Space of lipschitz functions: two metrics are topologically equivalent

1. Oct 8, 2013

mahler1

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