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The Toeplitz theorem says : If a closed operator is defined everywhere, then it is continuous.

So if a non continuous operator is defined everywhere, it is not closed. But why is it not closed? What condition fails for it to be closed? I thought if a (non continuous) operator $A$ is symmetric and defined on a dense domain in $H$, we can always close it (unambiguously) :

For $f$ where $Af$ is not defined : if $f_n \rightarrow f$ and $\lim Af_n$ exists (i.e. $A f_n \rightarrow f^*$), we define $\bar A f = f^*$ (It will be well defined because if $f_n\rightarrow f,g_n \rightarrow f,Af_n \rightarrow f^*, Ag_n \rightarrow g^*$ then $f^*=g^*$). For all other $f$, we define $\bar Af = Af$.

So $\bar A$ will then be a closed operator defined everywhere, and if it's non continuous that would contradict the theorem. Does that mean that $\bar A$ will be continuous?

user98231
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1 Answers1

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When you close a densely defined symmetric operator $A$, you get a closed operator $\bar{A}$, but $\bar{A}$ will (usually) still not be defined on all of $H$. There will simply exist some $f\in H$ such that there exists no sequence $(f_n)$ in the domain of $A$ converging to $f$ such that $\lim Af_n$ exists, so your definition of $\bar{A}$ fails for $f$. Indeed, Toeplitz's theorem says that if $A$ is unbounded, then some such $f$ must exist.

Eric Wofsey
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  • But what if we then set A¯f for that f (the one for which no lim A f_n exists) to some arbitrary value (say 0)? That should still leave A¯ closed (since no lim A f_n exists for any f_n->f, we can set A¯f to any value without affecting it being closed). Would it then fail to be symmetric? – user98231 Nov 19 '15 at 00:58
  • Yes, that would break symmetry. If $A$ is to remain symmetric, then $\bar{A}f$ needs to be defined so that $\langle \bar{A}f,g\rangle=\langle f,Ag\rangle$ whenever $g$ is in the domain of $A$. Since the domain of $A$ is dense in $H$, this means there is at most one possible value of $\bar{A}f$ that will make $\bar{A}$ symmetric. If that one possible value doesn't even exist (i.e., if the functional $g\mapsto\langle f,Ag\rangle$ is unbounded), then there is no way to define $\bar{A}f$. – Eric Wofsey Nov 19 '15 at 01:43