Left Shift Semigroup and Its Infinitesimal Generator

Left shift operator

Throughout we consider the Hilbert space \(L^2=L^2(\mathbb{R})\), the space of all complex-valued functions with real variable such that \(f \in L^2\) if and only if \[ \lVert f \rVert_2^2=\int_{-\infty}^{\infty}|f(t)|^2dm(t)<\infty \] where \(m\) denotes the ordinary Lebesgue measure (in fact it's legitimate to consider Riemann integral in this context).

For each \(t \geq 0\), we assign an bounded linear operator \(Q(t)\) such that \[ (Q(t)f)(s)=f(s+t). \] This is indeed bounded since we have \(\lVert Q(t)f \rVert_2 = \lVert f \rVert_2\) as the Lebesgue measure is translate-invariant. This is a left translation operator with a single step \(t\).

Properties of \(Q(t)\)

In Hilbert space

The inner product in \(L^2\) is defined by \[ (f,g)=\int_{-\infty}^{\infty}f(s)\overline{g(s)}dm(s), \quad f,g\in L^2. \] If we apply \(Q(t)\) on \(f\), we see \[ \begin{aligned} (Q(t)f,g) &= \int_{-\infty}^{\infty}f(s+t)\overline{g(s)}dm(s) \\ &= \int_{-\infty}^{\infty}f(u)\overline{g(u-t)}dm(u) \quad (u=s+t) \\ &= (f,Q(t)^{\ast}g) \end{aligned} \] where \(Q(t)^\ast\) is the adjoint of \(Q(t)\), which happens to be a left translation operator with a single step \(t\). Clearly we have \(Q(t)Q(t)^\ast=Q(t)^\ast Q(t)=I\), which indicates that \(Q(t)\) is unitary. Also we can check in a more manual way: \[ (Q(t)f,Q(t)g) = \int_{-\infty}^{\infty}f(s+t)\overline{g(s+t)}dm(s) = \int_{-\infty}^{\infty}f(s+t)\overline{g(s+t)}dm(s+t)=(f,g). \] By operator theory, since \(Q(t)\) is unitary and bounded, the spectrum of \(Q(t)\) lies in the unit circle \(S^1\).

As a semigroup

Note \(Q(0)=I\) and \[ Q(t+u)f(s)=f(s+t+u)=f[(s+t)+u]=Q(u)f(s+t)=Q(t)Q(u)f(s) \] for all \(f \in L^2\), which is to say that \(Q(t+u)=Q(t)Q(u)\). Therefore we say \(\{Q(t)\}\) is a semigroup. But what's more important is that it satisfies strong continuity near the origin: \[ \lim_{t \to 0}\lVert Q(t)f - f \rVert_2 = 0. \] This is not too hard to verify. It suffices to prove that \[ \lim_{t \to 0}\int_{-\infty}^{\infty} |f(s+t)-f(s)|^2dm(s) =0. \] Note \(C_c(\mathbb{R})\) (continuous function with compact support) is dense in \(L^2\), and for \(f \in C_c(\mathbb{R})\), it follows immediately from properties of continuous functions. Next pick \(f \in L^2\). Then for \(\varepsilon>0\) there exists some \(f_1 \in C_c(\mathbb{R})\) such that \(\lVert f-f_1 \rVert_2 < \frac{\varepsilon}{4}\) and \(\lVert f_1(s+t)-f_1(s)\rVert_2<\frac{\varepsilon}{2}\) for \(t\) small enough. If we put \(f_2=f-f_1\) we get \[ \begin{aligned} \lVert f(s+t)-f(s) \rVert_2 &\leq \lVert f_1(s+t)-f_1(s) \rVert_2+\lVert f_2(s+t)-f_2(s) \rVert \\ &< \frac{\varepsilon}{2}+2\lVert f_2(s)\rVert < \varepsilon. \end{aligned} \] The limit follows as \(\varepsilon \to 0\).

Infinitesimal generator of \(Q(t)\)

Recall that the infinitesimal generator of \(Q(t)\) is defined to be \[ A=\lim_{t \to 0}\frac{1}{t}[Q(t)-I] \] which is inspired by \(\frac{d}{dt}e^{tA}=A\) (thanks to von Neumann). Note if \(f \in L^2\) is differentiable, then \[ Af(s) = \lim_{t \to 0} \frac{f(s+t)-f(s)}{t} = f'(s). \] The infinitesimal generator of \(Q(t)\) being differentiation operator is quite intuitive. But we need to clarify it in \(L^2\) which is much larger. So what is the domain \(D(A)\)? We don't know yet but we can guess. When talking about differentiation in \(L^p\) space, it makes sense to extend our differentiation to absolute continuity. Also we need to make sure that \(Af \in L^2\), hence we put \[ D=\{f\in L^2:f \text{ absolutely continuous, }f' \in L^2\}. \] For every \(x \in D(A)\) and any fixed \(t\) we already have \[ \frac{d}{dt}Q(t)f(s)=f'(s+t)=Af(s+t) \] hence \(Af=f'\) for every \(x \in D(A)\) and it follows that \(D(A) \subset D\). In fact, \(A\) is the restriction of the differential operator on \(D(A)\). Conversely, By Hille-Yosida theorem, we see \(1 \in \rho(A)\) and also one can show that \(1 \in \rho(\frac{d}{dx})\). Therefore \[ (I-\frac{d}{dx})D(A)=(I-A)D(A)=L^2. \] But we also have \[ D=(I-\frac{d}{dx})^{-1}L^2. \] Thus \[ D = \left(1-\frac{d}{dx}\right)^{-1}\left(1-\frac{d}{dx}\right)D(A)=D(A). \] The fact that \((I-\frac{d}{dx})D=L^2\) can be realised by the equation \(f-f'=g\), where the existence of solution can be proved using Fourier transform. Note \(\hat{f'}(y)=iy\hat{f}(y)\), with some knowledge of distribution, the result can also be given by \[ D(A)=\left\{f\in L^2:\int_{-\infty}^{\infty}|y\hat{f}(y)|^2dy<\infty\right\}. \]

Spectrum of the generator

By the Hille-Yosida theorem, the half plane \(\{z:\Re z>0\} \subset \rho(A)\). But we can give a more precise result of it.

Pick any \(f \in D(A)\). It is directly verified that \[ (A-\lambda{I})f = f'-\lambda{f}. \] Put \(g=(A-\lambda{I})f\) then \[ \hat{g}(y)=iy\hat{f}(y)-\lambda{\hat{f}(y)}. \] Therefore \[ \hat{f}(y) = \frac{\hat{g}(y)}{iy-\lambda} \in L^2. \] Conversely, suppose \(h(y)=\frac{\hat{g}(y)}{iy-\lambda} \in L^2\), then \(\hat{g}(y)=iyh(y)-\lambda{h}(y)\). If we take its Fourier inverse, we see \(g \in R(A-\lambda{I})\).

If \(g \in L^2\), then clearly \(\hat{g} \in L^2\). It remains to discuss \(\hat{g}(y)/(iy-\lambda)\). Note \(iy\) is on the imaginary axis, hence if \(\lambda\) is not purely imaginary, then \(\hat{g}(y)/(iy-\lambda) \in L^2\). If \(\lambda\) is purely imaginary however, then we may have \(\hat{g}(y)/(iy-\lambda)\not\in L^2\). For example, we can take \(\hat{g}=\chi_{[s-1,s+1]}\) where \(\lambda = is\). Hence if \(\lambda\) is purely imaginary, \(R(A-{\lambda}I)\) is a proper subspace of \(L^2\). Therefore we conclude: \[ \sigma(A)= \{z \in \mathbb{C}:\Re z = 0\}. \] This is an exercise on W. Rudin's Functional Analysis. You can find related theorems in Chapter 13.

Left Shift Semigroup and Its Infinitesimal Generator

https://desvl.xyz/2021/05/26/left-shift/

Author

Desvl

Posted on

2021-05-26

Updated on

2021-10-11

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