Let $\phi:M\to N$. Show that $\phi*:B(N,\mathbb{R})\to B(M,\mathbb{R})$ where $\phi* = f(\phi)$ (contract spaces)

Question

Let $\phi:M\to N$, $\phi^*:B(N,\mathbb{R})\to B(M,\mathbb{R})$ where $\phi^* = f(\phi)$

Show that $\phi^*$ is a contraction mapping.

$B(M, \mathbb{R})$ denotes the space of all bounded funcitons from $N$ to $\mathbb{R}$ and the metric we're using is

$$d(f,g) = \sup\{|f(x) - g(x)|\}$$

So I must show that

$$d(\phi^*(f), \phi^*(g)) \le d(f,g)$$

right?

I think somehow I need to use that $\phi$ is a contraction but I don't have it in the hypotesis.

UPDATE: the metric definition was wrong, updated it

Answer 1

I don't think what you wrote down is the definition of a contraction mapping, but you can show it without additional hypothesis of $\phi$. $d(\phi^*(f),\phi^*(g))=\sup\{|\phi(f(x))-\phi(g(x))|\}\leq \sup\{|f(y)-g(y)|\}=d(f,g)$.