Let $\langle V, \langle\cdot, \cdot \rangle \rangle$ be an inner product space. We say $\varphi:\mathbb{Z}\to V$ is stationary if there exists a function $\gamma:\mathbb{Z}\to \mathbb{C}$ such that $$\langle\varphi(m),\varphi(n) \rangle = \gamma(m-n) \quad \forall m,n\in \mathbb{Z}$$
Set $V_{n}$ to be the subspace of $V$ generate by all finite linear combinations of $\varphi(m)$ for $m\leq n$. We define the one-step error by $$\sigma_{n}^{2} = \|\varphi(n+1) - \text{Proj}_{V_{n}}(\varphi(n+1))\|^2.$$
The question is to show that $\sigma^{2}_{n} = \sigma^{2}_{0}$ when $\varphi$ is stationary.
My attempt only goes so far as to notice that $$\text{Proj}_{V_{n}}(\varphi(n+1)) = \sum_{k\leq n}a_{k}\varphi(k)$$ for some coeffcients with only a finite number of $a$'s being non-zero. Then $$\sigma_{n}^{2} = \|\varphi(n+1) - \sum_{k\leq n}a_{k}\varphi(k)\|^2 = \|\varphi(1) - \sum_{l\leq 0}b_{l}\varphi(l)\|^2$$ by stationarity, were $b_{l} = a_{n+l}$.
My question is why this is equal to $$\sigma_{0}^{2} = \|\varphi(1) - \text{Proj}_{V_{0}}(\varphi(1))\|^2$$ It is clear that $\sum_{l\leq 0}b_{l}\varphi(l)\in V_{0}$ but not that it equals the projection of $\varphi(1)$