Modulation as a strongly continuous unitary group

On $H=L^2(\mathbb R,dx)$ , define

(M_t f)(x)=e^{itx}f(x),\qquad t\in\mathbb R.

The modulation group $(M_t)_{t\in\mathbb R}$ is a strongly continuous unitary representation of $(\mathbb R,+)$ , but it is not operator-norm continuous:

\|M_t-I\|_{\mathrm{op}}=2\qquad(t\ne0).

Two continuity calculations

For a fixed $f\in L^2$ ,

\|(M_t-I)f\|_2^2=\int_{\mathbb R}|e^{itx}-1|^2|f(x)|^2\,dx\longrightarrow0

by dominated convergence, using the integrable bound $4|f|^2$ . By contrast, the norm of a bounded multiplication operator is the essential supremum of its multiplier. For nonzero $t$ , the values of $|e^{itx}-1|$ approach $2$ on sets of positive measure, giving the operator-norm formula.

Generator and its domain

With the Stone convention $M_t=e^{itQ}$ ,

Qf=xf,\qquad\operatorname{Dom}(Q)=\{f\in L^2:xf\in L^2\}.

The norm derivative is $Kf=ixf$ on this domain. On it, $|(e^{itx}-1)/t|\leq |x|$ proves convergence of the difference quotient in $L^2$ ; conversely an $L^2$ derivative forces the pointwise candidate $ixf$ to lie in $L^2$ .

Finite versus infinite differentiability

The $C^k$ vectors are exactly $\{f:x^k f\in L^2\}$ . For example, $(1+x^2)^{-1}$ is a $C^1$ vector but not a $C^2$ vector. The smooth vectors for this one-parameter group have every polynomial weight in $L^2$ ; no spatial differentiability is required until translation operators are also included.