[proofplan] We show first that the optimal value is a finite extended-real number, not $-\infty$. The compact feasible sublevel set $L_\alpha$ prevents values of $f$ along feasible points from escaping to $-\infty$, because any such escaping sequence would have a convergent subsequence in $L_\alpha$ and lower semicontinuity would force $f$ to take the value $-\infty$. If the optimal value equals the compact sublevel threshold $\alpha$, then any point of $L_\alpha$ is already a minimizer. Otherwise the optimal value lies strictly below $\alpha$, so a minimizing sequence eventually lies in $L_\alpha$; compactness gives a convergent subsequence, and lower semicontinuity identifies its limit as an actual minimizer. [/proofplan] [step:Define the optimal value and place it below the compact sublevel threshold] Define the feasible objective value \begin{align*} p^* := \inf_{x \in C} f(x) \in [-\infty,+\infty]. \end{align*} Since $L_\alpha$ is nonempty, choose $x_\alpha \in L_\alpha$. By the definition of $L_\alpha$, we have $x_\alpha \in C$ and $f(x_\alpha) \leq \alpha$. Hence \begin{align*} p^* \leq f(x_\alpha) \leq \alpha. \end{align*} In particular, $p^* < +\infty$. [/step] [step:Rule out the possibility that the optimal value is $-\infty$] Assume, for contradiction, that $p^* = -\infty$. For each $k \in \mathbb{N}$, the definition of the infimum gives a point $x_k \in C$ such that \begin{align*} f(x_k) \leq -k. \end{align*} Choose $k_0 \in \mathbb{N}$ such that $k_0 \geq \max\{1,-\alpha\}$. Then for every $k \geq k_0$, \begin{align*} f(x_k) \leq -k \leq \alpha, \end{align*} so $x_k \in L_\alpha$. The sequence $(x_k)_{k \geq k_0}$ lies in the compact subset $L_\alpha \subset \mathbb{R}^n$. By the [Bolzano-Weierstrass Theorem](/theorems/628), applied to this compact Euclidean set, there exist a strictly increasing sequence of indices $(k_j)_{j \in \mathbb{N}}$ with $k_j \geq k_0$ and a point $x^* \in L_\alpha$ such that \begin{align*} x_{k_j} \to x^* \quad \text{in } \mathbb{R}^n. \end{align*} Since $f$ is lower semicontinuous on $\mathbb{R}^n$, the sequential lower-semicontinuity inequality gives \begin{align*} f(x^*) \leq \liminf_{j \to \infty} f(x_{k_j}). \end{align*} For every $j \in \mathbb{N}$, $f(x_{k_j}) \leq -k_j$, and $k_j \to \infty$. Hence \begin{align*} \liminf_{j \to \infty} f(x_{k_j}) = -\infty. \end{align*} Thus $f(x^*) \leq -\infty$, so $f(x^*) = -\infty$. This contradicts properness, since a proper extended-real-valued function never takes the value $-\infty$. Therefore \begin{align*} -\infty < p^* < +\infty. \end{align*} [guided] We must first exclude the case $p^* = -\infty$, because compactness can only produce a candidate [limit point](/page/Limit%20Point); lower semicontinuity then controls the value at that limit point. If the infimum were $-\infty$, the feasible set would contain points with arbitrarily negative objective value. Assume, for contradiction, that \begin{align*} p^* = \inf_{x \in C} f(x) = -\infty. \end{align*} By the defining property of the infimum, for every $k \in \mathbb{N}$ there exists $x_k \in C$ such that \begin{align*} f(x_k) \leq -k. \end{align*} We now force these points into the compact set $L_\alpha$. Choose $k_0 \in \mathbb{N}$ with $k_0 \geq \max\{1,-\alpha\}$. Then for every $k \geq k_0$, \begin{align*} -k \leq \alpha. \end{align*} Combining this with $f(x_k) \leq -k$ gives \begin{align*} f(x_k) \leq \alpha. \end{align*} Since also $x_k \in C$, the definition \begin{align*} L_\alpha = \{x \in C : f(x) \leq \alpha\} \end{align*} implies that $x_k \in L_\alpha$ for every $k \geq k_0$. Now compactness is used exactly once: the sequence $(x_k)_{k \geq k_0}$ is a sequence in the compact subset $L_\alpha$ of $\mathbb{R}^n$. By the [Bolzano-Weierstrass Theorem](/theorems/628), applied to the compact Euclidean set $L_\alpha$, it has a convergent subsequence whose limit remains in $L_\alpha$. Thus there are a strictly increasing index sequence $(k_j)_{j \in \mathbb{N}}$ and a point $x^* \in L_\alpha$ such that \begin{align*} x_{k_j} \to x^* \quad \text{in } \mathbb{R}^n. \end{align*} Lower semicontinuity now turns convergence of points into an inequality for function values. Since $f$ is lower semicontinuous at $x^*$, the sequential form of lower semicontinuity gives \begin{align*} f(x^*) \leq \liminf_{j \to \infty} f(x_{k_j}). \end{align*} But $f(x_{k_j}) \leq -k_j$ for every $j$, and $k_j \to \infty$, so \begin{align*} \liminf_{j \to \infty} f(x_{k_j}) = -\infty. \end{align*} Therefore \begin{align*} f(x^*) \leq -\infty. \end{align*} In the extended real line this means $f(x^*) = -\infty$. This is impossible because $f$ is proper, and properness for a function $f: \mathbb{R}^n \to (-\infty,+\infty]$ includes the condition that $f$ never assumes the value $-\infty$. The contradiction proves \begin{align*} p^* > -\infty. \end{align*} Together with the previously shown inequality $p^* \leq \alpha$, we have \begin{align*} -\infty < p^* < +\infty. \end{align*} [/guided] [/step] [step:Separate the equality case before building a compact minimizing tail] If $p^* = \alpha$, then the point $x_\alpha \in L_\alpha$ chosen above satisfies $x_\alpha \in C$ and \begin{align*} p^* \leq f(x_\alpha) \leq \alpha = p^*. \end{align*} Hence $f(x_\alpha) = p^*$, so $x_\alpha$ is a solution of the programme. It remains to consider the case \begin{align*} p^* < \alpha. \end{align*} Since $p^* \in \mathbb{R}$, for each $k \in \mathbb{N}$ the definition of infimum gives a point $y_k \in C$ such that \begin{align*} f(y_k) \leq p^* + \frac{1}{k}. \end{align*} Because $p^* \leq f(y_k)$ for every $k \in \mathbb{N}$, the [Squeeze Theorem](/theorems/627) gives \begin{align*} \lim_{k \to \infty} f(y_k) = p^*. \end{align*} Since $\alpha - p^* > 0$, choose $k_1 \in \mathbb{N}$ such that \begin{align*} \frac{1}{k_1} \leq \alpha - p^*. \end{align*} Then for every $k \geq k_1$, \begin{align*} f(y_k) \leq p^* + \frac{1}{k} \leq p^* + \frac{1}{k_1} \leq \alpha. \end{align*} Since $y_k \in C$, this gives $y_k \in L_\alpha$ for every $k \geq k_1$. [/step] [step:Extract a convergent subsequence in the strict case and use lower semicontinuity to identify a minimizer] Assume now that $p^* < \alpha$. The tail sequence $(y_k)_{k \geq k_1}$ lies in the compact set $L_\alpha$. By the [Bolzano-Weierstrass Theorem](/theorems/628), applied to this compact Euclidean set, there exist a strictly increasing sequence of indices $(m_j)_{j \in \mathbb{N}}$ with $m_j \geq k_1$ and a point $x^* \in L_\alpha$ such that \begin{align*} y_{m_j} \to x^* \quad \text{in } \mathbb{R}^n. \end{align*} Because $x^* \in L_\alpha$, we have $x^* \in C$. Lower semicontinuity of $f$ gives \begin{align*} f(x^*) \leq \liminf_{j \to \infty} f(y_{m_j}). \end{align*} Since $f(y_k) \to p^*$, the subsequence also satisfies \begin{align*} \lim_{j \to \infty} f(y_{m_j}) = p^*. \end{align*} Hence \begin{align*} f(x^*) \leq p^*. \end{align*} On the other hand, $x^* \in C$ and $p^* = \inf_{x \in C} f(x)$, so \begin{align*} p^* \leq f(x^*). \end{align*} Combining the two inequalities yields \begin{align*} f(x^*) = p^* = \inf_{x \in C} f(x). \end{align*} Thus $x^*$ is a solution of the programme. [guided] We are now in the strict case $p^* < \alpha$, and the preceding step constructed a minimizing sequence $(y_k)_{k \in \mathbb{N}}$ in $C$ whose tail lies inside the compact sublevel set $L_\alpha$. The point of forcing the tail into $L_\alpha$ is that compactness converts the minimizing sequence into a convergent subsequence. By the [Bolzano-Weierstrass Theorem](/theorems/628), applied to the compact Euclidean set $L_\alpha$, there are a strictly increasing sequence of indices $(m_j)_{j \in \mathbb{N}}$ with $m_j \geq k_1$ and a point $x^* \in L_\alpha$ such that \begin{align*} y_{m_j} \to x^* \quad \text{in } \mathbb{R}^n. \end{align*} Because $x^* \in L_\alpha$ and $L_\alpha = \{x \in C : f(x) \leq \alpha\}$, the point $x^*$ is feasible: $x^* \in C$. Lower semicontinuity is what prevents the limiting point from having objective value larger than the limiting objective values of the minimizing sequence. Since $f$ is lower semicontinuous on $\mathbb{R}^n$ and $y_{m_j} \to x^*$, the sequential lower-semicontinuity inequality gives \begin{align*} f(x^*) \leq \liminf_{j \to \infty} f(y_{m_j}). \end{align*} The full sequence satisfies $f(y_k) \to p^*$ by the [Squeeze Theorem](/theorems/627), so every subsequence has the same limit. Hence \begin{align*} \lim_{j \to \infty} f(y_{m_j}) = p^*. \end{align*} Therefore \begin{align*} f(x^*) \leq p^*. \end{align*} The reverse inequality comes from feasibility and the definition of the infimum: since $x^* \in C$ and $p^* = \inf_{x \in C} f(x)$, \begin{align*} p^* \leq f(x^*). \end{align*} Combining these inequalities gives \begin{align*} f(x^*) = p^* = \inf_{x \in C} f(x). \end{align*} Thus $x^*$ attains the infimum over $C$, so $x^*$ is a solution of the programme. [/guided] [/step]