[proofplan] The hypotheses are used through the existence of the central-path minimizer $x(t) \in C^\circ$ for each $t > 0$; uniqueness only makes this minimizer canonical, while the convergence argument uses only minimality and the lower bound on $\Phi$. We compare the central-path point $x(t)$ with interior feasible approximations of a fixed optimal solution. Since the optimal solution may lie on the boundary of $C$, convexity and nonempty interior allow us to approach it by points in $C^\circ$. The minimality of $x(t)$ for $F_t$ then gives an upper bound on $c \cdot x(t)$ whose error vanishes as $t \to \infty$. Passing to an accumulating sequence and using closedness of $C$ proves feasibility and optimality of the limit. [/proofplan] [step:Approximate an optimal solution by interior feasible points] Define the optimal set \begin{align*} S^* := \{x \in C : c \cdot x = p^*\}. \end{align*} Choose an optimal point $x^* \in S^*$, which exists by hypothesis. Since $C^\circ \neq \varnothing$, choose $y \in C^\circ$. For $a \in \mathbb{R}^n$ and $\rho > 0$, write $B(a,\rho) := \{u \in \mathbb{R}^n : |u-a| < \rho\}$ for the open Euclidean ball of centre $a$ and radius $\rho$. For each integer $k \geq 2$, define \begin{align*} z_k := \left(1 - \frac{1}{k}\right)x^* + \frac{1}{k}y. \end{align*} We claim that $z_k \in C^\circ$ and $z_k \to x^*$ in $\mathbb{R}^n$. Because $y \in C^\circ$, there exists $r > 0$ such that $B(y,r) \subset C$. Let $h \in \mathbb{R}^n$ satisfy \begin{align*} |h| < \frac{r}{k}. \end{align*} Then \begin{align*} z_k + h = \left(1 - \frac{1}{k}\right)x^* + \frac{1}{k}(y + kh). \end{align*} Since $|kh| < r$, we have $y + kh \in B(y,r) \subset C$. Also $x^* \in C$, and $C$ is convex, so $z_k + h \in C$. Hence \begin{align*} B\left(z_k,\frac{r}{k}\right) \subset C, \end{align*} proving $z_k \in C^\circ$. Finally, \begin{align*} z_k - x^* = \frac{1}{k}(y - x^*). \end{align*} Thus $z_k \to x^*$ as $k \to \infty$. [guided] The only reason for introducing the points $z_k$ is that the minimizing property of $x(t)$ is available only against competitors in $C^\circ$, while an optimal solution $x^*$ may lie on the boundary of $C$. We therefore move slightly from $x^*$ toward a fixed interior point. Define the optimal set \begin{align*} S^* := \{x \in C : c \cdot x = p^*\}. \end{align*} Choose $x^* \in S^*$ and choose $y \in C^\circ$. For $a \in \mathbb{R}^n$ and $\rho > 0$, write $B(a,\rho) := \{u \in \mathbb{R}^n : |u-a| < \rho\}$ for the open Euclidean ball of centre $a$ and radius $\rho$. For every integer $k \geq 2$, define \begin{align*} z_k := \left(1 - \frac{1}{k}\right)x^* + \frac{1}{k}y. \end{align*} This is a convex combination of $x^*$ and $y$, but we need more than membership in $C$: we need membership in $C^\circ$. Since $y$ is an interior point, there exists $r > 0$ such that $B(y,r) \subset C$. Let $h \in \mathbb{R}^n$ satisfy \begin{align*} |h| < \frac{r}{k}. \end{align*} Then $|kh| < r$, so $y + kh \in B(y,r) \subset C$. We rewrite $z_k + h$ as \begin{align*} z_k + h = \left(1 - \frac{1}{k}\right)x^* + \frac{1}{k}(y + kh). \end{align*} Both $x^*$ and $y + kh$ belong to $C$, and the coefficients displayed above are nonnegative and sum to $1$. By convexity of $C$, this implies $z_k + h \in C$. Since this holds for every $h$ satisfying the displayed bound, we have \begin{align*} B\left(z_k,\frac{r}{k}\right) \subset C, \end{align*} so $z_k \in C^\circ$. The convergence is direct: \begin{align*} z_k - x^* = \frac{1}{k}(y - x^*). \end{align*} Taking Euclidean norms gives \begin{align*} |z_k - x^*| = \frac{|y - x^*|}{k} \to 0, \end{align*} so $z_k \to x^*$ in $\mathbb{R}^n$. [/guided] [/step] [step:Use central-path minimality to bound the objective from above] Let $M \in \mathbb{R}$ be a lower bound for $\Phi$ on $C^\circ$, so $\Phi(x) \geq M$ for every $x \in C^\circ$. For each $t > 0$, define the central-path objective $F_t:C^\circ \to \mathbb{R}$ by \begin{align*} F_t(u) := t\, c \cdot u + \Phi(u). \end{align*} Fix $k \geq 2$. Since $x(t)$ minimizes $F_t$ over $C^\circ$ and $z_k \in C^\circ$, for every $t > 0$ we have \begin{align*} t\, c \cdot x(t) + \Phi(x(t)) \leq t\, c \cdot z_k + \Phi(z_k). \end{align*} Using $\Phi(x(t)) \geq M$ and dividing by $t > 0$ gives \begin{align*} c \cdot x(t) \leq c \cdot z_k + \frac{\Phi(z_k) - M}{t}. \end{align*} [/step] [step:Pass to an accumulating sequence and identify the objective value] Let $\bar{x} \in \mathbb{R}^n$ be an accumulation point of $x(t)$ as $t \to \infty$. Then there exists a sequence $(t_j)_{j=1}^\infty$ in $(0,\infty)$ such that $t_j \to \infty$ and $x(t_j) \to \bar{x}$ in $\mathbb{R}^n$. For each $j$, $x(t_j) \in C^\circ \subset C$. Since $C$ is closed and $x(t_j) \to \bar{x}$, we have $\bar{x} \in C$. Therefore, by the definition of $p^*$, $p^* \leq c \cdot \bar{x}$. Fix $k \geq 2$. Applying the estimate from the previous step with $t = t_j$ gives \begin{align*} c \cdot x(t_j) \leq c \cdot z_k + \frac{\Phi(z_k) - M}{t_j}. \end{align*} Because $k$ is fixed, the quantity $\Phi(z_k) - M$ is fixed, and therefore \begin{align*} \frac{\Phi(z_k)-M}{t_j} \to 0 \end{align*} as $j \to \infty$. Passing to the limit as $j \to \infty$, using continuity of the [linear map](/page/Linear%20Map) $x \mapsto c \cdot x$, gives $c \cdot \bar{x} \leq c \cdot z_k$. Now let $k \to \infty$. Since $z_k \to x^*$ and $x \mapsto c \cdot x$ is continuous, $c \cdot \bar{x} \leq c \cdot x^* = p^*$. Combining the two inequalities yields $c \cdot \bar{x} = p^*$. [/step] [step:Conclude that every accumulation point is optimal] We have shown that $\bar{x} \in C$ and $c \cdot \bar{x} = p^*$. Hence $\bar{x} \in S^*$ by the definition of the optimal set. Since $\bar{x}$ was an arbitrary accumulation point of $x(t)$ as $t \to \infty$, every accumulation point of the central path is an optimal solution. [/step]