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

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)

• DO you know the norm on a functional? If so, the distance between two functionals $\phi (t)$ and $\phi (s)$ is $||\phi (t)-\phi (s)||$. Oct 12, 2017 at 20:58
• @Behnam. What is the norm on $\mathcal S'(\mathbb R^3)$? Oct 12, 2017 at 21:00
• 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. Oct 12, 2017 at 21:02
• Which is not a normed space! Oct 12, 2017 at 21:03
• The "norms" in $\mathcal{S}$ are $\lVert f\rVert_{\alpha,\beta}=\sup_x|x^\alpha D^\beta f|$ for every multi-index. Oct 12, 2017 at 21:08

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

• Remark: For your definition of continuity it is essential that $\mathbb R$ is a metric space. Oct 12, 2017 at 21:12
• @amsmath. I hope that still is the case. :-D But could you explain why that is essential? Oct 13, 2017 at 7:31
• Well, if $\phi : T\to\mathcal S'(\mathbb R^3)$, where $T$ is just some topological space, you cannot characterize continuity by convergence. Oct 13, 2017 at 13:47
• @amsmath. Not by sequential convergence. But by as I understand, convergence in a net is sufficient. Please correct me if I'm wrong. Oct 13, 2017 at 15:50
• 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. Oct 13, 2017 at 22:27

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 $$A$$, $$\lim_{t\rightarrow t_0}||(t-t_0)^{-1}[\phi(t)-\phi(t_0)]-\psi||_A=0\ .$$