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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3Well, 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? – PtF Nov 15 '12 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.
Martin Argerami
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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} – PtF Nov 18 '12 at 10:42
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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? – PtF Nov 18 '12 at 10:42
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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 $\text{span}(e_2,e_3,\ldots)$. – Martin Argerami Nov 18 '12 at 14:16
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@MartinArgerami This is a very late comment, however I would like to ask why $\sum_2^\infty,\frac1k,e_k$ is not in $\text{span}(e_2,e_3,\ldots)$? – ahdahmani Nov 24 '20 at 09:01
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@Obvious. If you write $\sum_2^\infty\tfrac1k,e_k=\sum_{j=1}^mc_je_j$, doing the inner product with $e_{m+1}$ you get $\tfrac1{m+1}=0$, a contradiction. This can be done for any $m$, so the equality is not possible for any $m$. – Martin Argerami Nov 24 '20 at 10:44
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@MartinArgerami Why did you choose only finite number from $(e_2,e_3,\cdots)$? We are in a topological space here, and an infinite-dimensional basis makes sense, so we can write each element as an infinite linear combination. Isn't this true? – ahdahmani Nov 25 '20 at 09:06
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@MartinArgerami Great answer! May I ask how do we know that $T^*$ exists? Don't we need boundedness for this? Are we falling into a subtle circular argument like argued here? – Pellenthor Apr 25 '23 at 20:30
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Technically, yes But the usual meaning of "$Te_n=e_{n+1}$" is just a shorthand for $$T\Big(\sum_{k=1}^\infty x_ke_k\Big)=\sum_{k=1}^\infty x_ke_{k+1}.$$ With this latter definition there is no ambiguity. – Martin Argerami Apr 25 '23 at 20:38
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@MartinArgentami Why $\overline{\text{span}(e_2,\dots)}\supseteq\mathcal{R}(T)$? I have come to say the following, let $y\in\mathcal{R}(T)$, then $y=Tx$ for same $x\in H$, therefore $$x=\sum_{k=1}^\infty \langle x, e_k \rangle e_k.$$ Then $$y=Tx=\sum_{k=1}^\infty \langle x , e_k \rangle e_{k+1}.$$ From this can I conclude that $y\in \overline{\text{span}(e_2,\dots)?}$ – NatMath Jun 04 '23 at 07:55
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Yes. What you wrote says that $y$ is the limit of the sequence $$\Big{\sum_{k=2}^n\langle x,e_{k-1}\rangle,e_k\Big}_n\subset \operatorname{span}{e_2,e_3,\ldots}.$$ – Martin Argerami Jun 04 '23 at 11:08