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If $H$ is a Hilbert space, $D$ is a dense subspace of $H$, $T:D\to H$ is a $1-1$ bounded linear operator, is the extension of $T$ to $H$ still $1-1?$

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  • $\begingroup$ Not necessarily. $\endgroup$ – Daniel Fischer Jan 2 '17 at 16:42
  • $\begingroup$ @Nizar, please do not accept edits that perform only minor changes, such as adding pairs of dollar signs around some symbols. The edit that you have allowed above only turns "1-1" into "$1-1$". Such edits are to be discouraged. $\endgroup$ – Alex M. Jan 2 '17 at 16:42
  • $\begingroup$ @SteamyRoot, please do not accept edits that perform only minor changes, such as adding pairs of dollar signs around some symbols. The edit that you have allowed above only turns "1-1" into "$1-1$". Such edits are to be discouraged. $\endgroup$ – Alex M. Jan 2 '17 at 16:43
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Definitely not. You could construct example using an o.n. basis $\{e_n\}$ but let me give a different kind of example below.

Let $\mathcal{H} = L^2(\mathcal{D})$ be all square integrable complex value functions on the unit disk. Let $D\subset\mathcal{H}$ be the subspace of all complex polynomials in $z$ and $\overline{z}$, which is dense by the complex version of Stone-Weierstrauss. Finally, let $T$ be given by

$$T(f)(z) = \begin{cases} f(z) & if & Re(z) > 0 \\ 0 & if & Re(z) \leq 0 \end{cases} $$

This will be $1-1$ on polynomials but $T(f)=T(T(f))$ so it is not $1-1$ on the whole space.

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