Consider the map $$ \Lambda: \mathbb{R}\to\mathcal{L}(L^2(\mathbb{R}),L^2(\mathbb{R})) =: \mathcal{L}$$ defined by $\Lambda(s)(f)(x) = (1+x^2)^{is/2}f(x)$ where $f\in L^2$ and $s,x\in\mathbb{R}$. Recall that $\mathcal{L}$ has the norm topology (also called the uniform operator topology), but it also was the weak and strong operator topologies, defined as follows:
- The former is the weakest one such that all the maps $\mathcal{L}\to\mathbb{C}$, defined via $T \mapsto \ell(Tf)$, are continuous (where $\ell\in(L^2)^*,f\in L^2$ are arbitrary).
- The latter is the weakest one such that all the maps $\mathcal{L}\to L^2$, defined via $T\mapsto Tf$, are continuous (where $f\in L^2$ is arbitrary).
How can we determine if $\Lambda$ is continuous w.r.t. each of these three topologies? Letting $h_s(x):= (1+x^2)^{is/2}$, I think that continuity wrt the uniform, strong, and weak topolgies (respectively) is equivalent to:
- If $s\in \mathbb{R}$ is fixed, then as $t\to s$ we have $\int_\mathbb{R} |f|^2|h_s-h_t|^2\to 0$ where $f$ is allowed to depend on $t$. Rigorously, $\lim_{t\to s}\sup_{f\in L^2}\int_\mathbb{R}|f|^2 |h_s-h_t|^2 =0$.
- The same statement as above, but now $f$ is fixed as $t\to s$.
- If $s\in\mathbb{R}$ is fixed and $j\in L^1$ is fixed, then as $t\to s$ we have $\int_\mathbb{R} (h_s-h_t)\cdot j \to 0$.
All three are eluding me for the moment, unfortunately. Thanks for your help!
We claim it is not continuous wrt the norm topology, but it is continuous wrt the other two (it suffices to prove this for the strong operator topology). If $\Lambda$ were continuous wrt the uniform operator topology, then by definition $\forall s\in\mathbb{R}$ we'd have $$\lim_{t\to s} || \Lambda_s-\Lambda_t||_{L^2\to L^2} = 0.$$ Notice $\Lambda_s-\Lambda_t$ is simply a multiplication operator, that multiplies by $h_s-h_t$. We claim that this implies its operator norm equals $||h_s-h_t||_{\infty}$. This is a general fact: if $M: L^2(\mathbb{R})\to L^2(\mathbb{R})$ multiplies by some function $m$, then $||M||=||m||_\infty$. Indeed, $||M||\le ||m||_\infty$ is obvious and for any $\epsilon>0$ we can find a nonzero $f$ supported on $\{ |m| > ||m||_\infty-\epsilon\}$, so $||Mf||_2 \ge (||m||_\infty-\epsilon)||f||_2$ so $||M||\ge ||m||_\infty-\epsilon$. This means that $||\Lambda_s-\Lambda_t|| = ||h_s-h_t||_\infty = \sup_x (1+x^2)^{is/2} - (1+x^2)^{it/2}=2$, which is easy to verify (by setting $\xi=\log(1+x^2)$, which can be any positive real number, and solving $\xi = \frac{\pi(4n)}{s} = \frac{\pi(4n+2+\epsilon)}{t}$ for $n\in\mathbb{Z}$ and $\epsilon\ge 0$ really small. This makes $h_s(x)=1$ and $h_t(x)\approx -1$). Thus the above limit is in fact 2!
To see continuity wrt the strong operator norm, we claim it suffices to prove that all the compositions $$ \mathbb{R}\overset\Lambda\longrightarrow \mathcal{L} \overset{\text{ev}_f}\longrightarrow L^2(\mathbb{R})$$ are continuous, for any fixed $f\in L^2$, where $\text{ev}_f$ is the evaluation map at $f$. This is because the strong operator topology is the coarsest one such that all these evaluation maps are continuous. (And if the above compisitions were continuous, then $\Lambda^{-1}(U)$ is open for any $U\subset\mathcal{L}$ of the form $U=\text{ev}_f^{-1}(V)$ and $V\subset L^2$ open, but these $U$'s generate the strong operator topology and so $\Lambda^{-1}(U)$ must be open for any $U$). The above compositions simply take $s\mapsto h_sf$. So we must prove $$ \lim_{t\to s} \int_\mathbb{R} |f|^2|h_s-h_t|^2 =0,$$ with $f\in L^2$ fixed. It suffices to take a sequence $t_n\to s$. Notice $|h_s-h_{t_n}|^2\le 4$ and so the functions $g_n := |f|^2|h_s-h_{t_n}|^2$ are dominated by $4|f|^2\in L^1(\mathbb{R})$. Moreover, $g_n\to 0$ pointwise because for any $x$, $\lim_{t\to s} h_t(x)=h_s(x)$. Hence, by the Dominated Convergence Theorem, $$ \lim_{n\to\infty} \int_\mathbb{R} g_n = \int_\mathbb{R} \lim_{n\to\infty} g_n = \int_\mathbb{R} 0 = 0,$$ as desired (where the second $\lim$ denotes the pointwise limit).