Search results
Results From The WOW.Com Content Network
The definition of uniform continuity appears earlier in the work of Bolzano where he also proved that continuous functions on an open interval do not need to be uniformly continuous. In addition he also states that a continuous function on a closed interval is uniformly continuous, but he does not give a complete proof.
Proof of Heine–Cantor theorem. Suppose that and are two metric spaces with metrics and , respectively.Suppose further that a function : is continuous and is compact. We want to show that is uniformly continuous, that is, for every positive real number > there exists a positive real number > such that for all points , in the function domain, (,) < implies that ((), ()) <.
For z = 1/3, the inverse of the function x = 2 C 1/3 (y) is the Cantor function. That is, y = y(x) is the Cantor function. In general, for any z < 1/2, C z (y) looks like the Cantor function turned on its side, with the width of the steps getting wider as z approaches zero.
However, f is continuous if all functions are continuous and the sequence converges uniformly, by the uniform convergence theorem. This theorem can be used to show that the exponential functions , logarithms , square root function, and trigonometric functions are continuous.
If X has a standard uniform distribution, then by the inverse transform sampling method, Y = − λ −1 ln(X) has an exponential distribution with (rate) parameter λ. If X has a standard uniform distribution, then Y = X n has a beta distribution with parameters (1/n,1). As such, The Irwin–Hall distribution is the sum of n i.i.d. U(0,1 ...
The function f(x) = √ x defined on [0, 1] is not Lipschitz continuous. This function becomes infinitely steep as x approaches 0 since its derivative becomes infinite. However, it is uniformly continuous, [8] and both Hölder continuous of class C 0, α for α ≤ 1/2 and also absolutely continuous on [0, 1] (both of which imply the former).
There are examples of uniformly continuous functions that are not α –Hölder continuous for any α. For instance, the function defined on [0, 1/2] by f(0) = 0 and by f(x) = 1/log(x) otherwise is continuous, and therefore uniformly continuous by the Heine-Cantor theorem. It does not satisfy a Hölder condition of any order, however.
Note that as an immediate consequence, any uniformly continuous function on a convex subset of a normed space has a sublinear growth: there are constants a and b such that |f(x)| ≤ a|x|+b for all x. However, a uniformly continuous function on a general metric space admits a concave modulus of continuity if and only if the ratios ...