Talk:Weak derivative

in the theory of Lp spaces and Sobolev spaces, functions that are equal almost everywhere are identified.

Did you mean to write "identical" in the sentence above?

Jörgen

No, the usual idiom is that the functions are identified. They are not identical functions, however. For instance, the characteristic function of a point and the zero function are different functions, but they are identified with each other in any discussion to do with Lp spaces or Sobolev spaces. silly rabbit (talk) 21:05, 9 October 2008 (UTC) [reply ]

Identified in the sense of: are elements of the same equivalence class. 129.107.240.1 (talk) 18:26, 5 March 2009 (UTC) [reply ]

Can we get a few more? The given example of the absolute value function seems pretty trivial. We could, in this case, just see that the signum is the derivative almost everywhere, and thus see that it is a weak derivative. What would be much better is an example of a function which is not differentiable throughout any interval (or even further, not differentiable anywhere) but which has a weak derivative. 129.107.240.1 (talk) 18:26, 5 March 2009 (UTC) [reply ]

I added an example, however contrived it might be! --Paul Laroque (talk) 02:26, 13 April 2009 (UTC) [reply ]

I am an undergrad, so forgive me if this is a stupid question. It wasn't really clear from the article. The Lebesgue Integral of a weak derivative is equal to the original function(not phrased correctly i think, but should be clear enough)? If so, almost everywhere, or everywhere?--79.235.185.160 (talk) 12:40, 6 February 2010 (UTC) [reply ]

What are some of the applications of weak derivatives? Is it possible to build an optimisation theory based on weak differentiation? 203.167.251.186 (talk) 03:17, 24 June 2010 (UTC) [reply ]

I was wondering why a function that is only weakly differential must be in ${\displaystyle L^{1}}${\displaystyle L^{1}}. I suppose that if the function is not strongly differentiable then one couldn't use the chain rule to get the derivative of say the ${\displaystyle P^{2}}${\displaystyle P^{2}} norm for this function. However, perhaps a ${\displaystyle P^{2}}${\displaystyle P^{2}} norm may still be integrable if we remove problematic points by using an improper integral.

On another note, why are we using the term Lebesgue space. Isn't the term Lp space more common? S243a (talk) 18:48, 18 October 2013 (UTC) [reply ]

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