[proofplan] We first show that the infimum of the barrier objective is finite by ruling out a sequence along which the objective tends to $-\infty$. Then we take a minimizing sequence and use bounded sublevel sets to extract a convergent subsequence in $\mathbb{R}^n$. Closedness of $C$ places the limit in $C$, while the boundary blow-up of $\Phi$ prevents the limit from lying on $\partial C$; hence the limit lies in $C^\circ$ and attains the infimum by continuity. Finally, strict convexity gives uniqueness by evaluating the objective at the midpoint of two hypothetical minimizers. [/proofplan] [step:Rule out escape to negative infinity] Define the barrier objective $F_t:C^\circ \to \mathbb{R}$ by $F_t(x) := t\,c \cdot x + \Phi(x)$. Write $\partial C$ for the topological boundary of $C$; since $C$ is closed, $\partial C = C \setminus C^\circ$. Define \begin{align*} m := \inf_{x \in C^\circ} F_t(x) \in [-\infty,\infty). \end{align*} Since $C^\circ \neq \varnothing$ and $F_t$ is real-valued on $C^\circ$, we have $m < \infty$. We prove that $m > -\infty$. Assume for contradiction that $m = -\infty$. Then there exists a sequence $(x_k)_{k=1}^\infty$ in $C^\circ$ such that \begin{align*} F_t(x_k) \leq -k \end{align*} for every $k \in \mathbb{N}$. Set $\alpha := F_t(x_1)$. For all sufficiently large $k$, we have $F_t(x_k) \leq \alpha$, so the tail of $(x_k)$ lies in the bounded sublevel set \begin{align*} S_\alpha := \{x \in C^\circ : F_t(x) \leq \alpha\}. \end{align*} Thus a subsequence $(x_{k_j})_{j=1}^\infty$ is bounded in $\mathbb{R}^n$ and has a further convergent subsequence in $\mathbb{R}^n$ by the [Bolzano-Weierstrass theorem](/theorems/171). Relabel this further subsequence as $(x_{k_j})_{j=1}^\infty$, and let $x_* \in \mathbb{R}^n$ be its limit. Since each $x_{k_j} \in C^\circ \subset C$ and $C$ is closed, $x_* \in C$. If $x_* \in \partial C$, then the boundary blow-up hypothesis gives \begin{align*} \Phi(x_{k_j}) \to \infty. \end{align*} Since $(x_{k_j})_{j=1}^\infty$ converges in $\mathbb{R}^n$, the scalar sequence $(t\,c \cdot x_{k_j})_{j=1}^\infty$ is bounded and converges to $t\,c \cdot x_*$. Hence \begin{align*} F_t(x_{k_j}) = t\,c \cdot x_{k_j} + \Phi(x_{k_j}) \to \infty, \end{align*} contradicting $F_t(x_{k_j}) \leq -k_j$. If $x_* \in C^\circ$, then $F_t$ is continuous at $x_*$ because $\Phi$ is continuous and $x \mapsto t\,c \cdot x$ is linear. Therefore \begin{align*} F_t(x_{k_j}) \to F_t(x_*) \in \mathbb{R}, \end{align*} again contradicting $F_t(x_{k_j}) \leq -k_j$. Since $x_* \in C$ and the only alternatives are $x_* \in C^\circ$ or $x_* \in \partial C$, both cases are impossible. Therefore $m > -\infty$. [guided] The first point to settle is that the infimum is not $-\infty$. This matters because the usual minimizing-sequence argument only proves existence when the sequence is trying to converge to a finite target value. Define the barrier objective $F_t:C^\circ \to \mathbb{R}$ by $F_t(x) := t\,c \cdot x + \Phi(x)$. Write $\partial C$ for the topological boundary of $C$; since $C$ is closed, $\partial C = C \setminus C^\circ$. Define \begin{align*} m := \inf_{x \in C^\circ} F_t(x). \end{align*} Because $C^\circ$ is nonempty and $F_t$ takes real values on $C^\circ$, there is at least one point $x_0 \in C^\circ$ with $F_t(x_0) \in \mathbb{R}$. Hence $m \leq F_t(x_0) < \infty$. The only possible problem is $m = -\infty$. Assume $m = -\infty$. Then for each $k \in \mathbb{N}$ we can choose $x_k \in C^\circ$ such that \begin{align*} F_t(x_k) \leq -k. \end{align*} The bounded-sublevel hypothesis now becomes useful. Let $\alpha := F_t(x_1)$ and define \begin{align*} S_\alpha := \{x \in C^\circ : F_t(x) \leq \alpha\}. \end{align*} For every sufficiently large $k$, the inequality $F_t(x_k) \leq -k$ implies $F_t(x_k) \leq \alpha$, so the tail of the sequence lies in $S_\alpha$. By hypothesis, $S_\alpha$ is bounded in $\mathbb{R}^n$. Thus the tail of $(x_k)$ is bounded. A bounded sequence in $\mathbb{R}^n$ has a convergent subsequence by the [Bolzano-Weierstrass theorem](/theorems/171). Passing to a subsequence and relabelling, we may assume \begin{align*} x_{k_j} \to x_* \end{align*} for some $x_* \in \mathbb{R}^n$. Since every $x_{k_j}$ belongs to $C^\circ \subset C$ and $C$ is closed, the limit satisfies $x_* \in C$. Now there are two possibilities. First suppose $x_* \in \partial C$. Then the boundary blow-up hypothesis applies directly to the sequence $(x_{k_j})_{j=1}^\infty \subset C^\circ$, giving \begin{align*} \Phi(x_{k_j}) \to \infty. \end{align*} At the same time, the linear term remains bounded because $x_{k_j} \to x_*$ in $\mathbb{R}^n$: \begin{align*} t\,c \cdot x_{k_j} \to t\,c \cdot x_*. \end{align*} Therefore \begin{align*} F_t(x_{k_j}) = t\,c \cdot x_{k_j} + \Phi(x_{k_j}) \to \infty, \end{align*} which contradicts $F_t(x_{k_j}) \leq -k_j$. Second suppose $x_* \in C^\circ$. Since $\Phi$ is continuous on $C^\circ$ and the [linear map](/page/Linear%20Map) $x \mapsto t\,c \cdot x$ is continuous on $\mathbb{R}^n$, the map $F_t$ is continuous at $x_*$. Therefore \begin{align*} F_t(x_{k_j}) \to F_t(x_*) \in \mathbb{R}. \end{align*} This is again incompatible with $F_t(x_{k_j}) \leq -k_j$, which forces the same sequence to tend to $-\infty$. Because $x_* \in C$ and $C = C^\circ \cup \partial C$ for a [closed set](/page/Closed%20Set), the two cases exhaust all possibilities. The contradiction proves \begin{align*} -\infty < m < \infty. \end{align*} [/guided] [/step] [step:Extract a convergent subsequence from a minimizing sequence] Since $m \in \mathbb{R}$, choose a sequence $(y_k)_{k=1}^\infty$ in $C^\circ$ such that \begin{align*} F_t(y_k) \to m. \end{align*} Choose $k_0 \in \mathbb{N}$ such that $F_t(y_k) \leq m + 1$ for all $k \geq k_0$. Define the sublevel set \begin{align*} S_{m+1} := \{x \in C^\circ : F_t(x) \leq m + 1\}. \end{align*} By hypothesis, $S_{m+1}$ is bounded in $\mathbb{R}^n$, and the tail $(y_k)_{k=k_0}^\infty$ lies in $S_{m+1}$. Hence a subsequence of $(y_k)_{k=1}^\infty$ converges in $\mathbb{R}^n$ by the [Bolzano-Weierstrass theorem](/theorems/171). Relabel the convergent subsequence as $(y_{k_j})_{j=1}^\infty$, and write \begin{align*} y_{k_j} \to y_* \end{align*} for some $y_* \in \mathbb{R}^n$. Since $C$ is closed and $y_{k_j} \in C$ for every $j$, we have $y_* \in C$. [/step] [step:Use boundary blow-up to place the limit in the interior] We claim that $y_* \notin \partial C$. If $y_* \in \partial C$, then the boundary blow-up hypothesis applied to the sequence $(y_{k_j})_{j=1}^\infty \subset C^\circ$ gives \begin{align*} \Phi(y_{k_j}) \to \infty. \end{align*} Since $y_{k_j} \to y_*$ in $\mathbb{R}^n$, the linear terms satisfy \begin{align*} t\,c \cdot y_{k_j} \to t\,c \cdot y_*. \end{align*} Therefore \begin{align*} F_t(y_{k_j}) = t\,c \cdot y_{k_j} + \Phi(y_{k_j}) \to \infty. \end{align*} This contradicts $F_t(y_{k_j}) \to m \in \mathbb{R}$. Hence $y_* \notin \partial C$. Since $y_* \in C$ and $C$ is closed, it follows that $y_* \in C^\circ$. [/step] [step:Pass to the limit and attain the infimum] By hypothesis, $\Phi$ is continuous on $C^\circ$. Define the map $\ell: \mathbb{R}^n \to \mathbb{R}$ by $\ell(x) := t\,c \cdot x$. The map $\ell$ is linear and therefore continuous. Thus $F_t = \ell|_{C^\circ} + \Phi$ is continuous on $C^\circ$. Since $y_* \in C^\circ$ and $y_{k_j} \to y_*$, continuity gives \begin{align*} F_t(y_*) = \lim_{j \to \infty} F_t(y_{k_j}) = m. \end{align*} Thus $y_*$ is a minimizer of $F_t$ on $C^\circ$. [/step] [step:Use strict convexity to rule out two minimizers] Let $x_1, x_2 \in C^\circ$ be minimizers of $F_t$ on $C^\circ$. We prove that $x_1 = x_2$. Assume for contradiction that $x_1 \neq x_2$. Since $C$ is convex and $x_1,x_2 \in C^\circ$, the midpoint \begin{align*} x_m := \frac{x_1 + x_2}{2} \end{align*} belongs to $C^\circ$. For $a \in \mathbb{R}^n$ and $r>0$, write $B(a,r) := \{z \in \mathbb{R}^n : |z-a|<r\}$ for the open Euclidean ball. Indeed, choose $r_1,r_2 > 0$ such that $B(x_1,r_1) \subset C$ and $B(x_2,r_2) \subset C$. If \begin{align*} 0 < r < \frac{1}{2}\min\{r_1,r_2\}, \end{align*} then for every $h \in \mathbb{R}^n$ with $|h| < r$ we have $x_1 + 2h \in C$ and $x_2 \in C$, so convexity of $C$ gives \begin{align*} x_m + h = \frac{x_1 + 2h}{2} + \frac{x_2}{2} \in C. \end{align*} Hence $B(x_m,r) \subset C$, and therefore $x_m \in C^\circ$. The linear part of $F_t$ satisfies \begin{align*} t\,c \cdot x_m = \frac{1}{2}t\,c \cdot x_1 + \frac{1}{2}t\,c \cdot x_2. \end{align*} Strict convexity of $\Phi$ on $C^\circ$ and the assumption $x_1 \neq x_2$ give \begin{align*} \Phi(x_m) < \frac{1}{2}\Phi(x_1) + \frac{1}{2}\Phi(x_2). \end{align*} Adding these two relations yields \begin{align*} F_t(x_m) < \frac{1}{2}F_t(x_1) + \frac{1}{2}F_t(x_2). \end{align*} Since both $x_1$ and $x_2$ are minimizers, $F_t(x_1)=F_t(x_2)=m$, so \begin{align*} F_t(x_m) < m. \end{align*} This contradicts the definition of $m$ as the infimum of $F_t$ over $C^\circ$. Therefore $x_1=x_2$. Combining existence from the previous step with uniqueness, there is a unique minimizer $x(t) \in C^\circ$ of the barrier subproblem. [/step]