# An example of a non unitary isometry on $L^2(\mathbb{R})$?

I would like to see one or preferrably two isometries on $$L^2(\mathbb{R})$$ which are non surjective (equivalently non unitaries)?

Math.

• Transfer the unilateral (right) shift on $\ell^2$ to $L^2(\mathbb{R})$, for example. Jun 6 '19 at 14:29
• some more details are welcome. Thanks
– Math
Jun 6 '19 at 14:32

I'll give you uncountably many non-unitary isometries. For $$T>0$$, define $$U_T:L^2(\mathbb R)\to L^2(\mathbb R)$$ by \begin{align*} (U_Tf)(t)=\left\{\begin{array}{lcl} f(t-T) &:& t\in[T,\infty),\\ 0 &:& t\in[0,T),\\ f(t)&:&t\in(-\infty,0). \end{array}\right. \end{align*}

That is $$U_Tf (x) = \mathbb{1}_{(-\infty,0)}(x)f(x) + f(x-T)\mathbb{1}_{[T,\infty)}(x)$$

• Thanks a lot Aweygan. But I would like one "in one piece" if this is possible.
– Math
Jun 6 '19 at 14:45
• Please define what you mean. Jun 6 '19 at 14:46
• I mean just one expression for $t$ fully in $\mathbb{R}$ not in three parts as is your interesting example.
– Math
Jun 6 '19 at 14:48
• @Math: It is not difficult to put things in one line. Try to do it yourself. The pint being is that the operator $U_T$ is the summer of two operators: the first $S$ is multiplication by $\mathbb{1}_{(-\infty,0)}$ as in $Sf=\mathbb{1}_{(-\infty,0)}f$, and the second $W$ is the composition of multiplication by $\mathbb{1}_{[0,\infty)}$ with the translation (by T units) to the right as in $Wf =\tau_T(\mathbb{1}_{([0,\infty)}f)$, where $\tau_Tg=g(\cdot-T)$. Jun 6 '19 at 20:02
• Thanks a lot Oliver Diaz.
– Math
Jun 6 '19 at 20:37