Lesson plan
Localization theorem
In mathematics, particularly in integral calculus, the localization theorem allows, under certain conditions, to infer the nullity of a function given only information about its continuity and the value of its integral. Let be a real-valued function defined on some open interval Ω of the real line that is continuous in Ω. Let D be an arbitrary subinterval contained in Ω. The theorem states the following implication: \int_D F(x) \, \mathrm{d}x = 0 ~ \forall D \subset \Omega ~ \Rightarrow ~ F(x) = 0 ~ \forall x \in \Omega A simple proof is as follows: if there were a point x 0 within Ω for which , then the continuity of would require the existence of a neighborhood of x 0 in which the value of was nonzero, and in particular of the same sign than in x 0 . Since such a neighborhood N , which can be taken to be arbitrarily small, must however be of a nonzero width on the real line, the integral of over N would evaluate to a nonzero value. However, since x 0 is part of the open set Ω, all neighborhoods of x 0 smaller than the distance of x 0 to the frontier of Ω are included within it, and so the integral of over them must evaluate to zero. Having reached the contradiction that must be both zero and nonzero, the initial hypothesis must be wrong, and thus there is no x 0 in Ω for which .…
Article summary
In mathematics, particularly in integral calculus, the localization theorem allows, under certain conditions, to infer the nullity of a function given only information about its continuity and the value of its integral. Let be a real-valued function defined on some open interval Ω of the real line that is continuous in Ω. Let D be an arbitrary subinterval contained in Ω. The theorem states the following implication: \int_D F(x) \, \mathrm{d}x = 0 ~ \forall D \subset \Omega ~ \Rightarrow ~ F(x) = 0 ~ \forall x \in \Omega A simple proof is as follows: if there were a point x 0 within Ω for which , then the continuity of would require the existence of a neighborhood of x 0 in which the value of was nonzero, and in particular of the same sign than in x 0 . Since such a neighborhood N , which can be taken to be arbitrarily small, must however be of a nonzero width on the real line, the integral of over N would evaluate to a nonzero value. However, since x 0 is part of the open set Ω, all neighborhoods of x 0 smaller than the distance of x 0 to the frontier of Ω are included within it, and so the integral of over them must evaluate to zero. Having reached the contradiction that must be both zero and nonzero, the initial hypothesis must be wrong, and thus there is no x 0 in Ω for which .…
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Discussion questions
- Summarise the article in your own words. Which single claim would you verify first, and why?
- Which claim in this article is most likely to be revised as the sources update? How would you check its current status?
- If two sources disagreed about one of these facts, what would you need in order to trust one over the other?
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