[proofplan] We first prove [weak duality](/theorems/2549) directly from the definition of the dual cone in the finite-dimensional conic program stated above. For the primal Slater case, we encode perturbations of the equality constraint in a convex value function and separate the closure of the epigraph at the boundary point corresponding to the optimal value. The Slater point makes the separating hyperplane nonvertical after restricting to the relevant affine hull, so it can be normalised into a dual feasible multiplier attaining value $p^*$. The dual Slater case is proved by the analogous hypograph separation argument, using the bipolar identity for closed convex cones; the closed convexity of the proper cone $K$ is what supplies the bipolar step, while the two Slater hypotheses themselves supply the required interior points. [/proofplan]

[step:Prove weak duality from the dual cone inequality]Let $x \in E$ be primal feasible for the conic program, so $x \in K$ and $Ax=b$. Let $y \in F$ be dual feasible, so $c-A^*y \in K^*$. By the definition of the dual cone $K^*$, \begin{align*} (c-A^*y,x)_E \geq 0. \end{align*} Using the definition of the adjoint $A^*:F \to E$, we have \begin{align*} (A^*y,x)_E = (y,Ax)_F = (y,b)_F = (b,y)_F. \end{align*} Therefore \begin{align*} (c,x)_E - (b,y)_F = (c-A^*y,x)_E \geq 0. \end{align*} Taking the infimum over primal feasible $x$ and the supremum over dual feasible $y$, where $p^*$ and $d^*$ denote the primal and dual optimal values from the theorem statement, gives \begin{align*} d^* \leq p^*. \end{align*}[/step]

[guided]The dual cone condition is designed precisely to make every primal feasible point dominate every dual feasible point. Fix a primal feasible point $x \in E$ and a dual feasible point $y \in F$. The feasibility conditions are \begin{align*} x \in K,\qquad Ax=b,\qquad c-A^*y \in K^*. \end{align*} Since $c-A^*y$ belongs to $K^*$ and $x$ belongs to $K$, the defining inequality for the dual cone gives \begin{align*} (c-A^*y,x)_E \geq 0. \end{align*} Now expand this expression. The adjoint $A^*:F \to E$ is defined by the identity \begin{align*} (A^*y,x)_E = (y,Ax)_F \end{align*} for every $x \in E$ and $y \in F$. Since $Ax=b$, this becomes \begin{align*} (A^*y,x)_E = (y,b)_F = (b,y)_F. \end{align*} Substituting into the dual cone inequality yields \begin{align*} (c,x)_E - (b,y)_F \geq 0. \end{align*} Thus every primal feasible objective value is at least every dual feasible objective value. Taking the infimum over all primal feasible $x$ and the supremum over all dual feasible $y$ gives \begin{align*} d^* \leq p^*. \end{align*} This is weak duality.[/guided]

[step:Build the perturbation value function for the primal problem] Throughout the proof, $E$ and $F$ carry their given finite-dimensional inner-product norms, both denoted by $|\cdot|$; every subspace and product space is equipped with the induced norm and induced topology. Assume that there exists a primal Slater point $x_0 \in \operatorname{int}K$ with $Ax_0=b$ and that $p^* \in \mathbb{R}$. Define the perturbation value function \begin{align*} \varphi:F \to \mathbb{R}\cup\{+\infty\} \end{align*} by \begin{align*} \varphi(v) := \inf\{(c,x)_E : x \in K,\ Ax=b+v\} \end{align*} for $v\in F$. Then $\varphi(0)=p^*$. Its epigraph is the convex set \begin{align*} \operatorname{epi}\varphi = \{(v,t)\in F\times\mathbb{R} : \text{there exists }x\in K\text{ with }Ax=b+v\text{ and }(c,x)_E\leq t\}. \end{align*} Convexity follows because $K$ is convex, $A$ is linear, and the objective map $x \mapsto (c,x)_E$ is affine. The Slater point gives $0 \in \operatorname{int}(\operatorname{dom}\varphi)$ relative to $\operatorname{Range}A$. Indeed, since $x_0 \in \operatorname{int}K$, there exists $\varepsilon>0$ such that $x_0+h\in K$ whenever $h\in E$ and $|h|<\varepsilon$. On the finite-dimensional space $\operatorname{Range}A$, choose a linear right inverse \begin{align*} R:\operatorname{Range}A \to E \end{align*} with $A(Rw)=w$ for every $w\in\operatorname{Range}A$. Since $R$ is a [linear map](/page/Linear%20Map) between finite-dimensional normed spaces, it is continuous, so there exists $\delta>0$ such that $|Rv|<\varepsilon$ whenever $v\in\operatorname{Range}A$ and $|v|<\delta$. For every such $v$, the point $x_0+Rv$ belongs to $K$ and satisfies \begin{align*} A(x_0+Rv)=b+v. \end{align*} Hence $\varphi(v)<+\infty$ for every $v\in\operatorname{Range}A$ with $|v|<\delta$. Thus $0$ lies in the relative interior of $\operatorname{dom}\varphi$ in $\operatorname{Range}A$. Let $\varphi_A:\operatorname{Range}A\to\mathbb{R}\cup\{+\infty\}$ denote the restriction of $\varphi$ to $\operatorname{Range}A$. The function $\varphi_A$ is convex because $\varphi$ is convex, it is proper because $\varphi_A(0)=p^*\in\mathbb{R}$, and the preceding paragraph proves that its effective domain has nonempty relative interior containing $0$ when $\operatorname{Range}A$ is viewed as a finite-dimensional [vector space](/page/Vector%20Space). Therefore [Convex Functions Are Locally Lipschitz](/theorems/3086), applied to the proper convex function $\varphi_A$ on the finite-dimensional vector space $\operatorname{Range}A$, gives relative continuity of $\varphi_A$ at $0$; equivalently, $\varphi$ is relatively continuous at $0$ along $\operatorname{Range}A$. [/step]

[step:Separate the closed epigraph inside its affine hull]Let $W:=\operatorname{Range}A\times\mathbb{R}$, regarded as a finite-dimensional [inner product space](/page/Inner%20Product%20Space) with the product [inner product](/page/Inner%20Product) inherited from $F\times\mathbb{R}$. Define \begin{align*} C:=\overline{\operatorname{epi}\varphi\cap W}^{\,W}. \end{align*} The point $(0,p^*)$ belongs to $C$ because $\varphi(0)=p^*$ and, by the definition of the infimum, there are feasible points with objective values decreasing to $p^*$ if the infimum is not attained. For every $\alpha<p^*$, the point $(0,\alpha)$ does not belong to $C$: if $(0,\alpha)\in C$, then there would be a sequence $(v_k,t_k)\in\operatorname{epi}\varphi\cap W$ with $v_k\to0$ in $\operatorname{Range}A$ and $t_k\to\alpha$. Relative continuity of $\varphi$ at $0$ gives $\varphi(v_k)\to\varphi(0)=p^*$, while $\varphi(v_k)\leq t_k$ gives $p^*\leq\alpha$, contradicting $\alpha<p^*$. Hence $(0,p^*)$ is a boundary point of $C$ in $W$. By the [Supporting Hyperplane Theorem](/theorems/2551) for finite-dimensional closed convex sets, applied to the boundary point $(0,p^*)$ of $C$ in the space $W$, there exist $\eta\in\operatorname{Range}A$ and $\mu\in\mathbb{R}$, not both zero, such that \begin{align*} (\eta,v)_F+\mu t \geq \mu p^* \end{align*} for every $(v,t)\in C$. Since $(v,t+\rho)\in C$ whenever $(v,t)\in C$ and $\rho\geq0$, this inequality forces $\mu\geq0$. We claim that $\mu>0$. If $\mu=0$, then \begin{align*} (\eta,v)_F \geq 0 \end{align*} for every $v\in\operatorname{dom}\varphi\cap\operatorname{Range}A$. Since $0$ is an interior point of $\operatorname{dom}\varphi$ relative to $\operatorname{Range}A$, for every $w\in\operatorname{Range}A$ there exists $r_w>0$ such that $sw\in\operatorname{dom}\varphi$ whenever $|s|<r_w$. Applying the displayed inequality to $sw$ and to $-sw$ gives $(\eta,w)_F=0$. Hence $\eta=0$ because $\eta\in\operatorname{Range}A$, contradicting that $(\eta,\mu)$ is nonzero. Therefore $\mu>0$. Dividing the separating inequality by $\mu$, define \begin{align*} q := \frac{\eta}{\mu}\in F. \end{align*} Then \begin{align*} (q,v)_F+t \geq p^* \end{align*} for every $(v,t)\in\operatorname{epi}\varphi\cap W$, hence for every perturbation point generated by an element of $K$.[/step]

[guided]The point $(0,p^*)$ need not lie in the epigraph itself, because the primal infimum may fail to be attained. The correct object to support is therefore the closed epigraph in the relevant affine space. We set $W:=\operatorname{Range}A\times\mathbb{R}$ and \begin{align*} C:=\overline{\operatorname{epi}\varphi\cap W}^{\,W}. \end{align*} By the definition of $p^*=\varphi(0)$, feasible objective values can be chosen arbitrarily close to $p^*$ from above, so $(0,p^*)\in C$. On the other hand, no point $(0,\alpha)$ with $\alpha<p^*$ belongs to $C$. Indeed, if $(0,\alpha)\in C$, then there are points $(v_k,t_k)\in\operatorname{epi}\varphi\cap W$ with $v_k\to0$ in $\operatorname{Range}A$ and $t_k\to\alpha$. The Slater argument placed $0$ in the relative interior of $\operatorname{dom}\varphi$ relative to $\operatorname{Range}A$. The restricted function $\varphi_A:\operatorname{Range}A\to\mathbb{R}\cup\{+\infty\}$ is proper and convex, and $0$ lies in the relative interior of its effective domain, so [Convex Functions Are Locally Lipschitz](/theorems/3086) gives $\varphi(v_k)=\varphi_A(v_k)\to\varphi_A(0)=p^*$. Since $(v_k,t_k)$ lies in the epigraph, $\varphi(v_k)\leq t_k$ for every $k$, and passing to the limit gives $p^*\leq\alpha$, contradicting $\alpha<p^*$. Thus $(0,p^*)$ is a boundary point of $C$ in $W$. We now apply the [Supporting Hyperplane Theorem](/theorems/2551) for finite-dimensional closed convex sets to this boundary point inside $W$. The restriction to $W$ is essential: it removes normals that only see directions orthogonal to $\operatorname{Range}A$ and therefore carry no dual information. The theorem gives $\eta\in\operatorname{Range}A$ and $\mu\in\mathbb{R}$, not both zero, such that \begin{align*} (\eta,v)_F+\mu t \geq \mu p^* \end{align*} for all $(v,t)\in C$. The epigraph is upward closed in the $t$-direction. Thus if $(v,t)\in C$ and $\rho\geq0$, then $(v,t+\rho)\in C$. Substituting this into the supporting inequality gives \begin{align*} (\eta,v)_F+\mu t+\mu\rho \geq \mu p^*. \end{align*} Letting $\rho$ grow shows that $\mu$ cannot be negative, so $\mu\geq0$. It remains to rule out $\mu=0$. If $\mu=0$, then \begin{align*} (\eta,v)_F \geq 0 \end{align*} for every $v\in\operatorname{dom}\varphi\cap\operatorname{Range}A$. The Slater point proved that $0$ is an interior point of this domain relative to $\operatorname{Range}A$. Hence, for every $w\in\operatorname{Range}A$, both small positive and small negative multiples of $w$ lie in the domain. Applying the inequality to $sw$ and $-sw$ gives $(\eta,w)_F=0$. Because this holds for all $w\in\operatorname{Range}A$ and $\eta\in\operatorname{Range}A$, we get $\eta=0$. This contradicts the nonzero normal condition. Therefore $\mu>0$. Since $\mu>0$, we may normalise the hyperplane. Define \begin{align*} q := \frac{\eta}{\mu}\in F. \end{align*} Dividing by $\mu$ gives \begin{align*} (q,v)_F+t \geq p^* \end{align*} for every $(v,t)\in\operatorname{epi}\varphi\cap W$.[/guided]

[step:Extract an optimal dual multiplier from the supporting hyperplane]For every $x\in K$, define \begin{align*} v_x := Ax-b\in F,\qquad t_x := (c,x)_E\in\mathbb{R}. \end{align*} Then $(v_x,t_x)\in\operatorname{epi}\varphi\cap W$, so the normalised supporting inequality gives \begin{align*} (q,Ax-b)_F+(c,x)_E \geq p^*. \end{align*} Using the adjoint identity, this is \begin{align*} (c+A^*q,x)_E-(q,b)_F \geq p^* \end{align*} for every $x\in K$. Set \begin{align*} y^* := -q \in F. \end{align*} The preceding inequality becomes \begin{align*} (c-A^*y^*,x)_E \geq p^*-(b,y^*)_F \end{align*} for every $x\in K$. Taking $x=0$, which belongs to $K$ because $K$ is a cone, gives \begin{align*} 0 \geq p^*-(b,y^*)_F, \end{align*} so \begin{align*} (b,y^*)_F \geq p^*. \end{align*} For arbitrary $x\in K$ and arbitrary $\lambda>0$, apply the same inequality to $\lambda x\in K$: \begin{align*} \lambda(c-A^*y^*,x)_E \geq p^*-(b,y^*)_F. \end{align*} Dividing by $\lambda$ and letting $\lambda\to+\infty$ gives \begin{align*} (c-A^*y^*,x)_E \geq 0 \end{align*} for every $x\in K$. Hence $c-A^*y^*\in K^*$, so $y^*$ is dual feasible. By weak duality, every dual feasible point has value at most $p^*$. Since $y^*$ is dual feasible and $(b,y^*)_F\geq p^*$, we conclude \begin{align*} (b,y^*)_F=p^*=d^*. \end{align*} Thus the dual optimum is attained at $y^*$ and there is no duality gap.[/step]

[guided]Now we turn the supporting hyperplane into a dual certificate. For each $x\in K$, define the perturbation and objective value \begin{align*} v_x := Ax-b\in F,\qquad t_x := (c,x)_E\in\mathbb{R}. \end{align*} The pair $(v_x,t_x)$ lies in $\operatorname{epi}\varphi\cap W$ because $x$ is feasible for the perturbed equality constraint $Ax=b+v_x$ and has objective value $t_x$. Therefore the normalised supporting inequality gives \begin{align*} (q,Ax-b)_F+(c,x)_E \geq p^*. \end{align*} Using the adjoint identity $(q,Ax)_F=(A^*q,x)_E$, this becomes \begin{align*} (c+A^*q,x)_E-(q,b)_F \geq p^* \end{align*} for every $x\in K$. Define the candidate dual multiplier by \begin{align*} y^* := -q \in F. \end{align*} Then the inequality reads \begin{align*} (c-A^*y^*,x)_E \geq p^*-(b,y^*)_F \end{align*} for every $x\in K$. First test this inequality at $x=0$. Since $K$ is a cone, $0\in K$, and hence \begin{align*} 0 \geq p^*-(b,y^*)_F. \end{align*} Thus \begin{align*} (b,y^*)_F \geq p^*. \end{align*} Next we prove dual feasibility. Fix an arbitrary $x\in K$ and let $\lambda>0$. Since $K$ is a cone, $\lambda x\in K$. Applying the same inequality to $\lambda x$ gives \begin{align*} \lambda(c-A^*y^*,x)_E \geq p^*-(b,y^*)_F. \end{align*} Divide by $\lambda$ and let $\lambda\to+\infty$. The right-hand side divided by $\lambda$ tends to $0$, so \begin{align*} (c-A^*y^*,x)_E \geq 0. \end{align*} Because $x\in K$ was arbitrary, the definition of the dual cone gives $c-A^*y^*\in K^*$. Hence $y^*$ is dual feasible. Weak duality says every dual feasible value is at most $p^*$. Since $y^*$ is dual feasible and we already proved $(b,y^*)_F\geq p^*$, we get \begin{align*} (b,y^*)_F=p^*=d^*. \end{align*} Therefore the dual optimum is attained at $y^*$ and there is no duality gap.[/guided]

[step:Separate the dual hypograph to prove the dual Slater case]Assume now that there exists $y_0\in F$ with \begin{align*} s_0:=c-A^*y_0\in\operatorname{int}K^* \end{align*} and that $d^*\in\mathbb{R}$. Define the dual perturbation value function \begin{align*} \psi:E\to\mathbb{R}\cup\{-\infty,+\infty\} \end{align*} by \begin{align*} \psi(u):=\sup\{(b,y)_F : y\in F,\ c+u-A^*y\in K^*\} \end{align*} for $u\in E$. Then $\psi(0)=d^*$. Since $s_0\in\operatorname{int}K^*$, so $y_0$ is a dual Slater point, there exists $\varepsilon>0$ such that $s_0+r\in K^*$ whenever $r\in E$ and $|r|<\varepsilon$. Hence $y_0$ is feasible for $\psi(u)$ whenever $|u|<\varepsilon$, so $0$ is an interior point of $\operatorname{dom}\psi$. We also need $\psi$ to be finite above near $0$. If $|u|<\varepsilon$ and $y\in F$ satisfies $c+u-A^*y\in K^*$, define $s:=c+u-A^*y\in K^*$. Since $s_0-u\in K^*$ and $K^*$ is convex, the multiplier $\bar y:=(y+y_0)/2$ is feasible for the unperturbed dual problem because \begin{align*} c-A^*\bar y=\frac{1}{2}s+\frac{1}{2}(s_0-u)\in K^*. \end{align*} Since $d^*\in\mathbb{R}$ is the finite supremum of the unperturbed dual objective, $(b,\bar y)_F\leq d^*$. Therefore \begin{align*} (b,y)_F\leq 2d^*-(b,y_0)_F. \end{align*} Taking the supremum over all feasible $y$ for $\psi(u)$ gives $\psi(u)\leq 2d^*-(b,y_0)_F$ whenever $|u|<\varepsilon$. Thus $\psi$ is finite-valued on a neighbourhood of $0$. The function $\psi$ is concave: if $u_1,u_2\in E$, $0\leq\lambda\leq1$, and $y_i\in F$ satisfies $c+u_i-A^*y_i\in K^*$ for $i=1,2$, then convexity of $K^*$ gives \begin{align*} c+\lambda u_1+(1-\lambda)u_2-A^*(\lambda y_1+(1-\lambda)y_2)\in K^*, \end{align*} and the dual objective at $\lambda y_1+(1-\lambda)y_2$ is $\lambda(b,y_1)_F+(1-\lambda)(b,y_2)_F$. Taking suprema over feasible $y_1$ and $y_2$ proves concavity. Hence $-\psi$ is a proper convex function finite on a neighbourhood of $0$, and [Convex Functions Are Locally Lipschitz](/theorems/3086) gives upper semicontinuity of $\psi$ at $0$. Let $H\subset E\times\mathbb{R}$ be the hypograph of $\psi$: \begin{align*} H:=\{(u,t)\in E\times\mathbb{R}:\text{ there exists }y\in F\text{ with }c+u-A^*y\in K^*\text{ and }(b,y)_F\geq t\}. \end{align*} The set $H$ is convex. Let $D:=\overline{H}$ in $E\times\mathbb{R}$. The point $(0,d^*)$ lies on the boundary of $D$: it belongs to $D$ by the definition of the supremum. If $(0,\alpha)\in D$ for some $\alpha>d^*$, then there would be a sequence $(u_k,t_k)\in H$ with $u_k\to0$ and $t_k\to\alpha$. Since $t_k\leq\psi(u_k)$ and $\psi$ is upper semicontinuous at $0$, this would give $\alpha\leq\psi(0)=d^*$, a contradiction. Hence no point $(0,\alpha)$ with $\alpha>d^*$ belongs to $D$. By the [Supporting Hyperplane Theorem](/theorems/2551) for finite-dimensional closed convex sets, applied to the boundary point $(0,d^*)$ of $D$, there exist $z\in E$ and $\mu\in\mathbb{R}$, not both zero, such that \begin{align*} (z,u)_E+\mu t\leq \mu d^* \end{align*} for every $(u,t)\in D$. Since $H$ is downward closed in the $t$-direction, this inequality forces $\mu\geq0$. If $\mu=0$, then $(z,u)_E\leq0$ for all $u\in\operatorname{dom}\psi$; applying this to small positive and negative multiples of every vector in $E$ gives $z=0$, a contradiction. Therefore $\mu>0$. Define \begin{align*} w:=\frac{z}{\mu}\in E. \end{align*} Then \begin{align*} (w,u)_E+t\leq d^* \end{align*} for every $(u,t)\in H$. For arbitrary $y\in F$ and $s\in K^*$, set $u:=s-c+A^*y$ and $t:=(b,y)_F$. Then $(u,t)\in H$, so \begin{align*} (w,s-c+A^*y)_E+(b,y)_F\leq d^*. \end{align*} Using the adjoint identity gives \begin{align*} (w,s)_E-(w,c)_E+(Aw+b,y)_F\leq d^* \end{align*} for every $y\in F$ and every $s\in K^*$. Because $y\in F$ is arbitrary, the coefficient of $y$ must vanish; otherwise replacing $y$ by a large positive or negative multiple of that coefficient would violate the inequality. Thus $Aw+b=0$. Define \begin{align*} x^*:=-w\in E. \end{align*} The sign is consistent with the preceding equation: since $Aw+b=0$, we have $Aw=-b$, and therefore \begin{align*} Ax^*=A(-w)=-Aw=b. \end{align*} Since $s\in K^*$ is arbitrary and scalable, the inequality also implies $(w,s)_E\leq0$ for every $s\in K^*$, so $x^*=-w$ belongs to the dual cone of $K^*$. Because $K$ is closed and convex in finite dimensions, the [Finite-Dimensional Bipolar Theorem](/theorems/4108) for closed convex cones gives that the dual cone of $K^*$ is $K$. Hence $x^*\in K$. Taking $s=0$ in the inequality and using $Aw+b=0$ gives \begin{align*} -(w,c)_E\leq d^*. \end{align*} Equivalently, \begin{align*} (c,x^*)_E\leq d^*. \end{align*} The point $x^*$ is primal feasible, so $p^*\leq(c,x^*)_E$. Weak duality gives $d^*\leq p^*$. Therefore \begin{align*} d^*=p^*=(c,x^*)_E. \end{align*} Thus the primal optimum is attained at $x^*$ and there is no duality gap.[/step]

[guided]The dual Slater case should not be reduced to the first part by pretending that the free variable $y\in F$ is constrained by the cone $K^*$. Instead we repeat the separation argument directly on the dual value function. Define \begin{align*} \psi:E\to\mathbb{R}\cup\{-\infty,+\infty\} \end{align*} by \begin{align*} \psi(u):=\sup\{(b,y)_F : y\in F,\ c+u-A^*y\in K^*\}. \end{align*} The original dual value is $\psi(0)=d^*$. The Slater point gives an actual neighbourhood of feasible perturbations: since $s_0=c-A^*y_0\in\operatorname{int}K^*$, there is $\varepsilon>0$ such that $s_0+r\in K^*$ whenever $r\in E$ and $|r|<\varepsilon$. Thus $y_0$ is feasible for $\psi(u)$ for every $u\in E$ with $|u|<\varepsilon$, so every sufficiently small $u$ lies in $\operatorname{dom}\psi$. We must also rule out the possibility that $\psi(u)=+\infty$ near $0$. Fix $u\in E$ with $|u|<\varepsilon$, and suppose $y\in F$ is feasible for $\psi(u)$. Define \begin{align*} s:=c+u-A^*y\in K^*. \end{align*} Because $s_0-u\in K^*$ and $K^*$ is convex, the average multiplier $\bar y:=(y+y_0)/2$ satisfies \begin{align*} c-A^*\bar y=\frac{1}{2}s+\frac{1}{2}(s_0-u)\in K^*. \end{align*} Thus $\bar y$ is feasible for the original unperturbed dual problem. Since $d^*\in\mathbb{R}$ is the finite supremum of the original dual objective, we have $(b,\bar y)_F\leq d^*$. Expanding the average gives \begin{align*} (b,y)_F\leq 2d^*-(b,y_0)_F. \end{align*} This bound is independent of the feasible $y$, so $\psi(u)\leq 2d^*-(b,y_0)_F$ for every $|u|<\varepsilon$. Therefore $\psi$ is finite-valued on a neighbourhood of $0$. Since $\psi$ is concave, $-\psi$ is a proper convex function finite near $0$, and [Convex Functions Are Locally Lipschitz](/theorems/3086) gives upper semicontinuity of $\psi$ at $0$. Now form the hypograph \begin{align*} H:=\{(u,t)\in E\times\mathbb{R}:\text{ there exists }y\in F\text{ with }c+u-A^*y\in K^*\text{ and }(b,y)_F\geq t\}. \end{align*} This set is convex because $K^*$ is convex and the constraint and objective are affine in $(u,y,t)$. Let $D:=\overline{H}$. Since $d^*=\psi(0)$, the point $(0,d^*)$ lies in $D$, even if the dual supremum is not attained. No point $(0,\alpha)$ with $\alpha>d^*$ lies in $D$: if such a point lay in $D$, then there would be a sequence $(u_k,t_k)\in H$ with $u_k\to0$ and $t_k\to\alpha$. The Slater condition and the finite value $d^*\in\mathbb{R}$ proved above that $\psi$ is finite on a neighbourhood of $0$. Applying [Convex Functions Are Locally Lipschitz](/theorems/3086) to the convex function $-\psi$ gives upper semicontinuity of $\psi$ at $0$. Since $t_k\leq\psi(u_k)$, passing to the limit gives $\alpha\leq\psi(0)=d^*$, contradicting $\alpha>d^*$. Therefore $(0,d^*)$ is a boundary point of $D$. Apply the [Supporting Hyperplane Theorem](/theorems/2551) for finite-dimensional closed convex sets to $D$ at this boundary point. We obtain $z\in E$ and $\mu\in\mathbb{R}$, not both zero, such that \begin{align*} (z,u)_E+\mu t\leq \mu d^* \end{align*} for every $(u,t)\in D$. The hypograph is downward closed in the $t$-direction, so replacing $t$ by $t-\rho$ with $\rho\geq0$ forces $\mu\geq0$. If $\mu=0$, then the inequality says $(z,u)_E\leq0$ on a neighbourhood of $0$ in $E$. Testing both $su$ and $-su$ for small $s>0$ gives $(z,u)_E=0$ for every $u\in E$, so $z=0$, contradicting that the normal is nonzero. Hence $\mu>0$. Normalise by defining \begin{align*} w:=\frac{z}{\mu}\in E. \end{align*} Then \begin{align*} (w,u)_E+t\leq d^* \end{align*} for every $(u,t)\in H$. To extract a primal point, choose arbitrary $y\in F$ and $s\in K^*$, and set $u:=s-c+A^*y$ and $t:=(b,y)_F$. This pair lies in $H$, so \begin{align*} (w,s-c+A^*y)_E+(b,y)_F\leq d^*. \end{align*} Using $(w,A^*y)_E=(Aw,y)_F$, this becomes \begin{align*} (w,s)_E-(w,c)_E+(Aw+b,y)_F\leq d^*. \end{align*} Since $y\in F$ is unrestricted, the linear coefficient $Aw+b$ must be zero. Hence $Aw+b=0$. Define $x^*:=-w$. The sign comes from $Aw+b=0$: since $Aw=-b$, we have \begin{align*} Ax^*=A(-w)=-Aw=b. \end{align*} The same inequality, with $Aw+b=0$, holds for every $s\in K^*$. Since $s$ can be replaced by $\lambda s$ for every $\lambda>0$, it follows that $(w,s)_E\leq0$ for every $s\in K^*$. Equivalently, $(-w,s)_E\geq0$ for every $s\in K^*$, so $x^*=-w$ belongs to the dual cone of $K^*$. Because $K$ is closed and convex in finite dimensions, the [Finite-Dimensional Bipolar Theorem](/theorems/4108) for closed convex cones identifies this cone with $K$. Thus $x^*\in K$. Finally set $s=0$ in the inequality. Since $Aw+b=0$, we get \begin{align*} -(w,c)_E\leq d^*. \end{align*} This is \begin{align*} (c,x^*)_E\leq d^*. \end{align*} The point $x^*$ is primal feasible, so $p^*\leq(c,x^*)_E$. Weak duality gives $d^*\leq p^*$. Combining the inequalities yields \begin{align*} d^*=p^*=(c,x^*)_E. \end{align*} Therefore the primal optimum is attained at $x^*$ and there is no duality gap.[/guided]