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 \(\mathcal{M} = \mathcal{N}^{\perp}\) . ◻