Denk Metrikler-III

Question

$( X,d_1) , (X,d_2) \text{ metrik uzay olsun.}$

$d_1\overset{L}{\sim} d_2  \Rightarrow d_1\overset{D}{\sim} d_2 \text{ olduğunu gösteriniz.}$

$NOT:$

$ \text{L:Lipschitz denk metrik - D:Düzgün denk metrik}$

Answer 1

$d_1\overset{L}{\sim} d_2 $ olsun. Amacımız $$(\forall\epsilon>0)(\exists \delta>0) (\forall x,y \in X)(d_1(x,y)< \delta \Rightarrow d_2(x,y)< \epsilon)(d_2(x,y)<\delta\Rightarrow d_1(x,y)<\epsilon)$$ önermesinin doğru olduğunu göstermek.

$\epsilon > 0$  verilmiş olsun.

$\left.\begin{array}{rr} d_1\overset{L}{\sim} d_2 \Rightarrow (\exists \lambda, \mu>0) (\forall x,y \in X ) ( \lambda\cdot d_1(x,y) \leq d_2(x,y) \leq \mu\cdot d_1(x,y))  \\ \\ \left(\delta_1:= \frac{\epsilon}{\mu}\right) \left(\delta_2:=\epsilon\cdot \lambda\right) \end{array}\right\}\overset{\epsilon>0}{\Rightarrow}$

$\left.\begin{array}{rr}\Rightarrow  (\delta_1,\delta_2 >0) (\forall x,y \in X)(d_1(x,y)< \delta_1 \Rightarrow d_2(x,y)<\epsilon)(d_2(x,y)<\delta_2 \Rightarrow d_1(x,y)<\epsilon) \\ \\ \delta:=\min\{\delta_1,\delta_2\}\end{array}\right\}\Rightarrow$

 

$\Rightarrow (\exists \delta>0) (\forall x,y \in X)(d_1(x,y)< \delta\Rightarrow d_2(x,y)< \epsilon)(d_2(x,y)<\delta \Rightarrow d_1(x,y)<\epsilon).$