Linearized stability theorem for maps

Remarks

For part 1, choose a norm on $\R^k$ in which $\norm{J} < 1$; such a norm exists because the spectral radius is below $1$. By continuity of $Df$, the map $f$ is a contraction with ratio $q < 1$ on a small closed ball around $x^*$ in this norm. So iterates stay in the ball and $\norm{x_n - x^*} \le q^n \norm{x_0 - x^*}$. Part 2 needs a finer argument along the direction of the expanding eigenvalues. For the logistic map $f(x) = r x (1 - x)$ with $1 < r < 3$, the fixed point $x^* = 1 - 1/r$ has $f'(x^*) = 2 - r$, of modulus less than $1$, so it is asymptotically stable. When some eigenvalue has modulus exactly $1$ and none has modulus greater than $1$, the theorem says nothing. For example, $x \mapsto x - x^3$ and $x \mapsto x + x^3$ both have derivative $1$ at $0$, but $0$ is asymptotically stable for the first and unstable for the second.

Going deeper

For a student

This is the main tool for the local stability of equilibria of nonlinear difference equations. For a scalar equation it reduces to $\abs{f'(x^*)} < 1$ for stability and $\abs{f'(x^*)} > 1$ for instability. For higher-order scalar equations one writes the system in vector form and applies the theorem to the companion-type Jacobian, whose characteristic polynomial can be tested by the Jury criterion. When no eigenvalue has modulus $1$, the fixed point is hyperbolic. Then the Hartman–Grobman theorem gives the stronger conclusion that $f$ is locally topologically conjugate to $J$ when $f$ is a local diffeomorphism.