Let $(e_n)$ be a total orthonormal sequence in a separable Hilbert space $H$ and define the right shift operator to be the linear operator $T:H\rightarrow H$ such that $Te_n=e_{n+1}$, for $n=1, 2, \ldots.$ Find the range, null space, norm and Hilbert-adjoint operator of $T$.
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3$\begingroup$ Well, with this hypothesis I can write any $x\in H$ as, \begin{align*} \displaystyle x=\sum_{k=1}^\infty \langle x, e_k\rangle e_k, \end{align*} $\langle x, e_k\rangle$ are the Fourier coefficientes of $x\in H$ with respect to the orthonormal sequence $(e_n)$. For finding the $R(T)$ we aply $T$ in the equality above, \begin{align*} \displaystyle Tx=\sum_{k=1}^\infty \langle x, e_k\rangle Te_k=\sum_{k=1}^\infty \langle x, e_k\rangle e_{k+1}. \end{align*} This lead us to conjecture, \begin{align*} \displaystyle R(T)=\overline{\textrm{span}(e_2, e_3, \ldots)}. \end{align*} But how to prove this? $\endgroup$– PtFNov 15, 2012 at 21:06
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To see the inclusion $\overline{\textrm{span}(e_2, e_3, \ldots)}\subset R(T)$, you do the following. You just check that $e_{k+1}\in R(T)$ for all $k\in\mathbb N$, and this is trivial because $e_{k+1}=Te_k$. So $\textrm{span}(e_2,e_3,\ldots)\in R(T)$.
Then it only remains to check that $R(T)$ is closed. This follows from the fact that $T$ is an isometry, i.e. $\|Tx\|=\|x\|$ for all $x\in H$. First, noting that $T^*$ is the operator that sends $e_1$ to $0$ and $e_{k+1}$ to $e_k$, $$ \|Tx\|^2=\|\sum_{k=1}^\infty\langle Tx,e_k\rangle\,e_k\|^2=\sum_{k=1}^\infty|\langle Tx,e_k\rangle|^2=\sum_{k=1}^\infty|\langle x,T^*e_k\rangle|^2=\sum_{k=2}^\infty|\langle x,e_{k-1}\rangle|^2=\sum_{k=1}^\infty|\langle x,e_k\rangle|^2=\|x\|^2, $$ so $T$ is isometric. Now, if $Tx_j\to y$, then $\{Tx_j\}$ is a Cauchy sequence; as $T$ is isometric $\{x_j\}$ is a Cauchy sequence too. Let $x=\lim x_j$. Then $$ y=\lim Tx_j=T(\lim x_j)=Tx\in R(T). $$ So $R(T)$ is closed.
answered Nov 15, 2012 at 21:12
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$\begingroup$ That sounds great, everything is fine with your proof =D Thanks a lot ^^ $\endgroup$– PtFNov 15, 2012 at 21:24
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$\begingroup$ Hello @Argerami, I was wondering one thing. It's easy to show that both inclusions below hold, \begin{align*} \displaystyle R(T)\subset \textrm{span}(e_2, e_3, \ldots),\ \textrm{span}(e_2, e_3, \ldots)\subset R(T), \end{align*} so that I could conclude $R(T)=\textrm{span}(e_2, e_3, \ldots)$, isn't it right? If that holds I could take the closure on both sides to get, \begin{align*} \displaystyle \overline{R(T)}=\overline{\textrm{span}(e_2, e_3, \ldots)}. \end{align*} $\endgroup$– PtFNov 18, 2012 at 10:42
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$\begingroup$ But as you proved $\overline{R(T)}=R(T)$ for $R(T)$ is closed, however as $R(T)=\textrm{span}(e_2, e_3, \ldots)$ we have $\textrm{span}(e_2, e_3, \ldots)$ is equal to a closed set therefore it's itself closed so that $\overline{\textrm{span}(e_2, e_3, \ldots)}=\textrm{span}(e_2, e_3, \ldots)$. Conclusion, \begin{align*} \displaystyle R(T)=\textrm{span}(e_2, e_3, \ldots), \end{align*} as we had at the start. Is something wrong with this reasoning? $\endgroup$– PtFNov 18, 2012 at 10:42
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$\begingroup$ What is wrong is that the equality $R(T)=\text{span}(e_2,e_3,\ldots)$ does not hold. For example, $\sum_2^\infty\,\frac1k\,e_k$ is in $R(T)$ but not in $\text{span}(e_2,e_3,\ldots)$. $\endgroup$ Nov 18, 2012 at 14:16
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