- 2,559
- 4
The Euclidean metric, d, is defined by:
d(x, y) = \left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2}
Define metrics dp for each p in {1, 2, 3, ...} as follows:
d_p(x,y) = \left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p}
Prove that each dp induces the same topology as the Euclidean metric.
To do this, I want to show that for every \epsilon > 0 and for every x \in \mathbb{R}^n, there is are \delta _1,\, \delta _2 > 0 such that for every y \in \mathbb{R}^n:
\left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2} < \delta _1 \Rightarrow \left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p} < \epsilon
and
\left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p} < \delta _2 \Rightarrow \left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2} < \epsilon
Is this the right way to prove it? Where do I go from here? Induction on n, or p? Or maybe both? Or is there a way to do it without induction? Help would be very much appreciated!
d(x, y) = \left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2}
Define metrics dp for each p in {1, 2, 3, ...} as follows:
d_p(x,y) = \left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p}
Prove that each dp induces the same topology as the Euclidean metric.
To do this, I want to show that for every \epsilon > 0 and for every x \in \mathbb{R}^n, there is are \delta _1,\, \delta _2 > 0 such that for every y \in \mathbb{R}^n:
\left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2} < \delta _1 \Rightarrow \left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p} < \epsilon
and
\left [\sum _{i = 1} ^n |x_i - y_i|^p\right ]^{1/p} < \delta _2 \Rightarrow \left [\sum _{i = 1} ^n |x_i - y_i|^2\right ]^{1/2} < \epsilon
Is this the right way to prove it? Where do I go from here? Induction on n, or p? Or maybe both? Or is there a way to do it without induction? Help would be very much appreciated!