Lebl-Basic-Analysis-I · Problem 3.4.13

Continuous Functions

Mathematics · uniform-continuity

Problem

a) If f:[0,1]→Rf : [0,1] \to \mathbb{R} is given by f(x):=xmf(x) := x^m for an integer m≥0m \geq 0, show ff is Lipschitz and find the best (the smallest) Lipschitz constant KK (depending on mm of course). Hint: (x−y)(xm−1+xm−2y+xm−3y2+⋯+xym−2+ym−1)=xm−ym(x-y)(x^{m-1} + x^{m-2}y + x^{m-3}y^2 + \cdots + xy^{m-2} + y^{m-1}) = x^m - y^m. b) Using the previous exercise, show that if f:[0,1]→Rf : [0,1] \to \mathbb{R} is a polynomial, that is, f(x):=amxm+am−1xm−1+⋯+a0f(x) := a_m x^m + a_{m-1}x^{m-1} + \cdots + a_0, then ff is Lipschitz.

Related theory

Browse the problem library