# Prob. 10, Sec. 3.9 in Kreyszig's functional analysis book: The null space and adjoint of the right-shift operator

Let $(e_n)$ be a total orthonormal sequence in a separable Hilbert space $H$, let $T \colon H \to H$ be defined as follows:

Since span of $(e_n)$ is dense in $H$, for every $x \in H$, we have $$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n.$$ Now let $$Tx \colon= \sum_{n=1}^\infty \langle x, e_n \rangle e_{n+1}.$$ Thus, we have $$Te_n = e_{n+1} \ \mbox{ for each } \ n = 1, 2, 3, \ldots.$$ Moreover, $T$ is linear.

What is the null space of $T$?

What is the Hilbert adjoint operator $T^*$ of $T$?

We show that $T$ is bounded as follows:

Since the series $\sum \langle x, e_n \rangle e_n$ converges, we must have $$\sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2 < +\infty.$$ So the series $\sum \langle x, e_n \rangle e_{n+1}$ also converges.

For each $n = 1, 2, 3, \ldots$, let $$x_n \colon= \sum_{j=1}^n \langle x, e_j \rangle e_j.$$ Then using the orthonormality, we have $$\Vert x_n \Vert^2 = \sum_{j=1}^n \vert \langle x, e_j \rangle \vert^2,$$ and so $$\Vert x \Vert^2 = \lim_{n\to \infty} \Vert x_n \Vert^2 = \sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2.$$ Similarly, $$\Vert Tx \Vert^2 = \sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2.$$ Thus $$\Vert Tx \Vert = \Vert x \Vert \ \mbox{ for all } \ x \in H.$$ So $T$ is bounded with $\Vert T \Vert = 1$. Hence $T^*$ exists.

Is this procedure correct?

Suppose that $T x = 0$. Then $$\Vert T x \Vert^2 = \sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2 = 0,$$ whence $$\langle x, e_n \rangle = 0 \ \mbox{ for each } \ n=1, 2, 3, \ldots,$$ and so $x= 0$. So $T$ is injective.

Am I going right?

Now how to find $T^*$?

My effort:

Let $x \colon= \sum \langle x, e_n \rangle e_n, y \colon= \sum \langle y, e_n \rangle e_n \in H$. Then $\langle x, y \rangle = \sum \langle x, e_n \rangle \overline{ \langle y, e_n \rangle }$.

Moreover, by the definitoon of $T^*$, we have $$\langle Tx , y \rangle = \langle x, T^* y \rangle.$$ Or, $$\left\langle \sum_{n=1}^\infty \langle x,e_n \rangle e_{n+1}, \sum_{n=1}^\infty \langle y, e_n \rangle e_n \right\rangle = \left\langle \sum_{n=1}^\infty \langle x,e_n \rangle e_{n}, \sum_{n=1}^\infty \langle y, e_n \rangle T^* e_n \right\rangle.$$ Or, $$\sum_{n=1}^\infty \langle x,e_n \rangle \overline{ \langle y, e_{n+1} \rangle} = \sum_{m=1}^\infty \sum_{n=1}^\infty \langle x, e_m \rangle \overline{ \langle y, e_n \rangle } \langle e_m, T^* e_n \rangle.$$ What next?

What if $H$ is not separable?

For each $m, n \in \mathbb{N}$, we have $$\langle T e_m, e_n \rangle = \langle e_m, T^* e_n \rangle.$$ Or, $$\langle e_{m+1}, e_n \rangle = \langle e_m, T^* e_n \rangle.$$ Thus, for each $m, n \in \mathbb{N}$, we have $$\langle T^* e_n, e_m \rangle = \begin{cases} 0 & \ \mbox{ if } \ m+1 \not= n; \\ 1 & \ \mbox{ if } \ m+1 = n. \end{cases}$$ What next?

• $T^*$ is left shift operator. The kernel of $T$ is trivial. If $H$ is not separable, we can hardly speak of shift operators. By the way: To determine $T^*$ it is much easier if you just look at basis vectors in place of general elements, since a bounded operator is determined by what it does on a basis. – user42761 Jun 2 '15 at 13:50
• @Andre, why can we not speak of shift operators if $H$ is not separable? – Saaqib Mahmood Jun 2 '15 at 14:18
• We can. The problem is that when your basis is indexed by an uncountable set, to have a "next" element for each element in your basis you need to make set theoretical considerations. For instance, if your framework is ZFC, then every set has a well-order, and so you can certainly define a shift. – Martin Argerami Jun 2 '15 at 16:27
• Try not to include unnecessary information in your title. All it does is clutter the main page and may detract people from reading your question (wall of text and all). – Cameron Williams Jun 2 '15 at 16:39
• Well, @CameronWilliams, thank you for your advice, but I feel that having the exact reference of the question is going to be more useful to me in the future. – Saaqib Mahmood Jun 3 '15 at 2:52

Note that your computations show that $$\|Tx\|^2=\left\|\sum_n\langle x,e_n\rangle\,e_{n+1}\right\|^2=\sum_n|\langle x,e_n\rangle|^2=\|x\|^2,$$ so $\|Tx\|=\|x\|$ for all $x$ and $T$ is an isometry. In particular, $T$ is bounded and injective.
For the adjoint note that, since $\langle T^*e_n,e_m\rangle=1$ only when $m+1=n$ (and zero elsewhere), $$\langle T^*e_n,y\rangle=\sum_{m=1}^\infty\overline{\langle y,e_m\rangle}\,\langle T^*e_n,e_m\rangle=\begin{cases}0,&\ n=1,\\ \ \\\overline{\langle y,e_{n-1}\rangle},&\ n\ne1\end{cases}$$ so if we write $e_0=0$, $$\langle T^*e_n,y\rangle=\langle e_{n-1},y\rangle.$$ As we can do this for any $y$, we get that $T^*e_n=e_{n-1}$. So $$T^*x=\sum_{n=1}^\infty\langle x,e_n\rangle\,e_{n-1}.$$
• I've not been able to understand the following equation in your answer: $$\langle T^*e_n,y\rangle=\sum_{m=1}^\infty\overline{\langle y,e_m\rangle}\,\langle T^*e_n,e_m\rangle=\begin{cases}0,&\ n=1,\\ \ \\\overline{\langle y,e_{n-1}\rangle},&\ n\ne1\end{cases}$$ Can you please explain the second equality in particular? – Saaqib Mahmood Jun 3 '15 at 5:13
• It is just the fact that $\langle T^*e_n,e_m\rangle=1$ precisely when $m+1=n$. – Martin Argerami Jun 3 '15 at 14:35