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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.