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.