every feasible $u$,which proves the theorem.

For a minimization problem, reverse the signs in the optimization Lagrangian: take $L=f+{\sum }_i{\lambda }_i{g}_i+{\sum }_j{\mu }_j{h}_j$, with ${\lambda }_i\ge 0$. If a feasible $u^{\ast }$ globally minimizes this optimization Lagrangian and satisfies complementary slackness, thenThe hypothesis is a global extremum of the Lagrangian. Merely solving its stationarity equations is insufficient; no convexity assumption is needed when the global extremum itself has been proved.

For minimization, the optimization Lagrangian isIts coefficient of $z$ is $-1-\lambda <0$, so its infimum over unrestricted $z$ is $-\infty$ for every allowable Lagrange multiplier. There is no finite global minimizer of the optimization Lagrangian to which the Lagrangian sufficiency theorem could apply.

The primal minimization problem is also unbounded: set $x=0$, $y=\sqrt{5}$ and $z=t\ge 0$. These points are feasible, and $3y-z=3\sqrt{5}-t\to -\infty$. Hence

For any feasible $x,y$, increasing $z$ lowers the objective, so the additional upper bound makes $z^{\ast }=0$. Feasibility then requires $2x-y\le 2$. The lowest feasible value of $y$ on the circle occurs at the lower intersection with $2x-y=2$. Solving gives the candidateA global certificate avoids relying on the circle sketch. Introduce Lagrange multipliers for $g\le 0$, $z\le 0$ and $h=0$:ChooseThe coefficient of $z$ vanishes, and $2\mu x^{\ast }+2\lambda =0$, $2\mu y^{\ast }+3-\lambda =0$. Therefore completing the square yieldsThe candidate minimizes $L$ globally and satisfies both active inequality constraints, so complementary slackness and the Lagrangian sufficiency theorem give