Differentiability of tempered distributions over $\mathbb{R}^3$

Question

One of the topics I find interesting in physics is the time evolution of improper quantum states (physicists call them kets). My question to this mathematical community is:

In what exact sense is a mapping $\phi:\mathbb{R}\to \mathcal{S}'(\mathbb{R}^3)$, $t\mapsto\phi(t)$ continuous and furthermore differentiable in the variable/parameter "$t$"? ($\mathcal{S}'(\mathbb{R}^3)$ is of course the topological dual of the Schwartz test function space)

Answer 1

Definition: $\phi : \mathbb R \to \mathcal S'(\mathbb R^3)$ is continuous if $\phi(t) \to \phi(t_0)$ whenever $t \to t_0$.

But what does $\phi(t) \to \phi(t_0)$ mean, i.e. how is convergence in $\mathcal S'(\mathbb R^3)$ defined?

Definition: $u_\lambda \to u_0$ when $\lambda \to 0$ if $\langle u_\lambda, \rho \rangle \to \langle u_0, \rho \rangle$ for every $\rho \in \mathcal S(\mathbb R^3).$

With $\langle u, \rho \rangle$ I mean the application of the tempered distribution $u$ on the testfunction $\rho$.

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For differentiability we can just use the normal definition. $$ \phi'(t_0) = \lim_{h \to 0} \frac{\phi(t_0+h) - \phi(t_0)}{h}$$ meaning that $$\langle \phi'(t_0), \rho \rangle = \lim_{h \to 0} \frac{\langle \phi(t_0+h) - \phi(t_0), \rho \rangle}{h} = \lim_{h \to 0} \frac{\langle \phi(t_0+h), \rho \rangle - \langle \phi(t_0), \rho \rangle}{h} .$$

Answer 2

The topology one should use on $\mathscr{S}'(\mathbb{R}^d)$ is not the weak-$\ast$ topology but rather the strong topology. It is the locally convex topology defined by the family of seminorms $$ ||\phi||_A=\sup_{f\in A}|\phi(f)| $$ indexed by bounded subsets $A$ of Schwartz space $\mathscr{S}(\mathbb{R}^d)$. Saying that a subset $A$ is bounded means that for all multiindex $\alpha$ and all nonnegative integer $k$, $$ \sup_{f\in A}\sup_{x\in\mathbb{R}^d}\langle x\rangle^k|\partial^{\alpha}f(x)|\ <\ \infty\ . $$

Now continuity of $t\mapsto \phi(t)$ at $t_0$ means that for all bounded set $A$, $$ \lim_{t\rightarrow t_0}||\phi(t)-\phi(t_0)||_A=0\ . $$

Finally, differentiability at $t_0$ with derivative equal to some distribution $\psi$ in $\mathscr{S}'(\mathbb{R}^d)$ means that for all bounded $A$, $$ \lim_{t\rightarrow t_0}||(t-t_0)^{-1}[\phi(t)-\phi(t_0)]-\psi||_A=0\ . $$