# Extension of an operator defined on (not necessarily closed) subspaces

Is there always an extension of an operator $T:U\rightarrow W$, defined on (not necessarily closed) subspaces of the infinitedimensional Hilbert spaces $H\supseteq U,L\supseteq W$, to the operator $$T':cl(U)\rightarrow cl(W),$$ where $cl$ denotes the closure.

Is this extension unique if we require uniform continuity ?

What about the case where $U$ is finite-dimensional ?

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Well, in the finite-dimensional case every subspace is closed. – Kevin Carlson Jul 31 '12 at 21:28

• Kevin already mentioned that the case where $U$ is finite dimensional is trivial.
• You can show that if $T$ is continuous (which implies that it is uniformly continuous) then there is a unique continuous extension. This uses completeness of $L$.
• If $T$ is not continuous (or you do not care if the extension is continuous), and you are not asking for any particular properties of the extension, then this is a linear algebra problem. If $\mathrm{cl}(U)=U + V$ with $V$ a subspace of $\mathrm{cl}(U)$ such that $U\cap V=\{0\}$, write $x\in \mathrm{cl}(U)$ as $x=u+v$ with $u\in U$ and $v\in V$, and define $T'(x)=T(u)$. (The complementary subspace $V$ exists assuming the axiom of choice.) Such an extension is not unique unless $U$ is closed.