Space of lipschitz functions: two metrics are topologically equivalent

[mahler1]

Homework Statement .
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.

 

[mighty2000]

Any help or hint would be appreciated. Thank you.

To prove that ##Id:(Lip_{M}(ℝ),d_1)→(Lip_{M}(ℝ),d_∞)## is continuous, we can use a similar approach as before. For a given ##f \in (Lip_{M}(ℝ),d_∞)## and ##ε>0##, there exists ##δ_{ε,f}## such that if ##d_1(f,g)=\int_0^1 |f(x)-g(x)|dx<ε##, then ##d_∞(f,g)=Sup_{x \in [0,1]}|f(x)-g(x)|≤M\int_0^1 |f(x)-g(x)|dx<ε##. Therefore, we can take ##δ_{ε,f}=\frac{ε}{M}## and the identity map is continuous.

To show that the metrics are not equivalent, we need to find a function that is M-Lipschitz but not M-integrable. Consider the function ##f(x)=x## for ##x \in [0,1]##. This function is M-Lipschitz with ##M=1##, since ##|f(x)-f(y)|=|x-y|≤M|x-y|## for all ##x,y \in [0,1]##. However, ##\int_0^1 |f(x)-g(x)|dx=\int_0^1 |x-g(x)|dx## is not finite for any function ##g(x)##, since the integral will always be infinite. Therefore, this function is M-Lipschitz but not M-integrable, showing that the metrics are not equivalent.