9
votes
How should I understand the "$C^\infty$ functions" whose domain is the dual of $C^\infty(\mathbb{R}^n)$?
This is more like a long comment on the notion of smoothness than an actual answer, which has already been provided by Jochen Wengenroth. It tries to address the follow-up question the OP posted as a ...
- 7,817
8
votes
Accepted
How should I understand the "$C^\infty$ functions" whose domain is the dual of $C^\infty(\mathbb{R}^n)$?
The standard topology on $C^\infty(\Omega)$ is that of uniform convergence on compact subsets of $\Omega$ of all derivatives which is given by the seminorms $\|f\|_{K,n}=\sup\{|\partial^\alpha f(x)|: ...
- 16.4k
3
votes
Accepted
The notion of "Admissible" and "Permitted" in the context of convolution with distributions and hypocontinuity
The authors Yoshinaga and Ogata of On concolutions use Schwartz's notation for spaces of functions and distributions. The requirements for the sequence of smooth functions $\alpha_k$ with values in $[...
- 16.4k
2
votes
When is the periodisation of a function continuous?
You just need that $f$ decreases fast enough to make the series uniformly convergent on compact sets. E.g., it would be enough that
$|x|^p |f(x)|$ is bounded for some $p>1$. Then you can estimate ...
- 16.4k
2
votes
Accepted
Restore initial condition for distributions
$\newcommand{\R}{\mathbb R}$In accordance with comments by Willie Wong and the OP, let us extend $u$ by $a$ to the left of $0$:
\begin{equation*}
U(t) :=
\begin{cases}
u(t) & \text{ if }t\ge0,...
- 128k
1
vote
When is the periodisation of a function continuous?
Short answer: e.g. for Schwartz functions.
Long answer: The Fourier transform of "periodic" is "discrete" and the Fourier transform of "discrete" is "periodic". ...
- 11
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