Equivalent Metrics on $\mathbb{R^n}$ I am working on a problem and want to verify that my logic and reasoning is correct. This is my first time working with metric spaces.

Show that the following define equivalent metrics on $\mathbb{R}^n$.
  \begin{eqnarray} 
\varphi^*(x,y)&=&|x_1-y_1|+\cdots+|x_n-y_n|\\
\varphi^+(x,y)&=&\text{max}\{|x_1-y_1|,\cdots,|x_n-y_n|\} \end{eqnarray}

By definition, we know two metrics on a set are equivalent if there are two positive numbers, $c_1$ and $c_2$, such that $\forall x,y\in \mathbb{R}^n$,
\begin{eqnarray*}
c_1\cdot\varphi^*(x,y)\leq\varphi^+(x,y)\leq c_2\cdot\varphi^*(x,y).
\end{eqnarray*}
Claim 1: $c_1\cdot\varphi^*(x,y)\leq\varphi^+(x,y)$
\begin{eqnarray*}
c_1\cdot\varphi^*(x,y)&=&c_1[|x_1-y_1|+\cdots+|x_n-y_n|]\\
&=&c_1|x_1-y_1|+\cdots+c_1|x_n-y_n|]
\end{eqnarray*}
Claim 2: $\varphi^+(x,y)\leq c_2\cdot\varphi^*(x,y)$
\begin{eqnarray*}
c_2\cdot\varphi^*(x,y)&=&c_2[|x_1-y_1|+\cdots+|x_n-y_n|]\\
&=&c_2|x_1-y_1|+\cdots+c_2|x_n-y_n|]
\end{eqnarray*}
Questions:


*

*Is my reasoning sound so far?

*Are $c_1,c_2\in\mathbb{N}$, meaning they can't be fractions?

*I know $\mathbb{R}^n$ is a normed linear space. How does this fact help me specifically? I feel like this is the key in being able to finish proving my claims.
 A: I will answer for $n = 2$. Then you can generalize to higher values of $n$. The argument is similar.
Given two points $(x_1,y_1)$ and $(x_2,y_2)$, say I know $M = \phi^{*}(x,y) = |x_1 - y_1| + |x_2 - y_2|$. How large can $\phi^{+}(x,y) = \max(|x_1 - y_1|, |x_2 - y_2|)$ be? The sum of the two nonnegative numbers $|x_1 - y_1|$ and $|x_2 - y_2|$ is $M$. Therefore, neither of the two numbers can exceed $M$. Hence the maximum of the two numbers does not exceed $M$. In other words, we have
$$\phi^{+}(x,y) \leq \phi^{*}(x,y).$$
This shows that the second inequality is satisfied for $c_2 = 1$.
Now say instead that I know $N = \phi^{+}(x,y) = \max(|x_1 - y_1|, |x_2 - y_2|)$. How large can $\phi^{*}(x,y) = |x_1 - y_1| + |x_2 - y_2|$ be? The quantity $\phi^{*}(x,y)$ is the sum of two numbers neither of which exceeds $N$. Therefore this sum is at most $2N$. Thus
$$\phi^{*}(x,y) \leq 2\phi^{+}(x,y).$$
This shows that the first inequality is satisfied for $c_1 = 1/2$.
To answer your questions:


*

*No, your reasoning is incorrect. You need to say what $c_1$ and $c_2$ are and prove that they work.

*The numbers $c_1$ and $c_2$ are not necessarily whole numbers.

*There is a theorem that says that any two norms on a finite-dimensional vector space are equivalent. If you are permitted to use this theorem, then the statement follows immediately from the fact that $\phi^{*}$ and $\phi^{+}$ are norms. However, if the purpose of the problem is to give a direct proof, then I don't see how this can be used to simplify the solution significantly.
