Another Stanford Qual question that has been bugging me.
Let $\mathcal{L}(L^2(\mathbb{R}))$ be the set of bounded linear operators on $L^2(\mathbb{R})$. Define $\Lambda_sf(\xi)=(1+|\xi|^2)^{is\backslash2}f(\xi)$, for $s \in \mathbb{R}$.
Consider the map $\Lambda:\mathbb{R}\to \mathcal{L}(L^2(\mathbb{R}))$ given by $\Lambda(s)=\Lambda_s$. Prove or disprove the following:
a) $\Lambda$ is continuous when $\mathcal{L}(L^2(\mathbb{R}))$ is given norm topology.
b)$\Lambda$ is continuous when $\mathcal{L}(L^2(\mathbb{R}))$ is given strong(operator) topology.
c)$\Lambda$ is continuous when $\mathcal{L}(L^2(\mathbb{R}))$ is given weak(operator) topology.
$\textbf{Thoughts}$:
1) $\Lambda_s$ is bounded in norm topology since $(1+|\xi|^2)^{is\backslash2}=\exp(\log(1+|\xi|^2)\frac{is}{2})$ which is in turn bounded by 1. Thus $||\Lambda_s f||\leq ||f||_2 $.
2) To prove continuity(or discontinuity) of $\Lambda$ it is enough to consider continuity at 0.
3) $||\Lambda_sf-\Lambda_0f||_2=||\left[\exp(\log(1+|\xi|^2)\frac{is}{2})-1\right]f||_2$
Now for a fixed $s>0$ we can choose $\xi$ so that $\log(1+|\xi|^2)\frac{s}{2}$ is a multiple of $\frac{\pi}{2}$, hence in a neighborhood of these points, $|\exp(\log(1+|\xi|^2)\frac{is}{2})-1| \geq \frac{1}{2}$. And hand wavely speaking, we can choose $f$ so that $f$ is supported on the union of these neighborhoods and $||f||_2=1$.
Thus we have $||\Lambda_sf-\Lambda_0f||_2=||\left[\exp(\log(1+|\xi|^2)\frac{is}{2})-1\right]f||_2 \geq \frac{1}{2}$. Hence $\Lambda$ is not continuous in strong and norm topologies.
I think that the same type of example can be used to show that $\Lambda$ is not continuous in the weak operator topology.
All of this could be complete BS and I would appreciate if anyone could give me suggestions. Thanks in advance.
PS- The operator looks awful lot like the Fourier definition of Sobolev spaces except there is an $i$ there. Is there any sort of connection there, perhaps like heat and schrodinger's equation?
$\textbf{Edit}:$ As pointed out by amsmath that my counterexample works for the norm but not for strong and weak operator topologies. The reason being that the $f$ I was choosing depends on $s$ and for a fixed $f$ we can use Lebesgue dominated conv. theorem to show that it is actually continuous in the strong toplolgy.