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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.

Suppose $K$ is a compact operator and $f_n\rightharpoonup f$. The Uniform boundedness principle makes $(f_n)$ bounded. If $K{f}_n$ did not converge in norm to $Kf$, some subsequence would stay a fixed positive distance away. Compactness provides a further norm-convergent subsequence, say to $g$. On the other hand, for every $h$, $⟨K{f}_n,h⟩=⟨f_n,K^{\ast }h⟩\to ⟨Kf,h⟩$, so its norm limit must be $Kf$, a contradiction.

Conversely, if $K$ sends every weakly convergent sequence to a norm-convergent sequence, take any sequence in the closed unit ball. Weak compactness of that ball supplies a weakly convergent subsequence, whose images converge in norm by hypothesis. Thus every sequence in the image has a convergent subsequence in $H$. Its closure is also sequentially compact: approximate its $n$th member by an image point within $1/n$. Since $H$ is a metric space, that closure is compact. We concludeThis is the principle that compact operators send weak convergence to norm convergence.

If $L$ is bounded and self-adjoint on a complex Hilbert space and $K$ is compact and self-adjoint, thenA singular Weyl sequence for $L$ stays singular for $L+K$, because compact operators send weak convergence to norm convergence and therefore $K{f}_n\to 0$. Applying the same argument with $-K$ proves the reverse inclusion. Finite-multiplicity isolated eigenvalues can move under such perturbations; the essential spectrum of a bounded self-adjoint operator is unchanged.