advanced39 min studyLesson 809 in Mathematics

The subadditive ergodic theorem

Identify almost-sure growth rates when composition gives an inequality instead of an additive sum.

On this page

What you’ll understand

  • State a sufficient finite-limit version of Kingman's theorem with explicit lower control.
  • Derive the subadditive inequality for logarithmic cocycle norms.
  • Distinguish exponential growth rates from finite-time expansion and polynomial growth.

Birkhoff averages add one-step observations exactly. Many growth problems only admit an inequality: the cost of a combined journey is at most the sum of two segment costs, or the norm of a product is at most the product of its norms. Kingman's theorem extends the ergodic principle to this subadditive setting. Its conclusion is a long-time rate, not a formula for every finite trajectory. Integrability and lower-tail control determine whether that rate is finite and whether almost-sure convergence also holds in mean.

A precise sufficient version of Kingman

Let T preserve a probability measure and be ergodic. Let each real-valued function g-n be integrable, and assume the cocycle subadditivity inequality for every positive pair of lengths, outside a common null set. A convenient sufficient lower bound is g-n at least minus C times n for one finite deterministic constant C. Under these hypotheses, Kingman's theorem gives a finite constant limit of g-n divided by n almost everywhere. The limit equals the infimum of the normalized integrals. Invertibility and independence are not required.

The linear lower bound is stronger than necessary, but deliberately prevents the limit from being minus infinity and controls negative tails. Standard more general versions can allow a minus-infinite limit when lower control is absent. Under our stated assumptions convergence also holds in L1: subadditivity bounds the positive part of g-n divided by n by averages of shifts of the integrable positive part of g-one. These averages are uniformly integrable, while the negative parts are bounded by C. Together with almost-sure convergence this yields mean convergence. This additional argument explains the stronger conclusion rather than silently attaching it to every version.

gn+m(x)≤gn(x)+gm(Tnx),gn∈L1(μ),gn≥−Cn⟹gnn⟶γ=inf⁡n≥11n∫gn dμa.e. and in L1g_{n+m}(x)\le g_n(x)+g_m(T^nx),\qquad g_n\in L^1(\mu),\quad g_n\ge-Cn\Longrightarrow \frac{g_n}{n}\longrightarrow\gamma=\inf_{n\ge1}\frac1n\int g_n\,d\mu\quad\text{a.e. and in }L^1
The transformation is ergodic and probability preserving, and C is one finite constant for all n. This is a sufficient finite-limit formulation, not the weakest form of Kingman's theorem.↔ Scroll wide equations horizontally.

Integrals have a numerical subadditive limit

Integrate the subadditive inequality and use preservation to remove the shift from the second integral. The resulting sequence of integrals is numerically subadditive. Fekete's lemma therefore identifies its normalized limit with its infimum. Kingman's deeper assertion identifies the same number as the almost-everywhere trajectory rate. The infimum is over integrals, not over the pointwise numbers at a chosen starting state. Moving an infimum through an integral or claiming the limit equals a pointwise infimum would generally change the statement.

The proof cannot simply divide a trajectory into blocks and assume those blocks are independent. Even when T is ergodic, a power of T used to sample nonoverlapping blocks need not be ergodic. Correct proofs handle shifted block offsets or invariant conditional limits carefully. Without ergodicity the limiting rate is an invariant function rather than necessarily a global constant. Thus subadditivity controls concatenation, while ergodicity controls whether different invariant regions can retain different asymptotic rates.

A(n)(x)=A(Tn−1x)⋯A(x),gn(x)=ln⁡∥A(n)(x)∥,A(n+m)(x)=A(m)(Tnx)A(n)(x)A^{(n)}(x)=A(T^{n-1}x)\cdots A(x),\qquad g_n(x)=\ln\|A^{(n)}(x)\|,\qquad A^{(n+m)}(x)=A^{(m)}(T^nx)A^{(n)}(x)
Submultiplicativity of the operator norm gives Kingman's inequality for these logarithms. Natural logarithms are used for growth rates; any needed integrability and lower control must still be checked.↔ Scroll wide equations horizontally.

Zero exponential rate does not mean bounded growth

For matrix products the logarithm of an operator norm is subadditive because norms multiply as an upper bound. If matrices and their inverses have uniformly bounded norms, a deterministic linear lower bound follows, so the sufficient theorem above applies. More general integrable cocycles use the corresponding broader theorem. The growth exponent measures exponential scale only. A sequence of norms growing like a polynomial can have exponent zero while remaining unbounded, and strong one-step expansion can be canceled by later matrices.

The matrix instrument selects a single upper-triangular matrix and reports its determinant and singular stretches. Its animation interpolates geometry; it does not repeatedly multiply the selected matrix. To investigate a cocycle, compute its products separately and use the instrument only to check specified individual matrices. This distinction keeps finite-time singular values separate from a proven asymptotic rate. Long numerical records can suggest a limit, but without hypotheses and a theorem they do not establish almost-sure convergence or identify all exceptional trajectories.

A shear grows unboundedly with exponent zero

Use the one-point probability system, which is ergodic, and constant matrix A=(1101)A=\begin{pmatrix}1&1\\0&1\end{pmatrix}. Determine the logarithmic norm growth rate and check the normalized value for a two-step product.

  1. Direct multiplication gives An=(1n01)A^n=\begin{pmatrix}1&n\\0&1\end{pmatrix}. The operator norm is at least one, and the triangle inequality applied to identity plus n times the elementary shear gives the upper bound n+1n+1.

  2. Consequently 0≤n−1ln⁡∥An∥≤n−1ln⁡(n+1)→00\le n^{-1}\ln\|A^n\|\le n^{-1}\ln(n+1)\to0. Each logarithm is integrable on the one-point space and the lower-bound constant can be zero, so the stated Kingman hypotheses hold.

  3. The exact largest singular value is (n2+4+n)/2(\sqrt{n^2+4}+n)/2. At n equal to two it is 1+21+\sqrt2, giving normalized logarithm 12ln⁡(1+2)≈0.4406868\tfrac12\ln(1+\sqrt2)\approx0.4406868. This positive finite-time value does not contradict the zero limiting exponent.

The result

The shear has unbounded linear norm growth but zero exponential growth rate. Kingman's limit concerns normalized logarithms, not boundedness of the products.

Explore the model

Interactive lab

Interactive lab · matrix

Check two shear products without treating animation as iteration

Pause and Reset; set both a=1,b=1a=1,b=1. Predict determinant one and largest singular value (1+5)/2(1+\sqrt5)/2. Reset and set both a=1,b=2a=1,b=2, the separately computed square of the first shear; predict determinant one and stretch 1+21+\sqrt2. Compare the normalized natural logarithms on paper. Play interpolates a selected matrix's geometry, not its powers. Control changes retain time; Reset clears time and both controls.

Check two shear products without treating animation as iterationThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFinite products can expand substantially even when the limiting exponential rate is zero.

Check your understanding

Which infimum gives the constant Kingman limit under the stated ergodic hypotheses?

Try it yourself

On a uniform three-cycle, let a scalar cocycle take values 2,1/2,1/22,1/2,1/2 around the cycle. Compute its logarithmic growth exponent and verify a deterministic linear lower bound for the logarithms of its products.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A full cycle has the same product from every starting point. Each individual logarithm is bounded below by minus the natural logarithm of two.

Compare with the solution

Every three-step product is 2(1/2)(1/2)=1/22(1/2)(1/2)=1/2. Thus each full cycle contributes −ln⁡2-\ln2, and the bounded remainder disappears after division by n. The exponent is −(ln⁡2)/3-(\ln2)/3 for every starting point. Every one-step logarithm is at least −ln⁡2-\ln2, so the n-step logarithm is at least −nln⁡2-n\ln2. The three-cycle is ergodic under uniform probability, and all finite-space functions are integrable.

Take it with you

  • Kingman replaces additive sums with subadditive trajectory costs.
  • The stated linear lower control ensures a finite limit and supports L1 convergence.
  • A zero exponential exponent permits unbounded polynomial growth.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.