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Let $H$ be a Hilbert space and let $A : D(A) \subset H \to H$ be an essentially self-adjoint operator. Let $\overline A$ be the unique self-adjoint extension of $A$.

Question: Is it true that $\ker(A) = \ker(\overline A)$?

I don't really see any reason for this to be true, so I tried constructing a counterexample... but somehow this is not so easy to do!

For instance, I tried starting with the negative Laplacian $\Delta = - \frac{d^2}{dx^2}$ on $L^2(S^1)$ with domain $\mathrm{dom}(\Delta) = C^\infty(S^1)$. This is an essentially self-adjoint operator and the closure $\overline \Delta$ is easily understood if we look to the frequency domain. On $\ell^2(\mathbb{Z})$, we find $\overline \Delta$ is conjugate to the diagonal operator $D$ on $\ell^2(\mathbb{Z})$ given by $D(a_n) = (n^2a_n)$ and with domain $\{ (a_n) \in \ell^2(\mathbb{Z}) : (na_n) \in \ell^2(\mathbb{Z})\}$. From this, we conclude that $\Delta$ and $\overline \Delta$ have the same one-dimensional kernel. $$\ker(\Delta) = \ker(\overline \Delta) = \{\text{constant functions}\} \subset L^2(S^1).$$ Next I tried to replace $\Delta$ by its restriction $\Delta_S$ to some smaller subspace $S \subset C^\infty(S^1)$ such that $S$ intersects trivially with the constant functions, but still manages to be a core for $\Delta$, i.e. has $\overline \Delta_S = \overline \Delta$. However, every subspace $S$ like this I try turns out to give a non essentially self-adjoint $\Delta_S$. For instance, taking $S = \{f \in C^\infty(S^1) : f(p) = 0\}$ for some point $p$ fails. So does taking $S$ to be the smooth functions vanishing on a whole neighbourhood of some point $p$.

Any thoughts?

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    $\begingroup$ Let $\mathcal{M}$ be a dense subspace of $H$ and let $A=0$ on $\mathcal{D}(A)=\mathcal{M}$. $A$ is essentially selfadjoint and $\overline{A}$ is the $0$ operator on $H$. So $\mathcal{N}(A)\ne \mathcal{N}(\overline{A})$. $\endgroup$ Mar 1, 2016 at 18:11
  • $\begingroup$ @TrialAndError: +1, that gets to the heart of how silly this question is :p. Note also that $\mathcal{N}(A)$ can fail to be dense in $\mathcal{N}(\overline A)$, by the the discussion below. $\endgroup$
    – Mike F
    Mar 1, 2016 at 19:28
  • $\begingroup$ @MikeF The accepted answer says that the counter-example is "non-constructive", which sounds like the counter-example is in some sense pathological - but then the comments seem to suggest that there a "better" examples. Are those examples more convinving (in the sense that they do not suffer from the same issue)? Sorry, I am a beginner in functional analysis and couldn't understand the entire discussion, an explanation would be much appreciated. $\endgroup$
    – Filippo
    Apr 24, 2023 at 20:18
  • $\begingroup$ @Filippo: Does the comment by user "Disintegrating By Parts" above make sense to you? $\endgroup$
    – Mike F
    Apr 25, 2023 at 13:43
  • $\begingroup$ @MikeF Yes, it does! $\endgroup$
    – Filippo
    Apr 25, 2023 at 13:45

1 Answer 1

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Let $B$ be selfadjoint with $\ker B = \operatorname{span}\{x_0\}$. Denote by $\|\cdot\|_B$ the graph norm with respect to $B$ on $D(B)$. Choose a $\|\cdot\|_B$-dense subspace $D\subset D(B)$ such that $D(B) = D + \operatorname{span}\{x_0\}$. Such a subspace may arise as the kernel of a non-$\|\cdot\|_B$-continuous linear functional on $D(B)$ (Note: This is non-constructive. As far as I know, such a functional could not be contructed, so far. However, the existence of such functionals follows from the axiom of choice. See here: https://en.wikipedia.org/wiki/Discontinuous_linear_map#A_nonconstructive_example). Now, define $A := B|D$. Then $A$ is essentially selfadjoint with $\overline A = B$ and $\ker A = \{0\}$.

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    $\begingroup$ Hmm OK, so we could actually even do this beginning with a bounded self-adjoint operator. And we don't need the discontinuous functional. Suppose $A \in B(H)$ and $A^*=A$. For simplicity, assume $A$ has one-dimensional kernel $\mathbb{C} \cdot \xi$, $\xi \neq 0$. Now choose a dense subspace $S \subset H$ with $\xi \notin S$. Then, $A|_S$ is essentially-self-adjoint, but $\ker(A|_S) = \{0\}$. To be concrete about this, we could take $H = \ell^2$, $\xi = (1,1/2,1/3,\ldots)$, $A$ to be orthogonal projection onto $\xi^\bot$ and $S$ to be the dense subspace of finitely supported sequences. $\endgroup$
    – Mike F
    Mar 1, 2016 at 1:40
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    $\begingroup$ Yes, this is alright. Forget about my complicated example. For unbounded $B$, choose $D$ dense in $(D(B),\|\cdot\|_B)$ such that $x_0\notin D$. $\endgroup$ Mar 1, 2016 at 1:45
  • $\begingroup$ Great! It appears we have muddled our way to a satisfactory conclusion. Thanks for your help! $\endgroup$
    – Mike F
    Mar 1, 2016 at 1:51
  • $\begingroup$ I am happy I could help. :o) $\endgroup$ Mar 1, 2016 at 1:53

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