# 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
Commented Jun 2, 2015 at 13:50
• @Andre, why can we not speak of shift operators if $H$ is not separable? Commented Jun 2, 2015 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. Commented Jun 2, 2015 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). Commented Jun 2, 2015 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. Commented Jun 3, 2015 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? Commented Jun 3, 2015 at 5:13
• It is just the fact that $\langle T^*e_n,e_m\rangle=1$ precisely when $m+1=n$. Commented Jun 3, 2015 at 14:35