Geometric potential of an expanding map ($\varphi =-\log |T^{\prime }|$)

For a differentiable uniformly expanding interval map, the geometric potential of an expanding map isThe exponential of its length-$n$ orbit sum is the reciprocal derivative magnitude $1/|(T^n{)}^{\prime }(x)|$. This weight is geometric: an inverse branch contracts interval lengths by that factor.

The corresponding Perron-Frobenius operator transports a density by summing contributions from inverse branches:For a finite smooth full-branch expanding map, standard bounded-distortion hypotheses give a positive regular fixed density $h$, normalized by $\int hdx=1$. The identity $\mathcal{L}h=h$ means that $h(x)dx$ is an invariant measure. The cylinder averages of an invariant density explain how the geometric cylinder weights recover that density.

For the doubling map, there are two inverse branches andThe constant density $h=1$ is fixed, giving Lebesgue measure. Choosing an unequal-symbol potential instead produces a different Gibbs measure; an arbitrary Gibbs measure need not be absolutely continuous. Thus the potential, rather than the symbolic alphabet alone, determines whether the measure agrees with geometric length.

The Perron-Frobenius operator transports a density under a map. For a piecewise smooth expanding interval map $T$, relative to Lebesgue measure,Each inverse branch contributes its incoming density times its length-contraction factor. A change of variables gives ${\int }_A\mathcal{L}fdx={\int }_{T^{-1}A}fdx$, proving that probability is transported correctly.

For the doubling map, $\mathcal{L}f(x)=[f(x/2)+f((x+1)/2)]/2$. Thus the constant density is fixed. More generally, a normalized nonnegative fixed density determines an absolutely continuous invariant measure; a constant density need not be fixed for an arbitrary expanding map.

Let a uniformly expanding interval map have cylinders ${\Delta }_w$ whose maximum length at level $n$ is at most $C{\lambda }^{-n}$, with $\lambda >1$. Suppose an invariant probability density $h$ is Lipschitz continuous with constant $L$ and satisfies $h\ge h_{\ast }>0$. Its cylinder probabilities are $\nu ([w])={\int }_{{\Delta }_w}hdx$. Define the piecewise constant logarithmic averageThe average of $h$ over ${\Delta }_w$ differs from $h(x)$ by at most $L\mathrm{diam}({\Delta }_w)$. Since the logarithm has derivative at most $1/h_{\ast }$ on this range,Thus the cylinder data recover the logarithm of the density uniformly at a geometric rate.

Conversely, take consistent positive cylinder probabilities and suppose these $g_n$ converge uniformly to a finite function $g$. Put $h=e^g$. If $\Vert g_n-g{\Vert }_{\infty }\le {\delta }_n\to 0$, integrating over every level-$n$ cylinder compares its $h$-mass with its assigned probability within factors $e^{\pm {\delta }_n}$. Subdivide any fixed cylinder into these finer cylinders and sum: consistency gives the same multiplicative bounds for the fixed cylinder. Letting $n\to \infty$ proves ${\int }_{{\Delta }_w}hdx=\nu ([w])$.

Shrinking cylinders determine the Borel probability measure, so the geometric image of the symbolic law is exactly $h(x)dx$. Use a consistent convention at branch endpoints; bounded limiting cylinder densities imply zero mass at individual endpoints. If the symbolic law is shift invariant, the coding relation transfers that invariance to the interval map. This converse is much stronger than inspecting a density ratio along just one itinerary.