# Left Shift Operator Spectrum

Consider the Hilbert space $\mathcal{H}=l^2(\mathbb{Z})$ and define the left shift operator $\mathcal{L}:\mathcal{H} \rightarrow \mathcal{H}$ by $$\mathcal L (a_n) = (b_n) \qquad \text{ where } \qquad b_n=a_{n+1} ,$$ so $\mathcal L$ is the right-shift operator.

My first issue is with notation, what does $\mathcal L -zI$ mean for $z \in \mathbb{C}$? Does this mean $$\mathcal L-I((a_n)) = (a_{n+1}-a_n)?$$

Moreover, I'm supposed to show $\mathcal L-zI$ is not invertible for $|z| =1$ by showing that $$(a_n) = \frac{1}{|n|+1}$$ is in $\ell^2(\mathbb{Z})$, yet not in the image of $\mathcal L$ for $z=1$, and then adjust for other values of $z$. Though, I don't quite understand how this shows that $\mathcal L -zI$ isn't surjective.. was this perhaps a typo? Should I show $(a_n)$ isn't in the image of $\mathcal L-I$ and then adjust for values of $z$? It doesn't seem like it would make sense because surely it's in the image of $\mathcal L$, as the left shift operator is $\mathcal L$'s adjoint and inverse, with both being norm-preserving, so this wouldn't make sense.

I don't quite see how to begin with this either way, so any hints would be wonderful. Indeed, this is homework. (The next part involves the construction of an inverse to $\mathcal L -z I$ using the power series for $(1-x)^{-1}$.)

More elementary way:

a. If $\lvert\,\lambda\rvert>1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal L)^{-1}=\lambda^{-1}\sum_{n=0}^\infty (\lambda^{-1}\mathcal L)^n$ converges.

b. If $\lvert\,\lambda\rvert<1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal L)^{-1}=-{\mathcal L}^{-1}\sum_{n=0}^\infty (\lambda\mathcal L^{-1})^n$ converges.

c. If $\lambda=\mathrm{e}^{i\vartheta}$ then $\lambda\in\sigma(\mathcal L)$. This can be shown as follows. For every $\varepsilon>0$, find a vector $\boldsymbol{a}=(a_k)_{k\in\mathbb Z}$, such that $\sum_{k\in\mathbb Z}\lvert a_k\rvert^2=1$, while $$\|\mathcal L\boldsymbol{a}\|^2= \sum_{k\in\mathbb Z}\lvert a_{k+1}-\mathrm{e}^{i\vartheta}a_k\rvert^2<\varepsilon^2.$$ To achieve that, look for $\boldsymbol{a}=(a_k)_{k\in\mathbb Z}$, with elements the Fourier coefficients of a smooth $2\pi-$function, which is equal to 1 at $x=\vartheta$, and very small elsewhere. In particular, consider $p_n(x)=n^{-1/2}\sum_{\lvert k\rvert\le n}\mathrm{e}^{-ik\vartheta}\mathrm{e}^{ikx}$, which approximates the $\delta-$function at $x=\vartheta$.

• Does the notation $\mathcal{L}^n$ mean apply $\mathcal L$ $n$ times, or $(a_{n+1}^n)$? That is, how do I apply $\frac{\mathcal L^k}{\lambda^k}$ to $(a_n)$? – Anthony Peter Nov 13 '15 at 5:40
• @AnthonyPeter $\big({\mathcal L}^n(a_k)_{k\in\mathbb Z}\big)_j=a_{j+n}$. – Yiorgos S. Smyrlis Nov 13 '15 at 11:33

Answer. $S=\{z\in\mathbb C: \lvert\,z\rvert=1\}$.

Explanation. The space $\ell^2(\mathbb Z)$ is isometric to $L^2(\mathbb T)$, where $\mathbb T$ is the unit circle (equivalently, the domain of $2\pi-$periodic functions), and this is done by the transfomation $$\varPhi\big((a_k)_{k\in\mathbb Z}\big)=\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx},$$ as $\|(a_k)_{k\in\mathbb Z}\|_{\ell^2}=\big\|\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx}\big\|_{L^2}$. The shift $S$ in $L^2(\mathbb T)$ becomes a multiplication operator, by $\mathrm{e}^{ix}$, and hence, its the spectrum coincides with its essential range.

• Thank you for this answer. Unfortunately, we don't have the tools of the trigonometric functions available at this time in the course, still, +1 – Anthony Peter Nov 12 '15 at 9:24

HINT.If $a_{n+1}-z a_n=b_n$ then $a_2=z a_1+b_1$, $a_3=z a_2+b_2=z^2 a_1+z b_1+b_2$ ,$a_4=z a_3+b_3=z^3a_1+z^2b_1+z b_2+b_3$, etc. If $|z|=1$ and $b_n= z^{n}/(1+|n|)$, what happens to $|a_n-z^{n-1}a_1|$ for large positive $n$?

• I didn't mean to imply that there was anything wrong with anyone else's argument.I usually find the most difficult method. – DanielWainfleet Nov 12 '15 at 18:57
• Actually, could you slightly further your hint? – Anthony Peter Nov 13 '15 at 3:18
• Is it because $|a_n -z^{n-1}a_1|$ can grow arbitrarily large in modulus, and thus so can $a_n$? Since the series $|b_n|$ diverges? – Anthony Peter Nov 13 '15 at 3:41
• But if this were correct what's the need for considering $|a_n-z^{n-1}a_1|$ at all? – Anthony Peter Nov 13 '15 at 4:00
• @Anthony Peter . Hence $L-z I$ is not invertible because it's not a surjection because if $b_n=z^n/(|n|+1)$ there is no $a$ such that $(L-z I)a=b$. – DanielWainfleet Nov 13 '15 at 6:07

The operator $L$ maps $\{ a_n \}_{n=-\infty}^{\infty}$ to $\{ a_{n+1} \}_{n=-\infty}^{\infty}$. If $e_n$ is the standard basis element defined as a sequence of all $0$'s except for a $1$ in the $n$-th place, then $Le_n = e_{n-1}$, which is why $L$ would be considered to be a left shift. And this follows from the definition $L\{ a_n \} = \{ a_{n+1} \}$: the $k$ index spot in $\{ a_n \}$ becomes the $(k-1)$ index spot in $\{ a_{n+1} \}$. I prefer dealing with the basis elements because "left shift" makes sense in terms of the basis, and because I was taught to work in terms of the basis in Linear Algebra. The standard basis $\{ e_n \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $\ell^2(\mathbb{Z})$. In terms of this basis, $$L x = L \sum_{n=-\infty}^{\infty}(x,e_n)e_n = \sum_{n=-\infty}^{\infty}(x,e_n)e_{n-1}.$$ That makes $L$ look more like an operator from Linear Algebra, and it's rigorously correct.

The operator $(L-zI)$ means $$(L-zI)\{ a_n \} = L\{ a_n\} - z\{ a_n \} = \{ a_{n+1}-za_n \}.$$ You can alternatively view the action of $(L-zI)$ on the standard basis elements: $$(L-zI)e_n = e_{n-1}-ze_{n}.$$ Try showing that the following has no solution $x\in\ell^{2}$ for $|z|=1$: $$(L-zI)x = \sum_{n=-\infty}^{\infty}\frac{\overline{z}^n}{|n|+1}e_n$$ If it did, \begin{align} \sum_{n=-\infty}^{\infty}(x,e_n)(e_{n-1}-ze_n)&=\sum_{n=-\infty}^{\infty}\frac{\overline{z}^{n}}{|n|+1}e_n \\ \sum_{n=-\infty}^{\infty}\{(x,e_{n+1})-z(x,e_{n})\}e_n &= \sum_{n=-\infty}^{\infty}\frac{\overline{z}^{n}}{|n|+1}e_n \\ (x,e_{n+1})-z(x,e_{n})&=\frac{1}{|n|+1} \\ (x,e_n)-z(x,e_{n-1})&=\frac{\overline{z}}{|n-1|+1} \\ (x,e_{n-1})-z(x,e_{n-2})&=\frac{\overline{z}^{2}}{|n-2|+1} \end{align} Multiply these equations by powers of $z$ and add them to obtain a telescoping series. Then, knowing $(x,e_{n-k})\rightarrow 0$ as $k\rightarrow\infty$, $$(x,e_{n+1})=\sum_{k=0}^{\infty}\frac{1}{|n-k|+1}$$ This is a definite problem because the right side diverges. So $L-zI$ is not surjective for any unimodular $z$.