# Extending a compact operator to the entire Hilbert space

In a course I'm taking we defined compact operators as a linear mapping $H\rightarrow H$, where $H$ is a Hilbert space, that maps bounded sets to relative compact ones. The lecturer mentioned that the reason we defined it like this and not as $M\rightarrow H$ with $M\subseteq H$ is that every such compact operator could be extended from $M$ to $H$. In passing he mentioned that this had something to do with the Hahn-Banach theorem.

My questions are:

• What exactly had he meant, how can one construct an extension ? I can't see the connection with said theorem since it involves linear functionals and not maps.
• Is this extension unique ?
• Is the argument that any such operator defined on a subset can be extend to the whole space even adequate ? I'm thinking that there may be examples operators that are naturally defined only on a subset of $H$ and extending them to $H$ would be inelegant. Could you provide such an example for compact operators ?
(To see what I mean by such an example I'm going to give one for symmetric operators: the differential operator that maps a function from $L^2(\Omega)$ to its second derivative is naturally only defined in a subset of $L^2(\Omega)$, since second derivatives of $L^2$-functions do not necessarily have to lie in $L^2(\Omega)$ again - now it would be possible to extend this operator to $L^2(\Omega)$ by letting it be $0$ outside its natural domain, since that would preserve symmetry, but it would be inelegant.)
• Good questions. I am also interested in answers. Jun 9 '14 at 9:33

I don't see how the Hahn-Banach theorem is relevant here either.

A natural construction of a compact extension of a compact operator $T\colon M\to H$ to a compact operator $\tilde{T}\colon H\to H$ is to first (if necessary) extend the operator continuously to $\overline{T}\colon \overline{M} \to H$. That extension is unique, and as is also compact (a bounded set in $\overline{M}$ is contained in the closure of a bounded set in $M$, therefore is mapped to a relatively compact set). Then you use the orthogonal decomposition $H = \overline{M} \oplus M^\perp$. The trivial extension defines the operator as $0$ on $M^\perp$, but taking any compact operator $A\colon M^\perp \to H$ yields a compact $\tilde{T}$ by setting

$$\tilde{T}(x) = \overline{T}(P_{\overline{M}}x) + A(P_{M^\perp}x).$$

So such an extension is unique if and only if $M$ is dense.

Since the continuous extension of a continuous operator to the closure of its domain is always possible if the codomain is complete (that is far more general than operators on Hilbert spaces), the orthogonal decomposition in Hilbert spaces guarantees that every continuous linear map from a subspace of a Hilbert space to a complete (Hausdorff) topological vector space can be extended to a continuous linear map defined on the whole space.

Your example of the differential operator is an unbounded (discontinuous) operator. Such operators need not be extensible in a meaningful way to the whole space (it can be extended by taking an algebraic complement of its domain, but such an extension is usually not meaningful), but compact operators are continuous, hence an extension always exists.

• Thanks for you detailed answer, +1! I have 2 questions though: The first is concerned with your general extension argument from the second paragraph: Does this also have the property that it is unique of $M$ is dense or was this property specific to Hilbert spaces ? The second is pertaining to the third (last) paragraph, which I didn't quite understand: You seem to imply that the extension of unbounded operators to the whole space exists, but isn't meaningful whereas the extension of compact operators exists and also is meaningful - but what exactly do you mean by "meaningful" ?
– user36772
Jun 9 '14 at 10:06
• The extension of a continuous operator to the closure of its domain is unique, that is not particular to Hilbert spaces. That is because if you have two continuous maps $f,g\colon A\to B$, where $B$ is a Hausdorff space, the set $\{ x : f(x) = g(x)\}$ is closed. If $f$ and $g$ coincide on a dense set, they are therefore equal. Regarding the last paragraph, we can extend any linear operator defined on a subspace to the entire space by taking a basis of an algebraic complement of the domain and defining the extension arbitrarily on that basis. That is however not meaningful in the sense that ... Jun 9 '14 at 10:15
• ... that destroys any nice properties the original operator had, like symmetry, continuity, closeability, and what else there is. On a Hilbert space, you can extend any continuous (compact) operator defined on a subspace to a continuous (compact) operator defined on the whole space. The conservation of the continuity (compactness) is what makes the extension "meaningful". Jun 9 '14 at 10:18
• So what you're saying is that one can prove that no matter how one would extend an unbounded, symmetric, linear operator on a subset of a Hilbert space, it would always lose at least one of these properties ? If this is what you meant, could you pleaseprovide me with a reference to a proof or a short argument of it ?
– user36772
Jun 9 '14 at 10:22
• Ah, I should have inserted a "usually" in there. For "symmetric", it holds, because if you have a symmetric operator defined on the whole Hilbert space, that operator is continuous (Hellinger-Toeplitz theorem), so you cannot extend an unbounded symmetric operator to a symmetric operator on the whole space. But for other nice properties (not closeability, however, because of the closed graph theorem) an unbounded operator may have, it might be possible to have an (unbounded) extension to the entire space that also has the nice property. Jun 9 '14 at 10:30