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 407441
Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
0
votes
1
answer
135
views
A minimal surface with a local extremum in normal direction is a plane [closed]
I'm currently struggling with concluding a proof and need a hint. So the first part of the exercise
was that given an open subset $\Omega \subset \mathbb{R}^2$ and a harmonic function
$f: \Omega \to \ …
- 13
0
votes
1
answer
706
views
Proof: If a reproducing kernel exists for a Hilbert space, then it is unique
I really want to prove the statement in the title but I'm struggling with it. Here my current state:
Proof via contradiction. Let $\mathcal{H}$ be a RKHS with two reproducing kernels $k$ and $\hat{k}$ …
- 13