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Space of lipschitz functions: two metrics are topologically equivalent

  1. Oct 8, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##Lip_{M}(ℝ)##={##f: [0,1]→ℝ : |f(x)-f(y)|≤M|x-y|##}. Prove that ##(Lip_{M}(ℝ),d_∞)## and ##(Lip_{M}(ℝ),d_1)## are topologically equivalent but not equivalent. ##d_∞(f,g)=Sup_{x \in [0,1]}|f(x)-g(x)|## ##d_1(f,g)=\int_0^1 |f(x)-g(x)|dx##

    The attempt at a solution.
    I am having a hard time with this exercise, is similar to the last one I've posted but in a different space. As in that problem, I've tried to prove that the identity map is homeomorphic between the metric spaces ##(Lip_{M}(ℝ),d_∞)## and ##(Lip_{M}(ℝ),d_1)##. First, I want to prove that ##Id:(Lip_{M}(ℝ),d_∞)##→##(Lip_{M}(ℝ),d_1)## is continuous. So I take ##f \in (Lip_{M}(ℝ),d_1)## and for a given ##ε>0##, there exists ##δ_{ε,f}## such that if ##d_∞(f,g)=Sup_{x \in [0,1]}|f(x)-g(x)|< → d_1(f,g)=\int_0^1 |f(x)-g(x)|dx<ε##
    In that case, we have ##|f(x)-g(x)|≤Sup_{x \in [0,1]}|f(x)-g(x)|##, so ##\int_0^1 |f(x)-g(x)|dx≤\int_0^1 Sup_{x \in [0,1]}|f(x)-g(x)|dx=Sup_{x \in [0,1]}|f(x)-g(x)|##. If we consider ##δ_{ε,f}=ε##, then the identity is continuous at every element f of the space.

    It remains to prove that ##Id:(Lip_{M}(ℝ),d_1)→(Lip_{M}(ℝ),d_∞)## and in this part I got stuck, I suppose I must use the fact that the functions are "M-lipschitz" somewhere and I am still trying to find a counterexample to prove that the metrics are not equivalent.
     
    Last edited: Oct 8, 2013
  2. jcsd
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