I would like to see one or preferrably two isometries on $L^2(\mathbb{R})$ which are non surjective (equivalently non unitaries)?
Thanks in advance.
Math.
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$\begingroup$ Transfer the unilateral (right) shift on $\ell^2$ to $L^2(\mathbb{R})$, for example. $\endgroup$– user10354138Jun 6, 2019 at 14:29
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$\begingroup$ some more details are welcome. Thanks $\endgroup$– MathJun 6, 2019 at 14:32
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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 $$U_Tf (x) = \mathbb{1}_{(-\infty,0)}(x)f(x) + f(x-T)\mathbb{1}_{[T,\infty)}(x)$$
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$\begingroup$ Thanks a lot Aweygan. But I would like one "in one piece" if this is possible. $\endgroup$– MathJun 6, 2019 at 14:45
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$\begingroup$ I mean just one expression for $t$ fully in $\mathbb{R}$ not in three parts as is your interesting example. $\endgroup$– MathJun 6, 2019 at 14:48
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1$\begingroup$ @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 $\tau_Tg=g(\cdot-T)$. $\endgroup$ Jun 6, 2019 at 20:02
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