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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)

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  • $\begingroup$ DO you know the norm on a functional? If so, the distance between two functionals $\phi (t)$ and $\phi (s)$ is $||\phi (t)-\phi (s)||$. $\endgroup$ Oct 12, 2017 at 20:58
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    $\begingroup$ @Behnam. What is the norm on $\mathcal S'(\mathbb R^3)$? $\endgroup$
    – md2perpe
    Oct 12, 2017 at 21:00
  • $\begingroup$ I assume that $\mathcal{S}$ is the Schwartz space. To talk about continuity, you must assign a topology, or definition of closeness in the space $\mathcal{S}'$. A usual base for this sort of spaces is the collection of translation of sets $V_{f}(\delta)=\{u\in \mathcal{S}'\mid |\langle u,f \rangle|<\delta\}$, for $f\in \mathcal{S}$. This is the general philosophy, look for some topology. $\endgroup$
    – user90189
    Oct 12, 2017 at 21:02
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    $\begingroup$ Which is not a normed space! $\endgroup$
    – amsmath
    Oct 12, 2017 at 21:03
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    $\begingroup$ The "norms" in $\mathcal{S}$ are $\lVert f\rVert_{\alpha,\beta}=\sup_x|x^\alpha D^\beta f|$ for every multi-index. $\endgroup$
    – user90189
    Oct 12, 2017 at 21:08

2 Answers 2

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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 $\rho$.


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 $$\langle \phi'(t_0), \rho \rangle = \lim_{h \to 0} \frac{\langle \phi(t_0+h) - \phi(t_0), \rho \rangle}{h} = \lim_{h \to 0} \frac{\langle \phi(t_0+h), \rho \rangle - \langle \phi(t_0), \rho \rangle}{h} .$$

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  • $\begingroup$ Remark: For your definition of continuity it is essential that $\mathbb R$ is a metric space. $\endgroup$
    – amsmath
    Oct 12, 2017 at 21:12
  • $\begingroup$ @amsmath. I hope that still is the case. :-D But could you explain why that is essential? $\endgroup$
    – md2perpe
    Oct 13, 2017 at 7:31
  • $\begingroup$ Well, if $\phi : T\to\mathcal S'(\mathbb R^3)$, where $T$ is just some topological space, you cannot characterize continuity by convergence. $\endgroup$
    – amsmath
    Oct 13, 2017 at 13:47
  • $\begingroup$ @amsmath. Not by sequential convergence. But by as I understand, convergence in a net is sufficient. Please correct me if I'm wrong. $\endgroup$
    – md2perpe
    Oct 13, 2017 at 15:50
  • $\begingroup$ I am sorry I cannot address you with @. It just doesn't work... Well, as far as I can remember, you are right. However, you did not refer to net convergence at all in your answer. But in fact, you do not have to do this: A map $f : X\to Y$ between a metric space and a topological space is continuous if and only if it is sequentially continuous. That was what I tried to point you to. $\endgroup$
    – amsmath
    Oct 13, 2017 at 22:27
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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 $A$, $$ \lim_{t\rightarrow t_0}||(t-t_0)^{-1}[\phi(t)-\phi(t_0)]-\psi||_A=0\ . $$

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