[proofplan] We first use properness to choose a point where $f$ is finite, and then use coercivity to trap a useful sublevel set inside a Euclidean ball. Lower semicontinuity makes this sublevel set closed, so finite-dimensional compactness makes it compact. We then prove that the infimum is finite, choose a minimizing sequence lying in the compact sublevel set, extract a convergent subsequence, and pass to the limit using lower semicontinuity. [/proofplan] [step:Construct a nonempty compact sublevel set containing all near minimizers] Since $f$ is proper, its effective domain \begin{align*} \operatorname{dom} f := \{x \in \mathbb{R}^n : f(x) < +\infty\} \end{align*} is nonempty. Choose $x_0 \in \operatorname{dom} f$, and define the finite real number \begin{align*} c := f(x_0) + 1. \end{align*} By coercivity applied with the stricter threshold $a = c + 1$, there exists $R_{c+1} > 0$ such that $f(x) \geq c + 1$ whenever $|x| \geq R_{c+1}$. Define \begin{align*} R := \max\{R_{c+1}, |x_0|\}. \end{align*} Define the closed Euclidean ball centered at $0$ with radius $R$ by \begin{align*} \overline{B}(0,R) := \{x \in \mathbb{R}^n : |x| \leq R\}. \end{align*} Define the sublevel set \begin{align*} S := \{x \in \mathbb{R}^n : f(x) \leq c\}. \end{align*} Then $x_0 \in S$, so $S$ is nonempty. Also, if $x \in S$ and $|x| \geq R$, then $|x| \geq R_{c+1}$, hence $f(x) \geq c + 1$. This contradicts $f(x) \leq c$. Therefore every $x \in S$ satisfies $|x| < R$, and hence \begin{align*} S \subset \overline{B}(0,R). \end{align*} We next show that $S$ is closed. Let $(y_k)_{k \in \mathbb{N}}$ be a sequence in $S$ converging in $\mathbb{R}^n$ to a point $y \in \mathbb{R}^n$. Since $f$ is lower semicontinuous, \begin{align*} f(y) \leq \liminf_{k \to \infty} f(y_k). \end{align*} Because $y_k \in S$ for every $k \in \mathbb{N}$, we have $f(y_k) \leq c$ for every $k$, and hence \begin{align*} \liminf_{k \to \infty} f(y_k) \leq c. \end{align*} Thus $f(y) \leq c$, so $y \in S$. Therefore $S$ is closed. [claim:The sublevel set $S$ is compact] The set $S$ is compact. [/claim] [proof] Let $(z_k)_{k \in \mathbb{N}}$ be a sequence in $S$. Since $S \subset \overline{B}(0,R)$, each coordinate sequence $(z_{k,i})_{k \in \mathbb{N}}$ is bounded in $\mathbb{R}$ for every $i \in \{1,\dots,n\}$. Use the fact that every bounded sequence in $\mathbb{R}$ has a convergent subsequence successively on the coordinates: first extract a subsequence along which the first coordinate converges, then extract from that subsequence one along which the second coordinate converges, and continue through the $n$-th coordinate. This gives a subsequence $(z_{k_j})_{j \in \mathbb{N}}$ and [real numbers](/page/Real%20Numbers) $z_{*,1},\dots,z_{*,n}$ such that $z_{k_j,i} \to z_{*,i}$ for every $i \in \{1,\dots,n\}$. Define $z_* := (z_{*,1},\dots,z_{*,n}) \in \mathbb{R}^n$. Coordinatewise convergence in $\mathbb{R}^n$ is equivalent to Euclidean convergence, so $z_{k_j} \to z_*$. Since $S$ is closed and each $z_{k_j}$ lies in $S$, the limit $z_*$ lies in $S$. Thus every sequence in $S$ has a subsequence converging to a point of $S$. In metric spaces, compactness is equivalent to this [sequential compactness](/page/Sequential%20Compactness) property, so $S$ is compact. [/proof] [guided] The purpose of this step is to build one compact set that contains all points where $f$ is close enough to being minimal. Properness gives us at least one finite value. Since $f$ is proper, the effective domain \begin{align*} \operatorname{dom} f := \{x \in \mathbb{R}^n : f(x) < +\infty\} \end{align*} is nonempty, so we may choose $x_0 \in \operatorname{dom} f$. Define \begin{align*} c := f(x_0) + 1. \end{align*} This is a real number because $f(x_0) < +\infty$ and the codomain of $f$ excludes the value $-\infty$. Now apply coercivity with the stricter threshold $a = c + 1$. This gives a radius $R_{c+1} > 0$ such that every point sufficiently far from the origin has value at least $c+1$: \begin{align*} |x| \geq R_{c+1} \implies f(x) \geq c+1. \end{align*} Define \begin{align*} R := \max\{R_{c+1}, |x_0|\}. \end{align*} Define the closed Euclidean ball centered at $0$ with radius $R$ by \begin{align*} \overline{B}(0,R) := \{x \in \mathbb{R}^n : |x| \leq R\}. \end{align*} We then consider the sublevel set \begin{align*} S := \{x \in \mathbb{R}^n : f(x) \leq c\}. \end{align*} This set is nonempty because $f(x_0) \leq f(x_0)+1 = c$, so $x_0 \in S$. Why is $S$ bounded? The threshold must be chosen strictly above $c$. If $x \in S$ and $|x| \geq R$, then $|x| \geq R_{c+1}$, so coercivity gives $f(x) \geq c+1$. But membership in $S$ gives $f(x) \leq c$, a contradiction. Hence every point of $S$ satisfies $|x| < R$, and therefore \begin{align*} S \subset \overline{B}(0,R). \end{align*} Why is $S$ closed? Let $(y_k)_{k \in \mathbb{N}}$ be a sequence in $S$ converging to some $y \in \mathbb{R}^n$. Lower semicontinuity gives \begin{align*} f(y) \leq \liminf_{k \to \infty} f(y_k). \end{align*} Since every $y_k$ lies in $S$, each value satisfies $f(y_k) \leq c$, so \begin{align*} \liminf_{k \to \infty} f(y_k) \leq c. \end{align*} Combining the two inequalities gives $f(y) \leq c$, hence $y \in S$. Therefore $S$ contains limits of convergent sequences from $S$, so $S$ is closed. We have shown that $S$ is nonempty, closed, and bounded in $\mathbb{R}^n$. To prove compactness, take any sequence $(z_k)_{k \in \mathbb{N}}$ in $S$. Since $S \subset \overline{B}(0,R)$, each coordinate sequence $(z_{k,i})_{k \in \mathbb{N}}$ is bounded in $\mathbb{R}$. Every bounded sequence in $\mathbb{R}$ has a convergent subsequence, so we apply this one-dimensional subsequence property successively to the coordinates: first to the first coordinate, then to the second coordinate along the previously chosen subsequence, and so on through the $n$-th coordinate. This produces a subsequence $(z_{k_j})_{j \in \mathbb{N}}$ and a point $z_* = (z_{*,1},\dots,z_{*,n}) \in \mathbb{R}^n$ such that $z_{k_j,i} \to z_{*,i}$ for every coordinate $i$. Coordinatewise convergence is Euclidean convergence in $\mathbb{R}^n$, so $z_{k_j} \to z_*$. Since $S$ is closed, $z_* \in S$. Thus every sequence in $S$ has a subsequence converging to a point of $S$. In metric spaces, compactness is equivalent to this sequential compactness property, and therefore $S$ is compact. [/guided] [/step] [step:Show that the global infimum is finite] Define the infimum \begin{align*} m := \inf_{x \in \mathbb{R}^n} f(x). \end{align*} Since $x_0 \in \operatorname{dom} f$, we have \begin{align*} m \leq f(x_0) < +\infty. \end{align*} It remains to exclude $m = -\infty$. Suppose, for contradiction, that $m = -\infty$. Then for each $k \in \mathbb{N}$ there exists $y_k \in \mathbb{R}^n$ such that \begin{align*} f(y_k) \leq -k. \end{align*} Choose \begin{align*} k_0 := \max\{1, \lceil -c \rceil\}. \end{align*} Then $k \geq k_0$ implies $k \geq -c$, hence $-k \leq c$. Therefore $f(y_k) \leq -k \leq c$, so $y_k \in S$ for every $k \geq k_0$. Since $S$ is compact, the sequence $(y_k)_{k \geq k_0}$ has a subsequence $(y_{k_j})_{j \in \mathbb{N}}$ converging to some $y_* \in S$. Lower semicontinuity of $f$ at $y_*$ gives \begin{align*} f(y_*) \leq \liminf_{j \to \infty} f(y_{k_j}). \end{align*} Because $f(y_{k_j}) \leq -k_j$ and $k_j \to \infty$, for every real number $M$ there exists $j_M \in \mathbb{N}$ such that $f(y_{k_j}) \leq M$ whenever $j \geq j_M$. Hence \begin{align*} \liminf_{j \to \infty} f(y_{k_j}) = -\infty. \end{align*} Thus $f(y_*) \leq -\infty$, which contradicts the fact that $f$ takes values in $(-\infty,+\infty]$. Therefore \begin{align*} -\infty < m \leq f(x_0) < +\infty. \end{align*} [guided] We now prove that the number \begin{align*} m := \inf_{x \in \mathbb{R}^n} f(x) \end{align*} is a genuine real number. The upper bound is immediate from the point $x_0$ chosen earlier: since $x_0 \in \operatorname{dom} f$, we have $f(x_0) < +\infty$, and therefore \begin{align*} m \leq f(x_0) < +\infty. \end{align*} The only remaining danger is that the infimum might be $-\infty$. Suppose this happens. By the definition of infimum, for each $k \in \mathbb{N}$ there is a point $y_k \in \mathbb{R}^n$ with \begin{align*} f(y_k) \leq -k. \end{align*} Choose \begin{align*} k_0 := \max\{1, \lceil -c \rceil\}. \end{align*} If $k \geq k_0$, then $k \geq -c$, so $-k \leq c$. Hence \begin{align*} f(y_k) \leq -k \leq c, \end{align*} and therefore $y_k \in S$ for all $k \geq k_0$. Now compactness of $S$ becomes useful. Since the tail sequence $(y_k)_{k \geq k_0}$ lies in $S$, compactness gives a subsequence $(y_{k_j})_{j \in \mathbb{N}}$ converging to some $y_* \in S$. Lower semicontinuity at $y_*$ gives \begin{align*} f(y_*) \leq \liminf_{j \to \infty} f(y_{k_j}). \end{align*} But $f(y_{k_j}) \leq -k_j$ and $k_j \to \infty$. Thus, for every real number $M$, there exists $j_M \in \mathbb{N}$ such that $f(y_{k_j}) \leq M$ whenever $j \geq j_M$. Therefore \begin{align*} \liminf_{j \to \infty} f(y_{k_j}) = -\infty. \end{align*} Thus $f(y_*) \leq -\infty$, which is impossible because $f$ takes values in $(-\infty,+\infty]$ and never takes the value $-\infty$. Therefore $m \neq -\infty$, and we have \begin{align*} -\infty < m \leq f(x_0) < +\infty. \end{align*} [/guided] [/step] [step:Extract a convergent minimizing subsequence from the compact sublevel set] Since $m$ is finite and is the infimum of $f$ over $\mathbb{R}^n$, for each $k \in \mathbb{N}$ there exists $x_k \in \mathbb{R}^n$ such that \begin{align*} m \leq f(x_k) \leq m + \frac{1}{k}. \end{align*} Because $m \leq f(x_0)$, we have \begin{align*} m + \frac{1}{k} \leq f(x_0) + 1 = c \end{align*} for every $k \in \mathbb{N}$. Hence $x_k \in S$ for every $k \in \mathbb{N}$. Since $S$ is compact, there exists a subsequence $(x_{k_j})_{j \in \mathbb{N}}$ and a point $x_* \in S$ such that \begin{align*} x_{k_j} \to x_* \end{align*} in $\mathbb{R}^n$ as $j \to \infty$. [guided] Because $m$ is finite, we can choose points whose values approach $m$ from above. More precisely, for each $k \in \mathbb{N}$, the definition of infimum gives a point $x_k \in \mathbb{R}^n$ such that \begin{align*} m \leq f(x_k) \leq m + \frac{1}{k}. \end{align*} The lower bound is automatic from the definition of $m$ as the greatest lower bound of all values of $f$. We must check that these points stay in the compact set $S$. Since $m \leq f(x_0)$, for every $k \in \mathbb{N}$ we have \begin{align*} m + \frac{1}{k} \leq f(x_0) + 1 = c. \end{align*} Thus \begin{align*} f(x_k) \leq c, \end{align*} so $x_k \in S$ for every $k \in \mathbb{N}$. Now compactness of $S$ gives the needed convergence. The sequence $(x_k)_{k \in \mathbb{N}}$ lies entirely in $S$, and $S$ is compact, so there is a subsequence $(x_{k_j})_{j \in \mathbb{N}}$ and a point $x_* \in S$ such that \begin{align*} x_{k_j} \to x_* \end{align*} in $\mathbb{R}^n$ as $j \to \infty$. [/guided] [/step] [step:Pass to the limit using lower semicontinuity and identify the minimizer] Lower semicontinuity of $f$ at $x_*$ gives \begin{align*} f(x_*) \leq \liminf_{j \to \infty} f(x_{k_j}). \end{align*} From the construction of the minimizing sequence, \begin{align*} m \leq f(x_{k_j}) \leq m + \frac{1}{k_j}. \end{align*} Since $k_j \to \infty$, the [squeeze theorem](/theorems/627) gives \begin{align*} \lim_{j \to \infty} f(x_{k_j}) = m. \end{align*} Therefore \begin{align*} f(x_*) \leq m. \end{align*} By definition of $m$ as the infimum of all values of $f$ on $\mathbb{R}^n$, we also have \begin{align*} m \leq f(x_*). \end{align*} Consequently \begin{align*} f(x_*) = m = \inf_{x \in \mathbb{R}^n} f(x). \end{align*} Thus $f$ attains its minimum on $\mathbb{R}^n$. [guided] The final step is to show that the compactness limit $x_*$ is not merely a cluster point of almost-minimizers, but an actual minimizer. Lower semicontinuity of $f$ at $x_*$ gives \begin{align*} f(x_*) \leq \liminf_{j \to \infty} f(x_{k_j}). \end{align*} From the construction of the minimizing sequence, every term satisfies \begin{align*} m \leq f(x_{k_j}) \leq m + \frac{1}{k_j}. \end{align*} Since $k_j \to \infty$, the upper and lower bounds both converge to $m$. By the squeeze theorem, \begin{align*} \lim_{j \to \infty} f(x_{k_j}) = m. \end{align*} Therefore lower semicontinuity gives \begin{align*} f(x_*) \leq m. \end{align*} On the other hand, $m$ is the infimum of $f$ over all of $\mathbb{R}^n$, so every value of $f$ is at least $m$, including the value at $x_*$. Hence \begin{align*} m \leq f(x_*). \end{align*} Combining the two inequalities yields \begin{align*} f(x_*) = m = \inf_{x \in \mathbb{R}^n} f(x). \end{align*} Thus $x_*$ is a point where $f$ attains its minimum on $\mathbb{R}^n$. [/guided] [/step]