Shift Operator has no "square root"?

Question

Consider the left shift operator $T : \ell^1(\mathbb N) \to \ell^1(\mathbb N) $ by $$T(x_1,x_2..... )=(x_2, x_3 ........),$$ and also the right shift operator $S : \ell^1(\mathbb N) \to \ell^1(\mathbb N) $ by $$S(x_1,x_2..... )=(0, x_1, x_2 ........).$$ Can we find an linear operator $R: \ell^1(\mathbb N) \to \ell^1(\mathbb N) $ such that $T=R^2$ or $S=R^2$?

In my opinion, this is intuitively not true because one cannot "shift a vector by half position". But how to prove this? Can anyone help? Thanks!

Have you computed the kernel and co-kernels of $T$ and $S$? Once you know that, consider what they'd have to be for $R$. — Ryan Budney, Sep 05 '13 at 22:43
Isn't this a similar question like that for a fractional (here: half) derivative? (Just curious). From that a similar answer should then emerge... — Gottfried Helms, Sep 06 '13 at 08:25
"You cannot shift a vector by half a position". Not a very good intuition, since the bilateral shift does have a square root. — Martin Argerami, Nov 09 '19 at 05:59

Brian M. Scott · Accepted Answer · 2013-09-06T08:03:19.503

14

Suppose that $R^2=T$. Then $\ker R\subseteq\ker T=\Bbb Re_0$, where $e_0=\langle 1,0,0,\ldots\rangle$. Clearly $\ker R$ is non-trivial, so $\ker R=\ker T$. Moreover, $T$ is surjective, so $R$ must also be surjective. In particular, $e_0=Rx$ for some $x\in\ell^1(\Bbb N)\setminus\ker T$, and therefore $R^2x=Re_0=0\ne Tx$.

Now suppose that $R^2=S$, and let $V=\operatorname{ran}S$; clearly $\operatorname{ran}R\supseteq V$. $R$ cannot be surjective, so $\operatorname{ran}R=V$. But then $\operatorname{ran}\left(R\upharpoonright V\right)=\operatorname{ran}S=V$, so for each $x\in\ell^1(\Bbb N)\setminus V$ there must be a $y\in V$ such that $Rx=Ry$. However, $S$ is injective, so $R$ must also be injective.

edited Sep 06 '13 at 08:03

answered Sep 05 '13 at 23:34

Brian M. Scott

616,228

Could you explain why there is an $x\in\ell^1(\Bbb N)\setminus V$ such that $Rx=e_1$? – L. Xu Sep 06 '13 at 07:39
1

@L.Xu: I have no idea what I was thinking when I wrote that: I must have been hallucinating. Fortunately, the idea of showing getting a contradiction involving (non-)injectivity of $R$ does work; I’ve completely revised the paragraph. – Brian M. Scott Sep 06 '13 at 08:04
@BrianM.Scott I got stuck. How can we derive in the second paragraph that because $V\subset ranR$ and $R$ is not surjective then $ranR=V$? Thank you in advance! – user113576 Apr 22 '15 at 19:14
1

@user113576: If $x=\langle x_1,x_2,\ldots\rangle\in\ell^1(\Bbb N)\setminus V$, then $x_1\ne 0$, and $x-S(x)=\langle x_1,0,0,0,\ldots\rangle\in\operatorname{ran}R$. But the span of ${x-S(x)}\cup V$ is clearly all of $\ell^1(\Bbb N)$, so that would make $\operatorname{ran}R=\ell^1(\Bbb N)$. – Brian M. Scott Apr 22 '15 at 19:21

score 1 · Answer 2 · answered Sep 06 '13 at 02:02

I don't think it is obvious. For $T$, in theory, we might shift $x_2$ to position $n, x_n$ to position $3, x_3$ to position $n+1$ and so on. The problem comes with the "and so on". Let $x_2$ go to position $n$. Then we must have $x_n$ goes to position $1, x_1$ to position $n-1, x_{n+1}$ to position $2$, $x_3$ to position $n+1$ and so on. So $R(x_1,x_2,x_3,\dots )=(x_{n-1}x_n,x_{n+1},x_{n+2}\dots x_2,x_3,x_4,\dots)$. But now we have position $n-1$ holding $x_{2n-3}$ and $R(x_1,x_2,x_3,\dots )=(x_{n-1},x_n,x_{n+1},x_{n+2}\dots ,x_{2n-4},x_{2n-3},x_2,x_3,x_4,\dots,x_{n-4},x_{n-3},x_{n-2}\dots)$ and we have accounted for the first $2n-3$ entries so far. Unfortunately, position $2n-3$ will end up having $x_1$ in it, so this fails.

Shift Operator has no "square root"?

2 Answers2

Linked