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 is an isometry on a finite-dimensional inner product space, and if is the subspace of all solutions of , then the sequence defined by converges as to the perpendicular projection .

Proof. Let be the range of the linear transformation . If is in , then so that This implies that converges to zero when is in .

On the other hand, if is in , that is, , then , so that in this case certainly converges to .

We shall complete the proof by showing that . (This will imply that every vector is a sum of two vectors for which converges, so that converges everywhere. What we have already proved about the limit of in and in shows that always converges to the projection of in .) To show that , we observe that is in the orthogonal complement of if and only if for all . This in turn implies that that is, that is orthogonal to every vector , so that , , or . Reading the last computation from right to left shows that this necessary condition is also sufficient; we need only to recall the definition of to see that . ◻

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

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