What is a derivative in the distribution sense?

pellman
Messages
683
Reaction score
6
Never mind. I got this one. Couldn't figure out how to delete the post though.
 

Attachments

  • upload_2015-8-15_11-27-0.png
    upload_2015-8-15_11-27-0.png
    969 bytes · Views: 559
  • upload_2015-8-15_11-27-24.png
    upload_2015-8-15_11-27-24.png
    829 bytes · Views: 564
  • upload_2015-8-15_11-28-24.png
    upload_2015-8-15_11-28-24.png
    743 bytes · Views: 562
  • upload_2015-8-15_11-30-2.png
    upload_2015-8-15_11-30-2.png
    743 bytes · Views: 542
  • upload_2015-8-15_11-30-25.png
    upload_2015-8-15_11-30-25.png
    743 bytes · Views: 558
  • upload_2015-8-15_11-30-51.png
    upload_2015-8-15_11-30-51.png
    1.1 KB · Views: 554
Last edited:
Physics news on Phys.org
Let ##u\in L^2(\mathbb{R})## be a function such that for all smooth ##\psi\in \mathcal{C}_c^1(\mathbb{R})##, we have that
\int_{-\infty}^{+\infty} \psi(x) u(x)dx = -\int_{-\infty}^{+\infty} \psi'(x) \varphi(x)dx
Then ##u## is said to be the derivative of ##\varphi## in the distributional sense.
 
  • Like
Likes pellman
micromass said:
Let ##u\in L^2(\mathbb{R})## be a function such that for all smooth ##\psi\in \mathcal{C}_c^1(\mathbb{R})##, we have that
\int_{-\infty}^{+\infty} \psi(x) u(x)dx = -\int_{-\infty}^{+\infty} \psi'(x) \varphi(x)dx
Then ##u## is said to be the derivative of ##\varphi## in the distributional sense.

Thanks, micromass! I had just found the answer and was editing the OP. Still much appreciated though.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top