[proofplan]
We use the direct method in the [Hilbert space](/page/Hilbert%20Space) $H^1_0(U)$. A minimizing sequence is bounded because coercivity makes sublevel sets bounded; Hilbert-space weak compactness then gives a weakly convergent subsequence. Closed convex subsets of a Hilbert space are weakly sequentially closed, so the weak limit remains in $\mathcal{A}$, and weak lower semicontinuity turns the limit into a minimizer. Strict convexity rules out two distinct minimizers by evaluating the energy at a nontrivial convex combination.
[/proofplan]
[step:Choose a bounded minimizing sequence in the admissible set]
Let
\begin{align*}
m := \inf_{v \in \mathcal{A}} I[v] \in [-\infty,\infty).
\end{align*}
Since $\mathcal{A}$ is nonempty, choose $a \in \mathcal{A}$. If $m > -\infty$, choose a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal{A}$ such that
\begin{align*}
I[u_k] \leq m + \frac{1}{k}
\end{align*}
for every $k \in \mathbb{N}$. Then, after discarding finitely many terms if necessary, $I[u_k] \leq I[a] + 1$ for all $k$, so $(u_k)$ is contained in the sublevel set
\begin{align*}
\{v \in \mathcal{A} : I[v] \leq I[a] + 1\}.
\end{align*}
By coercivity on $\mathcal{A}$, this sublevel set is bounded in $H$.
If $m = -\infty$, choose $(u_k)_{k=1}^{\infty}$ in $\mathcal{A}$ such that $I[u_k] \leq -k$ for every $k \in \mathbb{N}$. Since also $I[u_k] \leq I[a] + 1$ for all sufficiently large $k$, the same coercivity argument shows that a tail of $(u_k)$ is bounded in $H$.
[/step]
[step:Pass to a weak limit that still lies in $\mathcal{A}$]
In both cases, we have a bounded sequence in the Hilbert space $H$. Since Hilbert spaces are reflexive and bounded sequences in reflexive spaces admit weakly convergent subsequences, there are a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u \in H$ such that
\begin{align*}
u_{k_j} \rightharpoonup u \quad \text{weakly in } H.
\end{align*}
Because $\mathcal{A}$ is norm-closed and convex, it is weakly sequentially closed in $H$. Since each $u_{k_j}$ belongs to $\mathcal{A}$, the weak limit satisfies $u \in \mathcal{A}$.
[guided]
The point of this step is to replace boundedness by convergence. The sequence constructed above is bounded in the Hilbert space $H = H^1_0(U)$, and Hilbert spaces are reflexive. The weak compactness principle for reflexive spaces therefore gives a subsequence, which we denote by $(u_{k_j})_{j=1}^{\infty}$, and an element $u \in H$ such that
\begin{align*}
u_{k_j} \rightharpoonup u \quad \text{weakly in } H.
\end{align*}
We must still check that the weak limit is admissible. This is exactly where convexity of $\mathcal{A}$ is used together with norm-closedness. A norm-closed convex subset of a Hilbert space is weakly sequentially closed: weak limits of sequences from the set remain in the set. Since every term $u_{k_j}$ lies in $\mathcal{A}$, this property gives
\begin{align*}
u \in \mathcal{A}.
\end{align*}
Without convexity, a norm-[closed set](/page/Closed%20Set) need not be weakly closed, so this step is not a formality; it is the structural reason that convex admissible classes are stable under the direct method.
[/guided]
[/step]
[step:Use weak lower semicontinuity to identify the weak limit as a minimizer]
First suppose $m > -\infty$. Since $I$ is weakly lower semicontinuous on $H$ and $u_{k_j} \rightharpoonup u$ weakly in $H$, we have
\begin{align*}
I[u] \leq \liminf_{j \to \infty} I[u_{k_j}].
\end{align*}
The minimizing property of $(u_k)$ gives
\begin{align*}
\liminf_{j \to \infty} I[u_{k_j}] \leq m.
\end{align*}
Because $u \in \mathcal{A}$ and $m$ is the infimum of $I$ over $\mathcal{A}$, we also have $m \leq I[u]$. Hence
\begin{align*}
I[u] = m = \inf_{v \in \mathcal{A}} I[v].
\end{align*}
It remains to exclude the case $m = -\infty$. In that case the chosen sequence satisfies $I[u_{k_j}] \leq -k_j$, and hence
\begin{align*}
\liminf_{j \to \infty} I[u_{k_j}] = -\infty.
\end{align*}
Weak lower semicontinuity would imply
\begin{align*}
I[u] \leq -\infty,
\end{align*}
which is impossible because $I[u] \in \mathbb{R}$. Therefore $m > -\infty$, and the preceding paragraph proves existence of a minimizer.
[/step]
[step:Use strict convexity to rule out two different minimizers]
Assume that $I$ is strictly convex on $\mathcal{A}$. Suppose, for contradiction, that $u_0,u_1 \in \mathcal{A}$ are two distinct minimizers. Thus $u_0 \neq u_1$ and
\begin{align*}
I[u_0] = I[u_1] = m.
\end{align*}
Fix $t \in (0,1)$ and define
\begin{align*}
u_t := (1-t)u_0 + t u_1.
\end{align*}
Since $\mathcal{A}$ is convex, $u_t \in \mathcal{A}$. Since $I$ is strictly convex on $\mathcal{A}$ and $u_0 \neq u_1$, we obtain
\begin{align*}
I[u_t] < (1-t)I[u_0] + tI[u_1].
\end{align*}
Using $I[u_0] = I[u_1] = m$, the right-hand side equals $m$, so
\begin{align*}
I[u_t] < m.
\end{align*}
This contradicts the definition of $m$ as the infimum of $I$ over $\mathcal{A}$, because $u_t \in \mathcal{A}$. Therefore the minimizer is unique.
[/step]
