Shift operator

The shift operator T^t (t ∈ R) takes a function f on R to its translation f_t ,

${\displaystyle T^{t}f(x)=f_{t}(x)=f(x+t)~.}$

A practical representation of the linear operator T^t in terms of the plain derivative ^d⁄_dx was introduced by Lagrange,

which may be interpreted operationally through its formal Taylor expansion in t; and whose action on the monomial xⁿ is evident by the binomial theorem, and hence on all series in x, and so all functions f(x) as above.[3] This, then, is a formal encoding of the Taylor expansion.

The operator thus provides the prototype[4] for Lie's celebrated advective flow for Abelian groups,

${\displaystyle e^{t\beta (x){\frac {d}{dx}}}f(x)=e^{t{\frac {d}{dh}}}F(h)=F(h+t)=f(h^{-1}(h(x)+t)),}$

where the canonical coordinates h (Abel functions) are defined, s.t.

${\displaystyle h'(x)\equiv {\frac {1}{\beta (x)}}~,\qquad f(x)\equiv F(h(x)).}$

For example, it easily follows that ${\displaystyle \beta (x)=x}$ yields scaling,

${\displaystyle e^{tx{\frac {d}{dx}}}f(x)=f(e^{t}x)}$,

hence ${\displaystyle e^{i\pi x{\frac {d}{dx}}}f(x)=f(-x)}$ (parity); likewise, ${\displaystyle \beta (x)=x^{2}}$ yields[5]

${\displaystyle e^{tx^{2}{\frac {d}{dx}}}f(x)=f\left({\frac {1-tx}{x}}\right)}$,

${\displaystyle \beta (x)=1/x}$ yields

${\displaystyle e^{{\frac {t}{x}}{\frac {d}{dx}}}f(x)=f({\sqrt {x^{2}+2t}}~)}$,

${\displaystyle \beta (x)=e^{x}}$ yields

${\displaystyle \exp \left(te^{x}{\frac {d}{dx}}\right)f(x)=f\left(\ln \left({\frac {1}{\exp(-x)-t}}\right)\right)}$,

etc.

The initial condition of the flow and the group property completely determine the entire Lie flow, providing a solution to the translation functional equation[6]

${\displaystyle f_{t}(f_{\tau }(x))=f_{t+\tau }(x).}$

The left shift operator acts on one-sided infinite sequence of numbers by

${\displaystyle S^{*}:(a_{1},a_{2},a_{3},\ldots )\mapsto (a_{2},a_{3},a_{4},\ldots )}$

and on two-sided infinite sequences by

${\displaystyle T:(a_{k})_{k=-\infty }^{\infty }\mapsto (a_{k+1})_{k=-\infty }^{\infty }.}$

The right shift operator acts on one-sided infinite sequence of numbers by

${\displaystyle S:(a_{1},a_{2},a_{3},\ldots )\mapsto (0,a_{1},a_{2},\ldots )}$

and on two-sided infinite sequences by

${\displaystyle T^{-1}:(a_{k})_{k=-\infty }^{\infty }\mapsto (a_{k-1})_{k=-\infty }^{\infty }.}$

The right and left shift operators acting on two-sided infinite sequences are called bilateral shifts.

In general, as illustrated above, if F is a function on an abelian group G, and h is an element of G, the shift operator T ^g maps F to[6][7]

${\displaystyle F_{g}(h)=F(h+g).}$

The shift operator acting on two-sided sequences is a unitary operator on ℓ₂(Z). The shift operator acting on functions of a real variable is a unitary operator on L₂(R).

In both cases, the (left) shift operator satisfies the following commutation relation with the Fourier transform:

${\displaystyle {\mathcal {F}}T^{t}=M^{t}{\mathcal {F}},}$

where M^t is the multiplication operator by exp(i t x). Therefore, the spectrum of T^t is the unit circle.

The one-sided shift S acting on ℓ₂(N) is a proper isometry with range equal to all vectors which vanish in the first coordinate. The operator S is a compression of T⁻¹, in the sense that

${\displaystyle T^{-1}y=Sx{\text{ for each }}x\in \ell ^{2}(\mathbb {N} ),\,}$

where y is the vector in ℓ₂(Z) with y_i = x_i for i ≥ 0 and y_i = 0 for i < 0. This observation is at the heart of the construction of many unitary dilations of isometries.

The spectrum of S is the unit disk. The shift S is one example of a Fredholm operator; it has Fredholm index −1.