Suppose \( x \) and \( y \) are two arbitrary points in \( U \). And \( L \) is the line segment connecting \( x \) and \( y \). Then by the multivariable mean value theorem, \( \| f(x) - f(y) \|_\infty \leq \| x - y \| _\infty \underset{c \in L}{\text{max}} \| f'(c) \|_\infty \). Then by setting \( k=sup( \| f' \|_\infty ) \), thus \( 0 \leq k<1 \). Then
$$ \| f(x) - f(y) \|_\infty \leq k \| x - y \| _\infty .$$
The last inequality states \( f \) as a contraction mapping with the norm \( \| \; \|_\infty \).
Define $$ x_0=x , $$ $$ x_{n+1} = f(x_n) $$
Thus $$ \| x_{n+1} - x_n \|_\infty = \| f(x_n) - f(x_{n-1}) \|_\infty \leq k \| x_n - x_{n-1} \|_\infty, $$ so it follows by induction that \( \| x_{n+1} - x_n \|_\infty \leq k \| x_n - x_{n-1} \|_\infty. \) Then if \( 0 \leq n \leq m \), then
\( \| x_n - x_m \|_\infty \)
\( \leq \| x_n - x_{n+1} \|_\infty +\dotsb+\| x_{m-1} - x_m \|_\infty \)
\( \leq ( k^n +\dotsb+ k^{m-1} )\| x_1 - x_0 \|_\infty \)
\( \leq k^n \| x_1 - x_0 \|_\infty ( 1 + k + k^2 + \dots ) \)
\( \leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k } \)
By equivalence of norms, \( m \| \; \| \leq \| \; \|_\infty \leq M \| \; \| \).
\( m \| x_n - x_m \| \leq \| x_n - x_m \|_\infty \leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k } \leq \frac{ k^n M \| x_1 - x_0 \| }{ 1 - k } \)
Therefore
\( \| x_n - x_m \| \leq \| x_n - x_m \|_\infty \leq \frac{ k^n \| x_1 - x_0 \|_\infty }{ 1 - k } \leq \frac{M}{m} \frac{ k^n \| x_1 - x_0 \| }{ 1 - k } \).
As \( n \to \infty \), \( \| x_n - x_m \| \to 0 \). Hence for an arbitrary norm, \( \{x_n\}_0^\infty \) does converge to one point if \( sup( \| f' \|_\infty )<1 \). By the same reasoning, it can be shown that if \( sup( \| f' \| )<1 \), then the sequence \( \{ x_n \}_0^\infty \) converges in terms of \( \| \; \|_\infty \). That is, regardless of the norm actually used, \( f \) is a contraction mapping.