[proofplan] We first show that the infimum cannot be $-\infty$: otherwise a sequence with energies tending to $-\infty$ would be bounded by coercivity, and reflexivity plus weak closedness of $K$ would produce a weak limit contradicting lower semicontinuity. Once the infimum is finite, we choose a minimizing sequence. Coercivity bounds this sequence, reflexivity gives a weakly convergent subsequence, closed convexity keeps the weak limit inside $K$, and sequential weak lower semicontinuity identifies that limit as a minimizer. [/proofplan] [step:Show that sublevel sequences are bounded by coercivity] Let $(w_j)_{j=1}^{\infty}$ be a sequence in $K$ such that the [real numbers](/page/Real%20Numbers) $I[w_j]$ are bounded above by a constant $M\in\mathbb R$ for all $j\in\mathbb N$. We claim that $(w_j)_{j=1}^{\infty}$ is bounded in $X$. If not, then for every $m\in\mathbb N$ there exists an index $j_m\in\mathbb N$ such that \begin{align*} \|w_{j_m}\|_X \ge m. \end{align*} Choosing the indices recursively with $j_{m+1}>j_m$, we obtain a subsequence $(w_{j_m})_{m=1}^{\infty}$ in $K$ satisfying \begin{align*} \|w_{j_m}\|_X \to \infty. \end{align*} By coercivity of $I$ on $K$, \begin{align*} I[w_{j_m}] \to \infty. \end{align*} This contradicts the uniform upper bound $I[w_{j_m}]\le M$. Hence every sequence in $K$ with energies bounded above is bounded in $X$. [guided] The role of coercivity is to prevent minimizing sequences from escaping to infinity in norm. We prove the slightly more general fact that any sequence whose energies stay below a fixed real number must be norm-bounded. Let $(w_j)_{j=1}^{\infty}$ be a sequence in $K$ and assume that there exists $M\in\mathbb R$ such that \begin{align*} I[w_j]\le M \end{align*} for every $j\in\mathbb N$. Suppose, toward a contradiction, that $(w_j)_{j=1}^{\infty}$ is not bounded in $X$. Then the set of norms $\{\|w_j\|_X:j\in\mathbb N\}$ is unbounded. Therefore, for each $m\in\mathbb N$, we can choose an index $j_m$ with $j_{m+1}>j_m$ and \begin{align*} \|w_{j_m}\|_X \ge m. \end{align*} This produces a subsequence $(w_{j_m})_{m=1}^{\infty}$ in $K$ such that \begin{align*} \|w_{j_m}\|_X\to\infty. \end{align*} Now we use exactly the stated coercivity hypothesis. Since the subsequence remains in $K$ and its $X$-norm tends to infinity, coercivity gives \begin{align*} I[w_{j_m}]\to\infty. \end{align*} But the original energy bound passes to the subsequence: \begin{align*} I[w_{j_m}]\le M \end{align*} for every $m\in\mathbb N$. A sequence of extended real numbers cannot both tend to $+\infty$ and remain bounded above by the finite number $M$. This contradiction proves that $(w_j)_{j=1}^{\infty}$ is bounded in $X$. [/guided] [/step] [step:Exclude the possibility that the infimum is $-\infty$] Define \begin{align*} \alpha:=\inf_{v\in K} I[v]. \end{align*} By hypothesis, $\alpha<\infty$. We prove that $\alpha>-\infty$. Assume for contradiction that $\alpha=-\infty$. Then for each $j\in\mathbb N$ there exists $u_j\in K$ such that \begin{align*} I[u_j]\le -j. \end{align*} In particular, $I[u_j]\le 0$ for every $j\in\mathbb N$, so the preceding step implies that $(u_j)_{j=1}^{\infty}$ is bounded in $X$. Since $X$ is reflexive, every bounded sequence in $X$ has a weakly convergent subsequence by the [weak sequential compactness](/theorems/214) of bounded sets in reflexive Banach spaces. Hence there exist a subsequence $(u_{j_k})_{k=1}^{\infty}$ and a point $u\in X$ such that \begin{align*} u_{j_k}\rightharpoonup u \end{align*} in $X$. Because $K$ is norm-closed and convex, it is weakly closed in $X$ by the [Hahn-Banach separation theorem](/theorems/974) for closed convex sets (citing a result not yet in the wiki: weak closedness of norm-closed convex subsets of Banach spaces). Since each $u_{j_k}$ lies in $K$, the weak limit $u$ lies in $K$. Sequential weak lower semicontinuity of $I$ gives \begin{align*} I[u]\le \liminf_{k\to\infty} I[u_{j_k}]. \end{align*} Since $j_k\to\infty$ and $I[u_{j_k}]\le -j_k$, we have \begin{align*} \liminf_{k\to\infty} I[u_{j_k}]=-\infty. \end{align*} Thus $I[u]\le -\infty$, which is impossible because $I$ takes values in $(-\infty,\infty]$. Therefore \begin{align*} -\infty<\alpha<\infty. \end{align*} [/step] [step:Choose a finite-valued minimizing sequence] Since $\alpha=\inf_{v\in K}I[v]$ is finite, for each $j\in\mathbb N$ there exists $u_j\in K$ such that \begin{align*} I[u_j]\le \alpha+\frac{1}{j}. \end{align*} The definition of infimum also gives \begin{align*} \alpha\le I[u_j] \end{align*} for every $j\in\mathbb N$. Hence every $I[u_j]$ is finite and \begin{align*} I[u_j]\to\alpha. \end{align*} Since $\alpha+1$ is a finite upper bound for $I[u_j]$ for all $j\ge 1$, the first step implies that $(u_j)_{j=1}^{\infty}$ is bounded in $X$. [/step] [step:Extract a weakly convergent subsequence whose limit remains admissible] By reflexivity of $X$ and boundedness of $(u_j)_{j=1}^{\infty}$ in $X$, there exist a subsequence $(u_{j_k})_{k=1}^{\infty}$ and a point $u_*\in X$ such that \begin{align*} u_{j_k}\rightharpoonup u_* \end{align*} in $X$. As above, the norm-closed convex set $K$ is weakly closed in $X$ by the Hahn-Banach separation theorem for closed convex sets (citing a result not yet in the wiki: weak closedness of norm-closed convex subsets of Banach spaces). Since $u_{j_k}\in K$ for every $k\in\mathbb N$, it follows that \begin{align*} u_*\in K. \end{align*} [/step] [step:Apply weak lower semicontinuity and compare with the infimum] Since $I$ is sequentially weakly lower semicontinuous on $K$, since $u_{j_k}\rightharpoonup u_*$ in $X$, and since $u_*\in K$, we obtain \begin{align*} I[u_*]\le \liminf_{k\to\infty} I[u_{j_k}]. \end{align*} The sequence $(u_j)_{j=1}^{\infty}$ was chosen so that $I[u_j]\to\alpha$, and therefore its subsequence also satisfies \begin{align*} \liminf_{k\to\infty} I[u_{j_k}]=\alpha. \end{align*} Thus \begin{align*} I[u_*]\le\alpha. \end{align*} On the other hand, $u_*\in K$ and $\alpha=\inf_{v\in K}I[v]$, so \begin{align*} \alpha\le I[u_*]. \end{align*} Combining the two inequalities gives \begin{align*} I[u_*]=\alpha=\inf_{v\in K}I[v]. \end{align*} Therefore $u_*$ is a minimizer of $I$ over $K$. [/step]