A strongly continuous unitary group need not be norm continuous

Multiplication by $e^{itx}$ is well defined on almost-everywhere classes, preserves the $L^2$ norm, and has inverse multiplication by $e^{-itx}$. The scalar exponential identities give $U(s)U(t)=U(s+t)$ and $U(0)=I$, so $U$ is a unitary group.

By the position-operator example $U(t)=e^{itQ}$, and by the generation lemma its generator is $iQ$ on $D(Q)=\{f:\int x^2\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty }\}$. These are the uses of AC inherited in [A6]. For positive integers $n$, $f_n={\mathbf{1}}_{[n,n+1]}$ has norm one, belongs to $D(Q)$, and satisfies $\mathrm{\parallel }Q{f}_n\mathrm{\parallel }\ge n$. Thus this generator is unbounded.

For $t\mathrm{\ne }0$, $\mathrm{\mid }{e}^{itx}-1\mathrm{\mid }\le 2$ gives $\mathrm{\parallel }U(t)-I\mathrm{\parallel }\le 2$. At $x_0=\pi \mathrm{/}t$ the absolute value equals $2$. Given $0<\epsilon <2$, continuity supplies $\delta >0$ such that it exceeds $2-\epsilon$ on the finite interval $J=(x_0-\delta ,x_0+\delta )$. The vector ${\mathbf{1}}_J\mathrm{/}\sqrt{2\delta }$ has norm one and its image under $U(t)-I$ has norm at least $2-\epsilon$. Letting $\epsilon \downarrow 0$ proves $\mathrm{\parallel }U(t)-I\mathrm{\parallel }=2$. At $t=0$ the norm is zero.

To prove the general assertion, let $S$ be a semigroup on a Banach space $X$ as defined in the statement, with $\mathrm{\parallel }S(h)-I\mathrm{\parallel }\to 0$ as $h\downarrow 0$. If $X=\{0\}$ its generator is the zero operator on all of $X$. Otherwise choose $a>0$ so that $\mathrm{\parallel }S(h)\mathrm{\parallel }\le 2$ for $0\le h\le a$. For $0\le s\le T$, write $s=ka+r$ with $0\le r<a$ and $k\le \lfloor T\mathrm{/}a\rfloor$; the semigroup law gives $\mathrm{\parallel }S(s)\mathrm{\parallel }\le M_T:=2^{\lfloor T\mathrm{/}a\rfloor +1}$. Consequently, for $0\le s\le t\le T$, $\mathrm{\parallel }S(t)-S(s)\mathrm{\parallel }\le M_T\mathrm{\parallel }S(t-s)-I\mathrm{\parallel }$. Thus $S$ is uniformly continuous in operator norm on every compact time interval.

For each $f\in H$, $\mathrm{\parallel }U(t)f-f{\mathrm{\parallel }}^2=\int \mathrm{\mid }{e}^{itx}-1{\mathrm{\mid }}^2\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx\to 0$ by dominated convergence, with majorant $4\mathrm{\mid }f{\mathrm{\mid }}^2$. This applies along every sequence $t\to 0$, hence gives continuity at zero. The group law and isometry give $\mathrm{\parallel }U(t)f-U(t_0)f\mathrm{\parallel }=\mathrm{\parallel }U(t-t_0)f-f\mathrm{\parallel }$, proving continuity at every $t_0$.

On a compact interval define the operator integral of $S$ by tagged Riemann sums. To justify existence, uniform continuity in step 1.4 bounds the difference between a sum and any refinement by the interval length times the modulus of continuity at the original mesh. Comparing two sums through their common refinement proves the Cauchy property as both meshes tend to zero. Completeness in [A5] gives the limit, independent of tags and partitions. The triangle inequality for sums gives $\mathrm{\parallel }{\int }_c^{d}F(s)ds\mathrm{\parallel }\le (d-c){\sup }_{[c,d]}\mathrm{\parallel }F(s)\mathrm{\parallel }$ for the continuous integrands used here. Linearity, subdivision, translation of intervals, and interchange with a fixed bounded operator follow first for sums and then for their limits. Put $B_{\tau }={\int }_0^{\tau }S(s)ds$. Then $\mathrm{\parallel }{\tau }^{-1}{B}_{\tau }-I\mathrm{\parallel }\le {\sup }_{0\le s\le \tau }\mathrm{\parallel }S(s)-I\mathrm{\parallel }<1$ for sufficiently small $\tau >0$.

For such $\tau$, let $C=I-{\tau }^{-1}{B}_{\tau }$ and $q=\mathrm{\parallel }C\mathrm{\parallel }<1$. The series $R={\sum }_{n=0}^{\mathrm{\infty }}{C}^n$ converges in operator norm: its tails are bounded by the geometric tails $\sum q^n$, and [A5] gives completeness. Telescoping finite sums and the product bound in [A4] give $(I-C)R=R(I-C)=I$. Thus $B_{\tau }$ is invertible with bounded inverse ${\tau }^{-1}R$.

For $h>0$, the semigroup law and the integral identities from step 2.2 yield $\frac{S(h)-I}{h}{B}_{\tau }=\frac{1}{h}({\int }_{\tau }^{\tau +h}S(s)ds-{\int }_0^{h}S(s)ds)\to S(\tau )-I$ in operator norm, by continuity at $\tau$ and at zero and the integral bound. Hence $B_{\tau }X\subseteq D(G)$ and $G{B}_{\tau }=S(\tau )-I$. Since $B_{\tau }$ is onto, $D(G)=X$ and $G=(S(\tau )-I)B_{\tau }^{-1}$ is bounded. There is no additional factor $1\mathrm{/}\tau$ in this last formula with the unnormalized integral $B_{\tau }$.

Restrict $U$ to nonnegative times. Its right generator extends the two-sided generator $iQ$: on $D(Q)$ the two-sided limit from step 1.2 in particular gives the right limit. If $U$ were norm continuous at zero, step 4.1 would make this right generator bounded on all of $H$, contradicting the unit vectors $f_n$ in step 1.2. This corroborates the direct computation in step 1.3 and completes all claims.