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We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every $f\in X^{\ast }$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\epsilon$. The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.

The canonical embedding into the bidual isIt is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. For $x\ne 0$, the Hahn-Banach theorem supplies a norming functional $f$ with $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, proving equality. Thus $J_X$ is an isometric embedding. The coordinates defining the weak-star topology on $X^{\ast \ast }$, restricted to $J_X(X)$, are precisely $J_{X}x\mapsto f(x)$ for $f\in X^{\ast }$. These are exactly the coordinates defining the weak topology on $X$, so $J_X$ is also a topological embedding for these topologies.

A normed vector space is a reflexive space if $J_X$ is onto $X^{\ast \ast }$. Such a space is automatically a Banach space, since it is isometric to the complete continuous dual space $X^{\ast \ast }$. The Banach-Alaoglu theorem states that the closed unit ball of a continuous dual space is compact for its weak-star topology. The Goldstine theorem states that $J_X(B_X)$ is dense in the weak-star topology in $B_{X^{\ast \ast }}$. ConsequentlyIf $J_X$ is onto, its weak-to-weak-star homeomorphism carries $B_X$ onto the Banach-Alaoglu theorem's compact ball $B_{X^{\ast \ast }}$. Conversely, if $B_X$ is weakly compact, $J_X(B_X)$ is compact and hence closed in the Hausdorff weak-star topology. Its density from the Goldstine theorem makes it all of $B_{X^{\ast \ast }}$, and scaling gives surjectivity of $J_X$.

Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the Hahn-Banach theorem, choose norming functionals $f_n$ with $\Vert f_n\Vert =1$ and $f_n(u_n)=1$. These continuous linear functionals separate points: for $x\ne 0$, choose $\Vert u_n-x/\Vert x\Vert \Vert <1/2$ to get $f_n(x)>\Vert x\Vert /2$. DefineThe series converges uniformly, separates distinct points, and obeys the triangle inequality because $\min (c,a+b)\le \min (c,a)+\min (c,b)$. Thus $d$ is a metric. Its topology is coarser than the weak topology: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$. On a weakly compact set $K$, the identity $(K,\text{weak})\to (K,d)$ is therefore a continuous bijection from a compact space into a Hausdorff space, hence a homeomorphism. This proves that $K$ is metrizable in its weak topology, an instance of countable separating family metrizes a weakly compact set.

If $X$ is also a reflexive space, its closed unit ball is weakly compact and therefore metrizable by this construction. Compactness in a metric space implies sequential compactness, so every sequence in $B_X$ has a weakly convergent subsequence. Scaling proves the claim for every bounded sequence.

For the converse, assume every bounded sequence in $X$ has a weakly convergent subsequence. Use the same metric $d$ on $B_X$. A weak limit of points of $B_X$ remains in $B_X$: $|f(x)|\le 1$ for every $\Vert f\Vert \le 1$, and the Hahn-Banach theorem recovers $\Vert x\Vert \le 1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.

We must still recover the entire weak topology, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the weak topology $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of metric limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies weak convergence. Every $f\in X^{\ast }$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a metric space. The weak topology and $d$-topology on $B_X$ coincide. The ball is weakly compact, so the criterion above provesThe zero space satisfies all statements directly. The converse did not assume completeness, nor use the full Eberlein-Šmulian theorem.

For a normed space $X$, let $X^{\ast }$ be its continuous dual space and $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$ its bidual space. Its canonical embedding into the bidual isThe definition of a reflexive space in this setting isThe Hahn-Banach theorem makes $J_X$ a linear map that is an isometry. For any normed space $Z$, its continuous dual space is a Banach space: a Cauchy sequence $({\ell }_n)$ in the operator norm has scalar limits $\ell (z)={\lim }_n{\ell }_n(z)$, and the uniform Cauchy bound proves both boundedness of $\ell$ and $\Vert {\ell }_n-\ell \Vert \to 0$. Apply this with $Z=X^{\ast }$ to see that $X^{\ast \ast }$ is complete. If $J_X$ is onto, the isometry transfers this completeness to $X$. Therefore