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Let $K$ be a nonempty norm-closed convex set in a real Hilbert space, and let $F:K\to \mathbb{R}$ be continuous and a convex function. If $K$ is bounded, or $F(x)\to \infty$ as $x\in K$ tends to infinity in norm, then $F$ attains its infimum. Every sublevel set is bounded, norm-closed, and convex, hence contains the weak limits of its sequences by norm-closed convex set is weakly closed. A minimizing sequence in one fixed sublevel set has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Its limit lies in every sublevel set above the infimum. Since $F$ is real-valued, this also rules out infimum $-\infty$ and proves attainment. This is an instance of the direct method in the calculus of variations.

For each real $a$, define the sublevel set $K_a=\{u\in K:F(u)\le a\}$. Continuity of $F$ and closure of $K$ make $K_a$ a closed set in $X$; the convex function inequality makes it a convex set. The preceding result therefore makes every $K_a$ a sequentially weakly closed set. Equivalently, $F$ has weak lower semicontinuity along sequences in $K$.

Every $K_a$ is bounded. This is immediate if $K$ is bounded. Otherwise, an unbounded $K_a$ would contain $u_n$ with $\Vert u_n\Vert \ge n$, contradicting the assumed coercive function condition $F(u_n)\to \infty$.

Choose $z\in K$ and put $\ell ={\inf }_{K}F\in [-\infty ,F(z)]$. Choose a minimizing sequence $(u_n)$ entirely in $K_{F(z)}$, with $F(u_n)\to \ell$; when $\ell =-\infty$, require $F(u_n)\le -n$ for sufficiently large $n$. It is bounded by the previous paragraph.

The weak subsequence of a bounded Hilbert-space sequence theorem gives $u_{n_j}\rightharpoonup u_0$. This applies even if $X$ is not separable: the closed linear span of the sequence is a separable Hilbert space; successive extraction makes every coefficient in a countable Hilbertian basis converge, the common norm bound gives square-summable limiting coefficients by the Bessel inequality, and approximation by finite linear combinations proves weak convergence. Orthogonal decomposition extends that conclusion to all test vectors in $X$. Since $K$ is a sequentially weakly closed set, $u_0\in K$.

For any real $a>\ell$, a tail of $(u_{n_j})$ lies in $K_a$, so $u_0\in K_a$. If $\ell =-\infty$, this would give $F(u_0)\le a$ for every real $a$, impossible because $F$ is real-valued. Thus $\ell$ is finite; letting $a\downarrow \ell$ gives $F(u_0)\le \ell$, while the definition of infimum gives the reverse inequality. ThereforeThis proves minimum of a coercive convex functional on a Hilbert space by the direct method in the calculus of variations, with bounded sublevel sets supplying the required weak sequential compactness in a Hilbert space.

Take the weakly convergent subsequence in (i), with limit $u\in H_0^1(B)$. For each $\varphi \in H_0^1(B)$, the weak formulation givesThe left side tends to $\int \nabla u\cdot \nabla \varphi$ by weak convergence in $H^1$. The zero-boundary Sobolev inequality gives $\varphi \in L^6(B)$, so (ii) makes the right side tend to zero. Taking $\varphi =u$ in the limiting identity yields ${\int }_B|\nabla u{|}^2=0$. The zero-boundary Poincaré inequality then gives $u=0$.

The same argument applies to a weakly convergent subsequence of any subsequence of $(u_m)$, because all remain bounded as in (i). To see why this proves convergence of the full sequence, suppose a continuous linear functional $F$ on $H^1(B)$ did not satisfy $F(u_m)\to 0$. Some subsequence would have $|F(u_{m_j})|\ge \epsilon >0$. Its boundedness gives a further weakly convergent subsequence, whose limit has just been proved to be zero; then $F$ tends to zero along it, a contradiction. ThereforeUniqueness of a possible limit alone is not enough without the weak sequential compactness in a Hilbert space used to extract a further subsequence.

Let $w_m=|E_m{|}^{5/6}{u}_m$. By linearity of the weak formulation, $w_m$ is the zero-boundary weak solution with forcing ${\chi }_{E_m}$. The estimate in (a)(iii) therefore givesThe change of variables $x=r_{m}y$ gives $|E_m|=r_m^3|E|>0$, so the division is legitimate. Since $H^1(B)$ is a Hilbert space, weak sequential compactness in a Hilbert space now gives a subsequence and some $u\in H^1(B)$ such thatIn fact $u\in H_0^1(B)$: the Sobolev trace operator is continuous and linear, so its zero kernel is preserved by weak convergence.

Suppose the compact operator $K$ acts on $H$, and $h_n\rightharpoonup h$. The Uniform boundedness principle makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $K{h}_n\rightharpoonup Kh$: for every $g$, $⟨K{h}_n,g⟩=⟨h_n,K^{\ast }g⟩\to ⟨Kh,g⟩$. Every norm-convergent subsequence of $(K{h}_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that compact operators send weak convergence to norm convergence.

Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a Hilbert space has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the unit ball is relatively sequentially compact and hence relatively compact. Complete continuity and compactness are equivalent on a Hilbert space.

Choose nested interior open sets $U\subset \subset V\subset \subset W\subset \subset \Omega$. For small $\sigma$, part (b) gives a smooth equation on $V$ of the formThe divergence-forcing interior H1 estimate isOne can obtain this estimate directly by testing the smooth equation against ${\eta }^2{u}_{\sigma }$ and applying Young inequality, with a cutoff function equal to one on $U$ and supported in $V$. The approximate identity and the $L^2$ convolution bound give, uniformly in $\sigma$,with analogous bounds for $(cu{)}_{\sigma }$ and $f_{\sigma }$. The coefficients need only be bounded on $W$. Therefore $u_{\sigma }$ is bounded in $H^1(U)$. It converges to $u$ in $L^2(U)$, and weak sequential compactness in a Hilbert space in this Sobolev space gives $u\in H^1(U)$. Since $U$ was arbitrary, $u\in H_{\mathrm{loc}}^1(\Omega )$.

Now expand the distributional derivative:Its right side belongs to $L_{\mathrm{loc}}^2$, so the interior $H^2$ elliptic regularity estimate gives $u\in H_{\mathrm{loc}}^2$. More generally, if $u\in H_{\mathrm{loc}}^m$ for an integer $m\ge 1$, multiplication by the smooth coefficients puts the right side in $H_{\mathrm{loc}}^{m-1}$. Interior elliptic regularity then gives $u\in H_{\mathrm{loc}}^{m+1}$. This elliptic regularity bootstrap proves $u\in H_{\mathrm{loc}}^k$ for every integer $k$.

For any nonnegative integer $\ell$, choose $k>\ell +n/2$. The Sobolev embedding theorem on compact interior subsets gives $u\in C_{\mathrm{loc}}^{\ell }$. Taking all $\ell$ proves $u\in C^{\infty }(\Omega )$, for its smooth representative. The first gain from $L^2$ to $H^1$ is the step supplied by the mollified equation; assuming $H^1$ in the initial definition would miss that step.