Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with fa.functional-analysis
Search options not deleted
user 47322
Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
0
votes
Is $C^{\infty}(\mathbb{R}^{m+n})$ a flat module over $C^{\infty}(\mathbb{R}^{m})$?
Let $X,Y$ be smooth manifolds; let $\Im = <r_1,\dots,r_d>_{C^\infty(X)}$ be a finitely generated ideal of $C^\infty(X)$. Then $\Im \otimes_{C^\infty(X)} C^\infty(X \times Y) = \{ \sum_{i=1}^d r_i \oti …
- 489
2
votes
Is $C^{\infty}(\mathbb{R}^{m+n})$ a flat module over $C^{\infty}(\mathbb{R}^{m})$?
PS: The answer below has gaps, and it is likely incorrect.
Yes, $C^{\infty}(\mathbb{R}^{m+n})$ is a flat $C^{\infty}(\mathbb{R}^{m})$ module. Or, following @Pietro 's comment, with more generalit …
- 489
1
vote
Accepted
Smooth Approximation of Indicator Function of Convex Sets in $\mathbb{R}^n$
Let's pursue Jochen's idea. We assume $A \ne \emptyset.$
Let
$$ \varphi(t) = \begin{cases}
e^{-\frac{1}{t}} &\text{if $ t>0$}\\
0 &\text{otherwise.}
\end{cases}$$
This func …
- 489
1
vote
Regularity of finite variation kernels in the (intersection) of the semimartingale spaces $H^p$
If the answer to Question 1 was affirmative, then it would be the case when M = 0 and b is deterministic, and it would imply L$^1$[0,t] $\subset$ L$^p$[0,t] (for any finite t >0), which is not true if …
- 489