Ergodic theorem

The routine work is out of the way; we go on to illustrate the general theory by considering some very special but quite important convergence problems.

Theorem 1. If U is an isometry on a finite-dimensional inner product space, and if \mathcal{M} is the subspace of all solutions of U x=x , then the sequence defined by V_{n}=\frac{1}{n}(1+U+\cdots+U^{n-1}) converges as n \to \infty to the perpendicular projection E=P_{\mathcal{M}} .

Proof. Let \mathcal{N} be the range of the linear transformation 1-U . If x=y-U y is in \mathcal{N} , then \begin{align} V_{n} x &= \frac{1}{n}(y-U y+U y-U^{2} y+\cdots+U^{n-1} y-U^{n} y) \\ &= \frac{1}{n}(y-U^{n} y), \end{align}so that \begin{align} \|V_{n} x\| &= \frac{1}{n}\|y-U^{n} y\|\\ &\leq \frac{1}{n}\big(\|y\|+\|U^{n} y\|\big) \\ &= \frac{2}{n}\|y\|. \end{align}This implies that V_n x converges to zero when x is in \mathcal{N} .

On the other hand, if x is in \mathcal{M} , that is, U x=x , then V_{n} x=x , so that in this case V_{n} x certainly converges to x .

We shall complete the proof by showing that \mathcal{N}^{\perp}=\mathcal{M} . (This will imply that every vector is a sum of two vectors for which (V_{n}) converges, so that (V_{n}) converges everywhere. What we have already proved about the limit of (V_{n}) in \mathcal{M} and in \mathcal{N} shows that (V_{n} x) always converges to the projection of x in \mathcal{M} .) To show that \mathcal{N}^{\perp}=\mathcal{M} , we observe that x is in the orthogonal complement of \mathcal{N} if and only if (x, y-U y)=0 for all y . This in turn implies that \begin{align} 0 &= (x, y-U y)\\ &= (x, y)-(x, U y)\\ &= (x, y)-(U^{*} x, y) \\ &= (x-U^{*} x, y), \end{align}that is, that x-U^{*} x=x-U^{-1} x is orthogonal to every vector y , so that x-U^{-1} x=0 , x=U^{-1} x , or U x=x . Reading the last computation from right to left shows that this necessary condition is also sufficient; we need only to recall the definition of \mathcal{M} to see that \mathcal{M} = \mathcal{N}^{\perp} . ◻

This very ingenious proof, which works with only very slight modifications in most of the important infinite-dimensional cases, is due to F. Riesz.