Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $T$.

share|improve this question
2  
What have you tried? –  Norbert Nov 15 '12 at 20:36
2  
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? –  PtF Nov 15 '12 at 21:06
add comment

1 Answer

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.

share|improve this answer
    
That sounds great, everything is fine with your proof =D Thanks a lot ^^ –  PtF Nov 15 '12 at 21:24
    
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
    
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
    
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
    
@Argerami you're right.. Thanks again.. –  PtF Nov 18 '12 at 15:51
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.