If a function f is uniformly continuous on R and f(x^2) = f(x) for all x, then prove that f is a constant function.
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Updated 6/4/2023 12:10:26 AM
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User: If a function f is uniformly continuous on R and f(x^2) = f(x) for all x, then prove that f is a constant function.

Weegy: In mathematics, a function ƒ is uniformly continuous if, roughly speaking, [ it is possible to guarantee that ƒ(x) and ƒ(y) be as close to each other as we please by requiring only that x and y are sufficiently close to each other; unlike ordinary continuity, the maximum distance between ƒ(x) and ƒ(y) cannot depend on x and y themselves. For instance, any isometry (distance-preserving map) between metric spaces is uniformly continuous. Every uniformly continuous function between metric spaces is continuous. Uniform continuity, unlike continuity, relies on the ability to compare the sizes of neighbourhoods of distinct points of a given space. In an arbitrary topological space this may not be possible. Instead, uniform continuity can be defined on a metric space where such comparisons are possible, or more generally on a uniform space. The equicontinuity of a set of functions is a generalization of the concept of uniform continuity. ]

Question
Updated 6/4/2023 12:10:26 AM
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Assume f is not a constant function. By contradiction, for any > 0, there exists > 0 such that |f(x^2) - f(x)| < for all x. However, choosing = |f(a) - f(b)|, where a and b are distinct, leads to a contradiction. Therefore, f must be a constant function.

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