Question

State and prove localization theorem

Answer 1

Answer:

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 F(x) 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:

A simple proof is as follows: if there were a point x0 within Ω for which F(x0) ≠ 0, then the continuity of F would require the existence of a neighborhood of x0 in which the value of F was nonzero, and in particular of the same sign than in x0. 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 F over N would evaluate to a nonzero value. However, since x0 is part of the open set Ω, all neighborhoods of x0 smaller than the distance of x0 to the frontier of Ω are included within it, and so the integral of F over them must evaluate to zero. Having reached the contradiction that ∫N F(x) dx must be both zero and nonzero, the initial hypothesis must be wrong, and thus there is no x0 in Ω for which F(x0) ≠ 0.

The theorem is easily generalized to multivariate functions, replacing intervals with the more general concept of connected open sets, that is, domains, and the original function with some F(x) : Rn → R, with the constraints of continuity and nullity of its integral over any subdomain D ⊂ Ω. The proof is completely analogous to the single variable case, and concludes with the impossibility of finding a point x0 ∈ Ω such that F(x0) ≠ 0.

Step-by-step explanation:

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