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 493038
Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
0
votes
1
answer
312
views
Deduce that a function is zero on interval $[0,M]$
I have been thinking about this for the last few days but I was not able to produce a definitive answer.
Take an integrable function $g$ that maps in $\mathbb{R}$ and with domain contained in $[0,M]$ …
- 141
0
votes
1
answer
239
views
Riemann-Liouville integral of $f$ is zero implies $f =0$ a.e
The Riemann-Liouville integral is defined by
$$
I^\alpha f(x)=\frac{1}{\Gamma(\alpha)} \int_a^x f(t)(x-t)^{\alpha-1} d t
$$
where $\Gamma$ is the gamma function and $a$ is an arbitrary but fixed base …
- 141